Inserting images through javascript code to the database

Question

txtfld.php

<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js">     </script>
<script>
$(function () {
var inputName = 'myInput';
$('select[name=myselect]').change(function () {
    $('.inputs').html('');

     var forms = $('<form/>').attr({name:'submitBtn',method:'post',id:'upload-image',enctype:"multipart/form-data",action:'upload-image.php'});
     $('.inputs').append(forms);
    for (var i = 0; i < $(this).val(); i++) {
        var inputs = $('<input />').attr({
            name: inputName + i,
            type: 'file'
        });
        $('<label>Input ' + (i + 1) + '</label>').appendTo( "#upload-image" );
        $(inputs).appendTo( "#upload-image" );
        $("<br>").appendTo( "#upload-image" );
    }
    var inputs = $('<input />').attr({name:'total',type: 'hidden',value:$(this).val()});
    $(inputs).appendTo( "#upload-image" );
    var inputs = $('<input />').attr({name:'submitBtn',type: 'submit',value:'upload'});
    $(inputs).appendTo( "#upload-image" );
});
});
</script>
</head>
<body>
<select name="myselect">
<option>Select</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
</select>
<br />
<div class='inputs'></div>
</body>

upload-image.php

<?php
if (isset($_POST['submitBtn'])) {
$fileNames = false;
for ($i = 0; $i < $_POST['total']; $i++) {
    if (!isset($_FILES[$name]["name"]))
        continue;
    $allowedExts = array("gif", "jpeg", "jpg", "png");
    $name = 'myInput' . $i;
    $temp = explode(".", $_FILES[$name]["name"]);
    $extension = end($temp);
    if ((($_FILES[$name]["type"] == "image/gif") || ($_FILES[$name]["type"] == "image/jpeg") || ($_FILES[$name]["type"] == "image/jpg") || ($_FILES[$name]["type"] == "image/pjpeg") || ($_FILES[$name]["type"] == "image/x-png") || ($_FILES[$name]["type"] == "image/png")) && ($_FILES[$name]["size"] < 20000) && in_array($extension, $allowedExts)) {
        if ($_FILES[$name]["error"] > 0) {
            echo "Return Code: " . $_FILES[$name]["error"] . "<br>";
        } else {


            if (file_exists("upload/" . $_FILES[$name]["name"])) {
                echo $_FILES[$name]["name"] . " already exists. ";
            } else {
                move_uploaded_file($_FILES[$name]["tmp_name"], "upload/" . $_FILES[$name]["name"]);
                echo "Stored in: " . "upload/" . $_FILES[$name]["name"];
                $fileNames[] = $_FILES[$name]["name"];
            }
        }
    } else {
        echo "Invalid file";
    }
 }
if (is_array($fileNames)) {
    $files = implode(",", $fileNames);

    $con = mysql_connect("localhost", "root", "") or die(mysql_error());
    $db = mysql_select_db("test1") or die(mysql_error());
    $query = "INSERT INTO image_java ('image') VALUES ('" . addslashes($files) . "')";
    $result = mysql_query($query) or die(mysql_error());
}
}
?>

As i got this code from stack overflow to insert image through java script to the database, But here i m getting this undefined variable $name error in upload-image.php page on the line 5.. As we can see, here $name has not been declared, so what should i post in the $name variable.. So i want to insert my images to the database into one field through this code or any other ???? Any help.. Thankss

Vicky Gonsalves · Accepted Answer · 2014-01-23 06:32:52Z

1

Replace all $_FILES[$name]["name"] with $_FILES["myInput".$i]["name"]

answered Jan 23, 2014 at 6:32

Vicky Gonsalves

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