876

Can I use the following jQuery code to perform file upload using POST method of an ajax request ?

$.ajax({
    type: "POST",
    timeout: 50000,
    url: url,
    data: dataString,
    success: function (data) {
        alert('success');
        return false;
    }
});

If it is possible, do I need to fill data part? Is it the correct way? I only POST the file to the server side.

I have been googling around, but what I found was a plugin while in my plan I do not want to use it. At least for the moment.

3

27 Answers 27

643

File upload is not possible through AJAX.
You can upload file, without refreshing page by using IFrame.
You can check further details here.


UPDATE

With XHR2, File upload through AJAX is supported. E.g. through FormData object, but unfortunately it is not supported by all/old browsers.

FormData support starts from following desktop browsers versions.

  • IE 10+
  • Firefox 4.0+
  • Chrome 7+
  • Safari 5+
  • Opera 12+

For more detail, see MDN link.

Sign up to request clarification or add additional context in comments.

10 Comments

Here is a list of the specific browsers that are not supported: caniuse.com/#search=FormData Also I have not tested this but here is a polyfill for FormData gist.github.com/3120320
Specifically, IE < 10 doesn't, for those too lazy to read the link.
@Synexis no we don't have to wait that long any more because all IE only has a 22% world wide market share and 27% in the U.S. and dropping fast. Chances are it is people over 70 years old. So rather then IE dictating what developers have to do IE will either have to shape up or get out of the race.
@DrewCalder Most IE users are office workers who don't have the choice on which browser to use because of company policies. I don't think age has much to do with it. I'm guessing most people > 70 get their offspring to install Chrome or FF instead :)
This link really helped me in understanding the bare minimum. I didn't have to use a xhr request. If you do use ajax make sure to set the enctype to "form/multipart"!
|
386

Iframes is no longer needed for uploading files through ajax. I've recently done it by myself. Check out these pages:

Using HTML5 file uploads with AJAX and jQuery

http://dev.w3.org/2006/webapi/FileAPI/#FileReader-interface

Updated the answer and cleaned it up. Use the getSize function to check size or use getType function to check types. Added progressbar html and css code.

var Upload = function (file) {
    this.file = file;
};

Upload.prototype.getType = function() {
    return this.file.type;
};
Upload.prototype.getSize = function() {
    return this.file.size;
};
Upload.prototype.getName = function() {
    return this.file.name;
};
Upload.prototype.doUpload = function () {
    var that = this;
    var formData = new FormData();

    // add assoc key values, this will be posts values
    formData.append("file", this.file, this.getName());
    formData.append("upload_file", true);

    $.ajax({
        type: "POST",
        url: "script",
        xhr: function () {
            var myXhr = $.ajaxSettings.xhr();
            if (myXhr.upload) {
                myXhr.upload.addEventListener('progress', that.progressHandling, false);
            }
            return myXhr;
        },
        success: function (data) {
            // your callback here
        },
        error: function (error) {
            // handle error
        },
        async: true,
        data: formData,
        cache: false,
        contentType: false,
        processData: false,
        timeout: 60000
    });
};

Upload.prototype.progressHandling = function (event) {
    var percent = 0;
    var position = event.loaded || event.position;
    var total = event.total;
    var progress_bar_id = "#progress-wrp";
    if (event.lengthComputable) {
        percent = Math.ceil(position / total * 100);
    }
    // update progressbars classes so it fits your code
    $(progress_bar_id + " .progress-bar").css("width", +percent + "%");
    $(progress_bar_id + " .status").text(percent + "%");
};

How to use the Upload class

//Change id to your id
$("#ingredient_file").on("change", function (e) {
    var file = $(this)[0].files[0];
    var upload = new Upload(file);

    // maby check size or type here with upload.getSize() and upload.getType()

    // execute upload
    upload.doUpload();
});

Progressbar html code

<div id="progress-wrp">
    <div class="progress-bar"></div>
    <div class="status">0%</div>
</div>

Progressbar css code

#progress-wrp {
  border: 1px solid #0099CC;
  padding: 1px;
  position: relative;
  height: 30px;
  border-radius: 3px;
  margin: 10px;
  text-align: left;
  background: #fff;
  box-shadow: inset 1px 3px 6px rgba(0, 0, 0, 0.12);
}

#progress-wrp .progress-bar {
  height: 100%;
  border-radius: 3px;
  background-color: #f39ac7;
  width: 0;
  box-shadow: inset 1px 1px 10px rgba(0, 0, 0, 0.11);
}

#progress-wrp .status {
  top: 3px;
  left: 50%;
  position: absolute;
  display: inline-block;
  color: #000000;
}

