Ajax return value displaying code

Question

I have a form where the user uploads an image and the image is then relayed through Ajax to PHP. After the image is uploaded on the server, PHP will echo back the $target_file to Ajax. $target_file is essentially the URL of the image. Ajax will take that variable and then display it through the "readonly" input text box in the same form. After testing what I have, the image is being uploaded successfully. The only problem is that when the image is uploaded successfully, the entire code of the PHP script is being shown in the "readonly" input box instead of the $target_file variable that should display the URL of the image.

Here's what I have:

HTML

<form class="image_upload" method="post" action="ajax.php">
    <input name="server_upload" type="file">
     <input type="text" name="image_url" readonly>
    <input type="submit" value="UPLOAD TO SERVER">
</form>

Ajax

$('.image_upload').on('submit', function(e) {
        e.preventDefault();
        e.stopImmediatePropagation();
        if ($('.image_upload').valid())
        {

            $.ajax({
              type: 'POST',
              url: 'ajax.php',
              data: new FormData(this),
              processData: false,
              contentType: false,
              success: function(data) {
                     $('input[name=image_url]').val(data);
              }
            })
     }
 })

PHP

if (isset($_FILES["server_upload"]))
    {
        $name = $_FILES["server_upload"]["name"];
        $tempName = $_FILES["server_upload"]["tmp_name"];
        $target_file = $_SERVER['DOCUMENT_ROOT'] . "/stories/media/images/$name";
        if (getimagesize($target_file) == true)
        {
            $ext = pathinfo($name, PATHINFO_EXTENSION);     
            $name = basename($name, "." . $ext);
            $name = $name . uniqid() . "." . $ext;
            $target_file = $_SERVER['DOCUMENT_ROOT'] . "/stories/media/images/$name";
        }

        move_uploaded_file($tempName, $target_file);
      echo $target_file;
    }

How can I fix this so the URL is shown? Why is it printing out all my PHP code?

T.J. Crowder · Accepted Answer · 2016-08-05 08:32:19Z

3

.php files are typically HTML files with embedded PHP in them. The embedded sections of PHP must be marked off as PHP by using <?php to start and ?> to end. So put those around your code so your code is run, rather than being treated as HTML text:

<?php
if (isset($_FILES["server_upload"]))
    {
        $name = $_FILES["server_upload"]["name"];
        $tempName = $_FILES["server_upload"]["tmp_name"];
        $target_file = $_SERVER['DOCUMENT_ROOT'] . "/stories/media/images/$name";
        if (getimagesize($target_file) == true)
        {
            $ext = pathinfo($name, PATHINFO_EXTENSION);     
            $name = basename($name, "." . $ext);
            $name = $name . uniqid() . "." . $ext;
            $target_file = $_SERVER['DOCUMENT_ROOT'] . "/stories/media/images/$name";
        }

        move_uploaded_file($tempName, $target_file);
      echo $target_file;
    }
?>

answered Aug 5, 2016 at 8:32

T.J. Crowder

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1 Comment

user2896120 Over a year ago

I can't believe I forgot to put that! It works correctly now! I'll accept your answer soon :)

simialbi · Accepted Answer · 2016-08-05 08:34:13Z

0

That occurs if you don't have some kind of webserver like apache or nginx which executes the php script or if the mentioned webserver do not know how to handle php files. In this case php files will be treated as plain text files.

answered Aug 5, 2016 at 8:34

simialbi

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