jQuery AJAX file upload PHP

JavascriptPhpJqueryAjaxUpload

Javascript Problem Overview


I want to implement a simple file upload in my intranet-page, with the smallest setup possible.

This is my HTML part:

<input id="sortpicture" type="file" name="sortpic" />
<button id="upload">Upload</button>

and this is my JS jquery script:

$("#upload").on("click", function() {
    var file_data = $("#sortpicture").prop("files")[0];   
    var form_data = new FormData();
    form_data.append("file", file_data);
    alert(form_data);
    $.ajax({
        url: "/uploads",
        dataType: 'script',
        cache: false,
        contentType: false,
        processData: false,
        data: form_data,                         
        type: 'post',
        success: function(){
        	alert("works"); 
        }
    });
});

There is a folder named "uploads" in the root directory of the website, with change permissions for "users" and "IIS_users".

When I select a file with the file-form and press the upload button, the first alert returns "[object FormData]". the second alert doesn't get called and the"uploads" folder is empty too!?

Can someone help my finding out whats wrong?

Also the next step should be, to rename the file with a server side generated name. Maybe someone can give me a solution for this, too.

Javascript Solutions


Solution 1 - Javascript

You need a script that runs on the server to move the file to the uploads directory. The jQuery ajax method (running on the client in the browser) sends the form data to the server, then a script running on the server handles the upload.

Your HTML is fine, but update your JS jQuery script to look like this:

(Look for comments after // <-- )

$('#upload').on('click', function() {
    var file_data = $('#sortpicture').prop('files')[0];   
    var form_data = new FormData();                  
    form_data.append('file', file_data);
    alert(form_data);                             
    $.ajax({
        url: 'upload.php', // <-- point to server-side PHP script 
        dataType: 'text',  // <-- what to expect back from the PHP script, if anything
        cache: false,
        contentType: false,
        processData: false,
        data: form_data,                         
        type: 'post',
        success: function(php_script_response){
            alert(php_script_response); // <-- display response from the PHP script, if any
        }
     });
});

And now for the server-side script, using PHP in this case.

upload.php: a PHP script that is located and runs on the server, and directs the file to the uploads directory:

<?php

    if ( 0 < $_FILES['file']['error'] ) {
        echo 'Error: ' . $_FILES['file']['error'] . '<br>';
    }
    else {
        move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
    }

?>

Also, a couple things about the destination directory:

  1. Make sure you have the correct server path, i.e., starting at the PHP script location what is the path to the uploads directory, and
  2. Make sure it's writeable.

And a little bit about the PHP function move_uploaded_file, used in the upload.php script:

move_uploaded_file(

    // this is where the file is temporarily stored on the server when uploaded
    // do not change this
    $_FILES['file']['tmp_name'],

    // this is where you want to put the file and what you want to name it
    // in this case we are putting in a directory called "uploads"
    // and giving it the original filename
    'uploads/' . $_FILES['file']['name']
);

$_FILES['file']['name'] is the name of the file as it is uploaded. You don't have to use that. You can give the file any name (server filesystem compatible) you want:

move_uploaded_file(
    $_FILES['file']['tmp_name'],
    'uploads/my_new_filename.whatever'
);

And finally, be aware of your PHP upload_max_filesize AND post_max_size configuration values, and be sure your test files do not exceed either. Here's some help how you check PHP configuration and how you set max filesize and post settings.

Solution 2 - Javascript

**1. index.php**
<body>
    <span id="msg" style="color:red"></span><br/>
    <input type="file" id="photo"><br/>
  <script type="text/javascript" src="jquery-3.2.1.min.js"></script>
  <script type="text/javascript">
    $(document).ready(function(){
      $(document).on('change','#photo',function(){
        var property = document.getElementById('photo').files[0];
        var image_name = property.name;
        var image_extension = image_name.split('.').pop().toLowerCase();

        if(jQuery.inArray(image_extension,['gif','jpg','jpeg','']) == -1){
          alert("Invalid image file");
        }

        var form_data = new FormData();
        form_data.append("file",property);
        $.ajax({
          url:'upload.php',
          method:'POST',
          data:form_data,
          contentType:false,
          cache:false,
          processData:false,
          beforeSend:function(){
            $('#msg').html('Loading......');
          },
          success:function(data){
            console.log(data);
            $('#msg').html(data);
          }
        });
      });
    });
  </script>
</body>

**2.upload.php**
<?php
if($_FILES['file']['name'] != ''){
	$test = explode('.', $_FILES['file']['name']);
	$extension = end($test); 	
	$name = rand(100,999).'.'.$extension;

	$location = 'uploads/'.$name;
	move_uploaded_file($_FILES['file']['tmp_name'], $location);

	echo '<img src="'.$location.'" height="100" width="100" />';
}

Solution 3 - Javascript

Use pure js

async function saveFile() 
{
    let formData = new FormData();           
    formData.append("file", sortpicture.files[0]);
    await fetch('/uploads', {method: "POST", body: formData});    
    alert('works');
}

<input id="sortpicture" type="file" name="sortpic" />
<button id="upload" onclick="saveFile()">Upload</button>
<br>Before click upload look on chrome>console>network (in this snipped we will see 404)

The filename is automatically included to request and server can read it, the 'content-type' is automatically set to 'multipart/form-data'. Here is more developed example with error handling 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));  
    
    try {
       let r = await fetch('/upload/image', {method: "POST", body: formData}); 
       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>

Solution 4 - Javascript

var formData = new FormData($("#YOUR_FORM_ID")[0]);
$.ajax({
    url: "upload.php",
    type: "POST",
    data : formData,
    processData: false,
    contentType: false,
    beforeSend: function() {
        
    },
    success: function(data){
        



    },
    error: function(xhr, ajaxOptions, thrownError) {
       console.log(thrownError + "\r\n" + xhr.statusText + "\r\n" + xhr.responseText);
    }
});

Solution 5 - Javascript

and this is the php file to receive the uplaoded files

<?
$data = array();
    //check with your logic
    if (isset($_FILES)) {
        $error = false;
        $files = array();
        
        $uploaddir = $target_dir;
        foreach ($_FILES as $file) {
            if (move_uploaded_file($file['tmp_name'], $uploaddir . basename( $file['name']))) {
                $files[] = $uploaddir . $file['name'];
            } else {
                $error = true;
            }
        }
        $data = ($error) ? array('error' => 'There was an error uploading your files') : array('files' => $files);
    } else {
        $data = array('success' => 'NO FILES ARE SENT','formData' => $_REQUEST);
    }
    
    echo json_encode($data);
?>

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionuser2355509View Question on Stackoverflow
Solution 1 - JavascriptbloodyKnucklesView Answer on Stackoverflow
Solution 2 - JavascriptHasib Kamal ChowdhuryView Answer on Stackoverflow
Solution 3 - JavascriptKamil KiełczewskiView Answer on Stackoverflow
Solution 4 - JavascriptAzharView Answer on Stackoverflow
Solution 5 - JavascriptkavehmbView Answer on Stackoverflow