13 Comments

You can more or less copy the code straight off and use it. Just change some id names and class names. Any customation is on your own.
Note that myXhr seems to be global as well as name, size and type. Also it's better to use "beforeSend" to augment the already created XMLHttpRequest object rather than using "xhr" to create one then change it.
I don't think we can use that as is @Ziinloader. You're using some local method that's not included: writer(catchFile). What is writer()?
What if the data also contains few fields along with file to upload?
@Ziinloader This is a tremendously useful example that I see you've come back to and maintained several times. Truly an answer worth much more than the one upvote I can give.
|
269

Ajax post and upload file is possible. I'm using jQuery $.ajax function to load my files. I tried to use the XHR object but could not get results on the server side with PHP.

var formData = new FormData();
formData.append('file', $('#file')[0].files[0]);

$.ajax({
       url : 'upload.php',
       type : 'POST',
       data : formData,
       processData: false,  // tell jQuery not to process the data
       contentType: false,  // tell jQuery not to set contentType
       success : function(data) {
           console.log(data);
           alert(data);
       }
});

As you can see, you must create a FormData object, empty or from (serialized? - $('#yourForm').serialize()) existing form, and then attach the input file.

Here is more information: - How to upload a file using jQuery.ajax and FormData - Uploading files via jQuery, object FormData is provided and no file name, GET request

For the PHP process you can use something like this:

//print_r($_FILES);
$fileName = $_FILES['file']['name'];
$fileType = $_FILES['file']['type'];
$fileError = $_FILES['file']['error'];
$fileContent = file_get_contents($_FILES['file']['tmp_name']);

if($fileError == UPLOAD_ERR_OK){
   //Processes your file here
}else{
   switch($fileError){
     case UPLOAD_ERR_INI_SIZE:   
          $message = 'Error al intentar subir un archivo que excede el tamaño permitido.';
          break;
     case UPLOAD_ERR_FORM_SIZE:  
          $message = 'Error al intentar subir un archivo que excede el tamaño permitido.';
          break;
     case UPLOAD_ERR_PARTIAL:    
          $message = 'Error: no terminó la acción de subir el archivo.';
          break;
     case UPLOAD_ERR_NO_FILE:    
          $message = 'Error: ningún archivo fue subido.';
          break;
     case UPLOAD_ERR_NO_TMP_DIR: 
          $message = 'Error: servidor no configurado para carga de archivos.';
          break;
     case UPLOAD_ERR_CANT_WRITE: 
          $message= 'Error: posible falla al grabar el archivo.';
          break;
     case  UPLOAD_ERR_EXTENSION: 
          $message = 'Error: carga de archivo no completada.';
          break;
     default: $message = 'Error: carga de archivo no completada.';
              break;
    }
      echo json_encode(array(
               'error' => true,
               'message' => $message
            ));
}

8 Comments

What jquery library do i need to reference to run this code?
formData.append('file', $('#file')[0].files[0]); returns undefined and console.log(formData) has nothing except _proto_
I got this to work...Pinch me, I'm on jQuery Ajax file upload heaven! var formData = new FormData(); formData.append('file', document.getElementById('file').files[0]); $.ajax({ url : $("form[name='uploadPhoto']").attr("action"), type : 'POST', data : formData, processData: false, // tell jQuery not to process the data contentType: false, // tell jQuery not to set contentType success : function(data) { console.log(data); alert(data); } });
@RaymondWachaga That's encoding type, not encryption type. :)
I was really struggling with this... After hours of researching and so forth, I found this that really helped. Thanks mate! The first part of your solution worked like a charm for me. That's exactly what I needed :-)
|
136

Simple Upload Form

<script>

   //form Submit
   $("form").submit(function(evt){   

      evt.preventDefault();
      var formData = new FormData($(this)[0]);

      $.ajax({
          url: 'fileUpload',
          type: 'POST',
          data: formData,
          async: false,
          cache: false,
          contentType: false,
          enctype: 'multipart/form-data',
          processData: false,
          success: function (response) {
         
             alert(response);
          }
       });

       return false;

    });

</script>
<!--Upload Form-->
<form>
  <table>
    <tr>
      <td colspan="2">File Upload</td>
    </tr>
    <tr>
      <th>Select File </th>
      <td><input id="csv" name="csv" type="file" /></td>
    </tr>
    <tr>
      <td colspan="2">
        <input type="submit" value="submit"/> 
      </td>
    </tr>
  </table>
</form>

5 Comments

sir what are the js that were used on this example is there a specific jquery plugin for this one..i have a question i was pointed here can you please check my question.. I want to upload multiple file or images in that project here is the link stackoverflow.com/questions/28644200/…
$(this)[0] is this
What is the parameter on the server for posted file? Can you please post server part.
@FrenkyB and others - the files on the server (in PHP) are not stored in the $_POST variable - they are stored in the $_FILES variable. In this case, you would access it with $_FILES["csv"] because "csv" is the name attribute of the input tag.
perfectly right.
75

I'm pretty late for this but I was looking for an ajax based image uploading solution and the answer I was looking for was kinda scattered throughout this post. The solution I settled on involved the FormData object. I assembled a basic form of the code I put together. You can see it demonstrates how to add a custom field to the form with fd.append() as well as how to handle response data when the ajax request is done.

