How to get single value from array after using print_r

Question

I have an array $_FILES it can store multiple images.

When I use print_r($_FILES) after storing two images in this array it gives me following output:

Array ( [IMG_20170304_224950_jpg] => Array ( [name] => IMG_20170304_224950.jpg [type] => image/jpeg [tmp_name] => C:\wamp64\tmp\php4584.tmp [error] => 0 [size] => 4098228 ) [IMG_20170305_075610_jpg] => Array ( [name] => IMG_20170305_075610.jpg [type] => image/jpeg [tmp_name] => C:\wamp64\tmp\php4602.tmp [error] => 0 [size] => 5314040 ) )

In this I want to echo both of two images.

This is how my code looks like:

index.php

     <input id="avatar" type="file" name="avatar[]" multiple />
     <button id="upload" value="Upload" type="button">upload</button> 
    <div class="preview">
    </div>

   <div class="return_php"></div>

  <script   src="https://code.jquery.com/jquery-3.1.0.min.js" ></script>
 <script>
 $(document).ready(function(){
    var form_data = new FormData(); 
    var number = 0;

    /* WHEN YOU UPLOAD ONE OR MULTIPLE FILES */
    $(document).on('change','#avatar',function(){
        console.log($("#avatar").prop("files").length);
        len_files = $("#avatar").prop("files").length;
        for (var i = 0; i < len_files; i++) {
            var file_data = $("#avatar").prop("files")[i];
            form_data.append(file_data.name, file_data);
            number++;
            var construc = '<img width="200px" height="200px" src="' +  window.URL.createObjectURL(file_data) + '" alt="'  +  file_data.name  + '" />';
            $('.preview').append(construc); 
        }
    }); 

    /* WHEN YOU CLICK ON THE IMG IN ORDER TO DELETE IT */
    $(document).on('click','img',function(){

        var filename = $(this).attr('alt');
        var newfilename = filename.replace(/\./gi, "_");
        form_data.delete($(this).attr('alt'))
        $(this).remove()

    });

    /* UPLOAD CLICK */
    $(document).on("click", "#upload", function() {
        $.ajax({
                    url: "target.php",
                    dataType: 'script',
                    cache: false,
                    contentType: false,
                    processData: false,
                    data: form_data,                         // Setting the data attribute of ajax with form_data
                    type: 'post',
                    success: function(data){
                        $('.return_php').html(data);
                    }
           })
    })
});

target.php

<?php


 var_dump($_FILES);
 print_r ($_FILES);

foreach ($array as $key => $image) {
  /* Now you can get each image name */
 echo $image['name'];
 }
 ?>

but what are you want to print? the image name? a <img tag with images files? — Oscar Lidenbrock
– Oscar Lidenbrock, Commented Sep 9, 2018 at 10:48

Oscar Lidenbrock · Accepted Answer · 2018-09-09 16:49:54Z

1

You can try this code to print the image name:

foreach ($array as $key => $image) {
  /* Now you can get each image name */
  echo $image['name'];
}

EDIT: This is a upload file example

index.html

<form action="upload.php" method="POST" enctype="multipart/form-data">
    <label for="fileToUpload">Select image to upload:</label>
    <input type="file" name="uploadFile" id="fileToUpload">
    <input type="submit" value="Upload File" name="submit">
</form>

upload.php

if (isset($_FILES['uploadFile'])) {
  print "Your file information:";
  var_dump($_FILES['uploadFile']);
} else {
  print "You need to upload a file.";
}

edited Sep 9, 2018 at 16:49

answered Sep 9, 2018 at 10:58

Oscar Lidenbrock

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9 Comments

Oscar Lidenbrock Over a year ago

No problem, glad to help you if i can.

Nitin Verma Over a year ago

what if i want to upload the image?

Oscar Lidenbrock Over a year ago

You need to do a HTML form with a <input type="file" name="..." /> and get de upload image in your php with the $_FILES array. Try to search "upload file with php" in google, it's easy.

Nitin Verma Over a year ago

are you here sir? if have seen this comment plz see my code below and tell me how can i upload file cause i am not able to use input name. i don't understand ajax so much ,should i use form_data instead of avatar

Oscar Lidenbrock Over a year ago

You don't need to use Ajax, try to add a <input type="file" name="uploadFile" /> in your form and then, in your php code you can access to file with $_FILES['uploadFile'], put a var_dump($_FILES); to see the variables.

|

Mani Ratna · Accepted Answer · 2018-09-09 11:08:00Z

0

you can try this

foreach($array as $val)
{ 
    echo $val['name'];
}

answered Sep 9, 2018 at 11:08

Mani Ratna

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1 Comment

Clément Prévost Over a year ago

This may be an answer the question. But I agree that the question itself is not clear and lacks basic php knowledge. Please explain in details why your solution might work for @nitin-verma and also try to explain what print_r does and what $_FILES is. All this information could help him understand what went wrong.

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