Upload html:

<!DOCTYPE html>
<html>
<head>
    <title>Image Upload Form</title>
    <script src="//code.jquery.com/jquery-1.9.1.js"></script>
    <script type="text/javascript">
        function submitForm() {
            console.log("submit event");
            var fd = new FormData(document.getElementById("fileinfo"));
            fd.append("label", "WEBUPLOAD");
            $.ajax({
              url: "upload.php",
              type: "POST",
              data: fd,
              processData: false,  // tell jQuery not to process the data
              contentType: false   // tell jQuery not to set contentType
            }).done(function( data ) {
                console.log("PHP Output:");
                console.log( data );
            });
            return false;
        }
    </script>
</head>

<body>
    <form method="post" id="fileinfo" name="fileinfo" onsubmit="return submitForm();">
        <label>Select a file:</label><br>
        <input type="file" name="file" required />
        <input type="submit" value="Upload" />
    </form>
    <div id="output"></div>
</body>
</html>

In case you are working with php here's a way to handle the upload that includes making use of both of the custom fields demonstrated in the above html.

Upload.php

<?php
if ($_POST["label"]) {
    $label = $_POST["label"];
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 200000)
&& in_array($extension, $allowedExts)) {
    if ($_FILES["file"]["error"] > 0) {
        echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
    } else {
        $filename = $label.$_FILES["file"]["name"];
        echo "Upload: " . $_FILES["file"]["name"] . "<br>";
        echo "Type: " . $_FILES["file"]["type"] . "<br>";
        echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
        echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";

        if (file_exists("uploads/" . $filename)) {
            echo $filename . " already exists. ";
        } else {
            move_uploaded_file($_FILES["file"]["tmp_name"],
            "uploads/" . $filename);
            echo "Stored in: " . "uploads/" . $filename;
        }
    }
} else {
    echo "Invalid file";
}
?>

6 Comments

i am getting Cross origin requests are only supported for protocol schemes: http, data, chrome, chrome-extension, https, why is this so sir i copy paste your code as it is
@HogRider - if you Google your error message, this is the first result: stackoverflow.com/questions/10752055/… Are you accessing your web pages locally through file://, rather than using a web server? As an aside, it is not best practice to simply copy-and-paste code blindly without first understanding it. I would recommend you go through the code line-by-line to gain an understanding of what is happening before putting the code to use.
@colincameron thank you for clarify a few things i did go through the it line by line and i dont really understand much so i asked the question so that someone might clarify my doubts. I am using local by way xampp to be exact. Can i ask a question that maybe you can clarify?
@Brownman Revival : I know that its too late for the reply.. You got a cross origin error because you opened the html file as file than running it from the server.
How do I apply form action based on selection in this code?
|
35

An AJAX upload is indeed possible with XMLHttpRequest(). No iframes necessary. Upload progress can be shown.

For details see: Answer https://stackoverflow.com/a/4943774/873282 to question jQuery Upload Progress and AJAX file upload.

2 Comments

Unfortunately IE < 10 don't support this.
When you merely want to refer to another page as an answer, you can vote to close as a dupkucate or leave a comment under the question. This post is Not An Answer. A post of this kind looks like an attempt to farm rep.
30

Using pure js it is easier

async function saveFile(inp) 
{
    let formData = new FormData();           
    formData.append("file", inp.files[0]);
    await fetch('/upload/somedata', {method: "POST", body: formData});    
    alert('success');
}
<input type="file" onchange="saveFile(this)" >

  • In server side you can read original file name (and other info) which is automatically included to request.
  • You do NOT need to set header "Content-Type" to "multipart/form-data" browser will set it automatically
  • This solutions should work on all major browsers.

Here is more developed snippet with error handling, timeout and additional json sending

async function saveFile(inp) 
{
    let user = { name:'john', age:34 };
    let formData = new FormData();
    let photo = inp.files[0];      
         
    formData.append("photo", photo);
    formData.append("user", JSON.stringify(user));  
    
    const ctrl = new AbortController() // timeout
    setTimeout(() => ctrl.abort(), 50000);

    try {
       let r = await fetch('/upload/image', 
         {method: "POST", body: formData, signal: ctrl.signal}); 
       console.log('HTTP response code:',r.status); 
       alert('success');
    } catch(e) {
       console.log('Huston we have problem...:', e);
    }
    
}
<input type="file" onchange="saveFile(this)" >
<br><br>
Before selecting the file Open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>

1 Comment

Best answer, also, the code is shorter.
29

Here's how I got this working:

HTML

<input type="file" id="file">
<button id='process-file-button'>Process</button>

JS

$('#process-file-button').on('click', function (e) {
    let files = new FormData(), // you can consider this as 'data bag'
        url = 'yourUrl';

    files.append('fileName', $('#file')[0].files[0]); // append selected file to the bag named 'file'

    $.ajax({
        type: 'post',
        url: url,
        processData: false,
        contentType: false,
        data: files,
        success: function (response) {
            console.log(response);
        },
        error: function (err) {
            console.log(err);
        }
    });
});

PHP

if (isset($_FILES) && !empty($_FILES)) {
    $file = $_FILES['fileName'];
    $name = $file['name'];
    $path = $file['tmp_name'];


    // process your file

}

2 Comments

What helped the most for me about this was the $('#file')[0].files[0] which is some sort of strange JS workaround without requiring a proper <form>
This is the complete solution, the PHP bit helps too.
21

Use FormData. It works really well :-) ...

var jform = new FormData();
jform.append('user',$('#user').val());
jform.append('image',$('#image').get(0).files[0]); // Here's the important bit

$.ajax({
    url: '/your-form-processing-page-url-here',
    type: 'POST',
    data: jform,
    dataType: 'json',
    mimeType: 'multipart/form-data', // this too
    contentType: false,
    cache: false,
    processData: false,
    success: function(data, status, jqXHR){
        alert('Hooray! All is well.');
        console.log(data);
        console.log(status);
        console.log(jqXHR);

    },
    error: function(jqXHR,status,error){
        // Hopefully we should never reach here
        console.log(jqXHR);
        console.log(status);
        console.log(error);
    }
});

4 Comments

this is what : ('user',$('#user').val());
textbox with id="user" is appended to the form @rahim.nagori
A more direct approach: var jform = new FormData($('form').get(0));
nice, thanks, will try that next time i use FormData
19

In case you want to do it like that:

$.upload( form.action, new FormData( myForm))
.progress( function( progressEvent, upload) {
    if( progressEvent.lengthComputable) {
        var percent = Math.round( progressEvent.loaded * 100 / progressEvent.total) + '%';
        if( upload) {
            console.log( percent + ' uploaded');
        } else {
            console.log( percent + ' downloaded');
        }
    }
})
.done( function() {
    console.log( 'Finished upload');                    
});

than

https://github.com/lgersman/jquery.orangevolt-ampere/blob/master/src/jquery.upload.js

might be your solution.

3 Comments

Where is the upload method in $ object, the above link is not existing
Thanks for posting your answer! Please be sure to read the FAQ on Self-Promotion carefully. Also note that it is required that you post a disclaimer every time you link to your own site/product.
18
$("#submit_car").click(function() {
  var formData = new FormData($('#car_cost_form')[0]);
  $.ajax({
     url: 'car_costs.php',
     data: formData,
     contentType: false,
     processData: false,
     cache: false,
     type: 'POST',
     success: function(data) {
       // ...
     },
  });
});

edit: Note contentype and process data You can simply use this to upload files via Ajax...... submit input cannot be outside form element :)

2 Comments

Using this method, you can post form but not with 'file' type fields. This question is specifically about file upload.
It is method: 'POST' not type: 'POST'
17

2019 update:

html

<form class="fr" method='POST' enctype="multipart/form-data"> {% csrf_token %}
<textarea name='text'>
<input name='example_image'>
<button type="submit">
</form>

js

$(document).on('submit', '.fr', function(){

    $.ajax({ 
        type: 'post', 
        url: url, <--- you insert proper URL path to call your views.py function here.
        enctype: 'multipart/form-data',
        processData: false,
        contentType: false,
        data: new FormData(this) ,
        success: function(data) {
             console.log(data);
        }
        });
        return false;

    });

views.py

form = ThisForm(request.POST, request.FILES)

if form.is_valid():
    text = form.cleaned_data.get("text")
    example_image = request.FILES['example_image']

1 Comment

How dow this improve any of the answers already given? Also this answer mentions a views.py file which is Django and has nothing to do with the question.
16
  • Use a hidden iframe and set your form's target to that iframe's name. This way, when the form is submitted, only the iframe will be refreshed.
  • Have an event handler registered for the iframe's load event to parse the response.

3 Comments

Avoid iframes where possible
@BhargavNanekalva what is your concern?
I think it will make a clicking noise when submitted... it's more like workaround that is not suitable for 2020, but will work in IE 7
10

Using FormData is the way to go as indicated by many answers. here is a bit of code that works great for this purpose. I also agree with the comment of nesting ajax blocks to complete complex circumstances. By including e.PreventDefault(); in my experience makes the code more cross browser compatible.

    $('#UploadB1').click(function(e){        
    e.preventDefault();

    if (!fileupload.valid()) {
        return false;            
    }

    var myformData = new FormData();        
    myformData.append('file', $('#uploadFile')[0].files[0]);

    $("#UpdateMessage5").html("Uploading file ....");
    $("#UpdateMessage5").css("background","url(../include/images/loaderIcon.gif) no-repeat right");

    myformData.append('mode', 'fileUpload');
    myformData.append('myid', $('#myid').val());
    myformData.append('type', $('#fileType').val());
    //formData.append('myfile', file, file.name); 

    $.ajax({
        url: 'include/fetch.php',
        method: 'post',
        processData: false,
        contentType: false,
        cache: false,
        data: myformData,
        enctype: 'multipart/form-data',
        success: function(response){
            $("#UpdateMessage5").html(response); //.delay(2000).hide(1); 
            $("#UpdateMessage5").css("background","");

            console.log("file successfully submitted");
        },error: function(){
            console.log("not okay");
        }
    });
});

1 Comment

this runs the form thru jquery validate ... if (!fileupload.valid()) { return false; }
9

I have implemented a multiple file select with instant preview and upload after removing unwanted files from preview via ajax.

Detailed documentation can be found here: http://anasthecoder.blogspot.ae/2014/12/multi-file-select-preview-without.html

Demo: http://jsfiddle.net/anas/6v8Kz/7/embedded/result/

jsFiddle: http://jsfiddle.net/anas/6v8Kz/7/

Javascript:

    $(document).ready(function(){
    $('form').submit(function(ev){
        $('.overlay').show();
        $(window).scrollTop(0);
        return upload_images_selected(ev, ev.target);
    })
})
function add_new_file_uploader(addBtn) {
    var currentRow = $(addBtn).parent().parent();
    var newRow = $(currentRow).clone();
    $(newRow).find('.previewImage, .imagePreviewTable').hide();
    $(newRow).find('.removeButton').show();
    $(newRow).find('table.imagePreviewTable').find('tr').remove();
    $(newRow).find('input.multipleImageFileInput').val('');
    $(addBtn).parent().parent().parent().append(newRow);
}

function remove_file_uploader(removeBtn) {
    $(removeBtn).parent().parent().remove();
}

function show_image_preview(file_selector) {
    //files selected using current file selector
    var files = file_selector.files;
    //Container of image previews
    var imageContainer = $(file_selector).next('table.imagePreviewTable');
    //Number of images selected
    var number_of_images = files.length;
    //Build image preview row
    var imagePreviewRow = $('<tr class="imagePreviewRow_0"><td valign=top style="width: 510px;"></td>' +
        '<td valign=top><input type="button" value="X" title="Remove Image" class="removeImageButton" imageIndex="0" onclick="remove_selected_image(this)" /></td>' +
        '</tr> ');
    //Add image preview row
    $(imageContainer).html(imagePreviewRow);
    if (number_of_images > 1) {
        for (var i =1; i<number_of_images; i++) {
            /**
             *Generate class name of the respective image container appending index of selected images, 
             *sothat we can match images selected and the one which is previewed
             */
            var newImagePreviewRow = $(imagePreviewRow).clone().removeClass('imagePreviewRow_0').addClass('imagePreviewRow_'+i);
            $(newImagePreviewRow).find('input[type="button"]').attr('imageIndex', i);
            $(imageContainer).append(newImagePreviewRow);
        }
    }
    for (var i = 0; i < files.length; i++) {
        var file = files[i];
        /**
         * Allow only images
         */
        var imageType = /image.*/;
        if (!file.type.match(imageType)) {
          continue;
        }

        /**
         * Create an image dom object dynamically
         */
        var img = document.createElement("img");

        /**
         * Get preview area of the image
         */
        var preview = $(imageContainer).find('tr.imagePreviewRow_'+i).find('td:first');

        /**
         * Append preview of selected image to the corresponding container
         */
        preview.append(img); 

        /**
         * Set style of appended preview(Can be done via css also)
         */
        preview.find('img').addClass('previewImage').css({'max-width': '500px', 'max-height': '500px'});

        /**
         * Initialize file reader
         */
        var reader = new FileReader();
        /**
         * Onload event of file reader assign target image to the preview
         */
        reader.onload = (function(aImg) { return function(e) { aImg.src = e.target.result; }; })(img);
        /**
         * Initiate read
         */
        reader.readAsDataURL(file);
    }
    /**
     * Show preview
     */
    $(imageContainer).show();
}

function remove_selected_image(close_button)
{
    /**
     * Remove this image from preview
     */
    var imageIndex = $(close_button).attr('imageindex');
    $(close_button).parents('.imagePreviewRow_' + imageIndex).remove();
}

function upload_images_selected(event, formObj)
{
    event.preventDefault();
    //Get number of images
    var imageCount = $('.previewImage').length;
    //Get all multi select inputs
    var fileInputs = document.querySelectorAll('.multipleImageFileInput');
    //Url where the image is to be uploaded
    var url= "/upload-directory/";
    //Get number of inputs
    var number_of_inputs = $(fileInputs).length; 
    var inputCount = 0;

    //Iterate through each file selector input
    $(fileInputs).each(function(index, input){

        fileList = input.files;
        // Create a new FormData object.
        var formData = new FormData();
        //Extra parameters can be added to the form data object
        formData.append('bulk_upload', '1');
        formData.append('username', $('input[name="username"]').val());
        //Iterate throug each images selected by each file selector and find if the image is present in the preview
        for (var i = 0; i < fileList.length; i++) {
            if ($(input).next('.imagePreviewTable').find('.imagePreviewRow_'+i).length != 0) {
                var file = fileList[i];
                // Check the file type.
                if (!file.type.match('image.*')) {
                    continue;
                }
                // Add the file to the request.
                formData.append('image_uploader_multiple[' +(inputCount++)+ ']', file, file.name);
            }
        }
        // Set up the request.
        var xhr = new XMLHttpRequest();
        xhr.open('POST', url, true);
        xhr.onload = function () {
            if (xhr.status === 200) {
                var jsonResponse = JSON.parse(xhr.responseText);
                if (jsonResponse.status == 1) {
                    $(jsonResponse.file_info).each(function(){
                        //Iterate through response and find data corresponding to each file uploaded
                        var uploaded_file_name = this.original;
                        var saved_file_name = this.target;
                        var file_name_input = '<input type="hidden" class="image_name" name="image_names[]" value="' +saved_file_name+ '" />';
                        file_info_container.append(file_name_input);

                        imageCount--;
                    })
                    //Decrement count of inputs to find all images selected by all multi select are uploaded
                    number_of_inputs--;
                    if(number_of_inputs == 0) {
                        //All images selected by each file selector is uploaded
                        //Do necessary acteion post upload
                        $('.overlay').hide();
                    }
                } else {
                    if (typeof jsonResponse.error_field_name != 'undefined') {
                        //Do appropriate error action
                    } else {
                        alert(jsonResponse.message);
                    }
                    $('.overlay').hide();
                    event.preventDefault();
                    return false;
                }
            } else {
                /*alert('Something went wrong!');*/
                $('.overlay').hide();
                event.preventDefault();
            }
        };
        xhr.send(formData);
    })

    return false;
}

1 Comment

@Bhargav: Please see the blog post for explanations: anasthecoder.blogspot.com/2014/12/…. If you still have any query, get back to me Thanks
8

Yes you can, just use javascript to get the file, making sure you read the file as a data URL. Parse out the stuff before base64 to actually get the base 64 encoded data and then if you are using php or really any back end language you can decode the base 64 data and save into a file like shown below

Javascript:
var reader = new FileReader();
reader.onloadend = function ()
{
  dataToBeSent = reader.result.split("base64,")[1];
  $.post(url, {data:dataToBeSent});
}
reader.readAsDataURL(this.files[0]);


PHP:
    file_put_contents('my.pdf', base64_decode($_POST["data"]));

Of course you will probably want to do some validation like checking which file type you are dealing with and stuff like that but this is the idea.

1 Comment

file_put_contents($fname, file_get_contents($_POST['data'])); file_get_contents deals with the decoding and data:// header
8

To get all your form inputs, including the type="file" you need to use FormData object. you will be able to see the formData content in the debugger -> network ->Headers after you will submit the form.

var url = "YOUR_URL";

var form = $('#YOUR_FORM_ID')[0];
var formData = new FormData(form);


$.ajax(url, {
    method: 'post',
    processData: false,
    contentType: false,
    data: formData
}).done(function(data){
    if (data.success){ 
        alert("Files uploaded");
    } else {
        alert("Error while uploading the files");
    }
}).fail(function(data){
    console.log(data);
    alert("Error while uploading the files");
});

Comments

8
<html>
    <head>
        <title>Ajax file upload</title>
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
        <script>
            $(document).ready(function (e) {
            $("#uploadimage").on('submit', (function(e) {
            e.preventDefault();
                    $.ajax({
                    url: "upload.php", // Url to which the request is send
                            type: "POST", // Type of request to be send, called as method
                            data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
                            contentType: false, // The content type used when sending data to the server.
                            cache: false, // To unable request pages to be cached
                            processData:false, // To send DOMDocument or non processed data file it is set to false
                            success: function(data)   // A function to be called if request succeeds
                            {
                            alert(data);
                            }
                    });
            }));
        </script>
    </head>
    <body>
        <div class="main">
            <h1>Ajax Image Upload</h1><br/>
            <hr>
            <form id="uploadimage" action="" method="post" enctype="multipart/form-data">
                <div id="image_preview"><img id="previewing" src="noimage.png" /></div>
                <hr id="line">
                <div id="selectImage">
                    <label>Select Your Image</label><br/>
                    <input type="file" name="file" id="file" required />
                    <input type="submit" value="Upload" class="submit" />
                </div>
            </form>
        </div>
    </body>
</html>

Comments

7

to upload a file which is submitted by user as a part of form using jquery please follow the below code :

var formData = new FormData();
formData.append("userfile", fileInputElement.files[0]);

Then send the form data object to server.

We can also append a File or Blob directly to the FormData object.

data.append("myfile", myBlob, "filename.txt");

Comments

6

You can use method ajaxSubmit as follow :) when you select a file that need upload to server, form be submit to server :)

$(document).ready(function () {
    var options = {
    target: '#output',   // target element(s) to be updated with server response
    timeout: 30000,
    error: function (jqXHR, textStatus) {
            $('#output').html('have any error');
            return false;
        }
    },
    success: afterSuccess,  // post-submit callback
    resetForm: true
            // reset the form after successful submit
};

$('#idOfInputFile').on('change', function () {
    $('#idOfForm').ajaxSubmit(options);
    // always return false to prevent standard browser submit and page navigation
    return false;
});
});

2 Comments

I believe you are talking about jquery form plugin. It is really the best option here, apart from the lack of details in your answer.
@fotanus you're right! that script must use jquery form plugin to submit use method ajaxSubmit that define in jquery form plugin
5

If you want to upload file using AJAX here is code which you can use for file uploading.

$(document).ready(function() {
    var options = { 
                beforeSubmit:  showRequest,
        success:       showResponse,
        dataType: 'json' 
        }; 
    $('body').delegate('#image','change', function(){
        $('#upload').ajaxForm(options).submit();        
    }); 
});     
function showRequest(formData, jqForm, options) { 
    $("#validation-errors").hide().empty();
    $("#output").css('display','none');
    return true; 
} 
function showResponse(response, statusText, xhr, $form)  { 
    if(response.success == false)
    {
        var arr = response.errors;
        $.each(arr, function(index, value)
        {
            if (value.length != 0)
            {
                $("#validation-errors").append('<div class="alert alert-error"><strong>'+ value +'</strong><div>');
            }
        });
        $("#validation-errors").show();
    } else {
         $("#output").html("<img src='"+response.file+"' />");
         $("#output").css('display','block');
    }
}

Here is the HTML for Upload the file

<form class="form-horizontal" id="upload" enctype="multipart/form-data" method="post" action="upload/image'" autocomplete="off">
    <input type="file" name="image" id="image" /> 
</form>

Comments

5
<input class="form-control cu-b-border" type="file" id="formFile">
<img id="myImg" src="#">

In js

<script>
    var formData = new FormData();
    formData.append('file', $('#formFile')[0].files[0]);
    $.ajax({
        type: "POST",
        url: '/GetData/UploadImage',
        data: formData,
        processData: false, // tell jQuery not to process the data
        contentType: false, // tell jQuery not to set contentType
        success: function (data) {
            console.log(data);
            $('#myImg').attr('src', data);
        },
        error: function (xhr, ajaxOptions, thrownError) {
        }
    })
</script>

In controller

public ActionResult UploadImage(HttpPostedFileBase file)
        {
            string filePath = "";
            if (file != null)
            {
                string path = "/uploads/Temp/";
                if (!Directory.Exists(Server.MapPath("~" + path)))
                {
                    Directory.CreateDirectory(Server.MapPath("~" + path));
                }
                filePath = FileUpload.SaveUploadedFile(file, path);
            }
            
            return Json(filePath, JsonRequestBehavior.AllowGet);
        }

Comments

4
var dataform = new FormData($("#myform")[0]);
//console.log(dataform);
$.ajax({
    url: 'url',
    type: 'POST',
    data: dataform,
    async: false,
    success: function(res) {
        response data;
    },
    cache: false,
    contentType: false,
    processData: false
});

1 Comment

you can improve your answer by adding some details
3

Here was an idea I was thinking of:

Have an iframe on page and have a referencer.

Have a form in which you move the input type file element to.

Form:  A processing page AND a target of the FRAME.

The result will post to the iframe, and then you can just send the fetched data up a level to the image tag you want with something like:

data:image/png;base64,asdfasdfasdfasdfa

and the page loads.

I believe it works for me, and depending you might be able to do something like:

.aftersubmit(function(){
    stopPropagation(); // or some other code which would prevent a refresh.
});

1 Comment

I don't see how this improves any other answer given before. Also it's Propagation not propigation! ;)
2

This is my code that it works

var formData = new FormData();
var files = $('input[type=file]');
for (var i = 0; i < files.length; i++) {
if (files[i].value == "" || files[i].value == null) {
 return false;
}
else {
 formData.append(files[i].name, files[i].files[0]);
}
}
var formSerializeArray = $("#Form").serializeArray();
for (var i = 0; i < formSerializeArray.length; i++) {
  formData.append(formSerializeArray[i].name, formSerializeArray[i].value)
}
$.ajax({
 type: 'POST',
 data: formData,
 contentType: false,
 processData: false,
 cache: false,
 url: '/Controller/Action',
 success: function (response) {
 if (response.Success == true) {
    return true;
 }
 else {
    return false;
 }
 },
 error: function () {
   return false;
 },
 failure: function () {
   return false;
 }
 });

Comments

1

A 2023 solution using jQuery3 and .Net

HTML:

<form>
  <input type=file /> <button class=upload>Upload</button>
</form>

JS:

$('button.upload').closest('form').on('submit', ev => {
  ev.preventDefault();
  const form = $(ev.target);
  const files = form.find('input[type=file]')[0].files;
  const formData = new FormData();
  formData.append('upload', files[0]);

  $.ajax({
    url: '/manage/upload',
    type: 'POST',
    data: formData,
    encType: 'multipart/form-data',  
    contentType: false,
    processData: false
  })
  .done(r => {
    debugger;
  });

  return false;
});

Asp.Net MVC Controller:

[HttpPost]
public async Task<JsonResult> Upload(HttpPostedFileBase upload)
{
  string s = null;
  using (upload.InputStream)
    using (var reader = new StreamReader(stream, true))
      s = await reader.ReadToEndAsync();

  // Do something with the uploaded contents of the file
  return Json(true);
}

Notes about this code:

formData.upload('<name goes here>',... - in this case, 'upload' - must match the parameter name in your .Net Controller Action (HttpPostedFileBase upload), So if you change the key in the AJAX call from 'upload' to 'file', you need to change the Action parameter like HttpPostedFileBase file if you want this to be easy. It can be done without matching these form body keys, but, it's a bunch of extra work.

This code assumes you have only 1 file upload control in your form. You can use more upload inputs in a form but it's more complicated - but it's very easy to use this code with 1 input type=file per form, and multiple such form/input pairs. The form lacking any attributes at all is intentional and totally fine, given it's never meant to submit anywhere on its own.

The url property of the options passed in js to the ajax call should be the path to the upload action on your Controller, wherever that is. There's nothing special here, just a plain web path.

The Controller Action here happens to consume the file on the premise that it's a text document, but, you don't need to do this. I'm just including how to read it in a small number of lines of code as one thing you might conveniently do with it.

I've left out any error handling to cut to the chase, and present the least possible code to get this to work.

The arrow function in the submit handler does not offer a this object, so if you were thinking of using $(this) or this.files... in that handler instead of $(ev.target), don't do it.

Comments

0

$("#form-id").submit(function (e) {
   e.preventDefault();
});

$("#form-id").submit(function (e) {

     var formObj = $(this);
     var formURL = formObj.attr("action");
     var formData = new FormData(this);
          $.ajax({
             url: formURL,
             type: 'POST',
             data: formData,
             processData: false,
             contentType: false,
             async: true,
             cache: false,
             enctype: "multipart/form-data",
             dataType: "json",
             success: function (data) {
                 if (data.success) {
                        alert(data.success)
                 } 
                                
                 if  (data.error) {
                        alert(data.error)
                 } 
             }
         });
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>

<form class="form-horizontal" id="form-id" action="masterFileController" enctype="multipart/form-data">
    <button class="btn-success btn" type="submit" id="btn-save" >Submit</button>
</form>

servlet responce as "out.print("your responce");"

Comments

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