Reputation: 1883
I am trying to create an image upload field in my application based on this question:
Send FormData and String Data Together Through JQuery AJAX?
and this tutorial:
http://www.formget.com/ajax-image-upload-php/
I have heard it is quite difficult, this is what i have tried.
HTML
<form method="POST" action="" id="logo_upload">
<input type="file" name="logo_location" id="logo_location" enctype="multipart/form-data">
<button type="submit" name="file_test" id="file_test">Test Upload</button>
</form>
JQuery
$('#logo_upload').submit(function(e) {
e.preventDefault();
var formData = new FormData();
var file_data = $('#logo_location')[0].files[0];
formData.append("file", file_data[0]);
$.ajax({
url: "../../../controllers/ajax_controller.php?action=image_upload",
type: 'POST',
data: formData ,
cache: false,
contentType: false,
processData: false,
id: id
});
});
PHP
var_dump($_FILES);
var_dump($_POST);
As you can see, I haven't got to the uploading side of things yet.
Result
<pre class='xdebug-var-dump' dir='ltr'>
<b>array</b> <i>(size=0)</i>
<i><font color='#888a85'>empty</font></i>
</pre>
<pre class='xdebug-var-dump' dir='ltr'>
<b>array</b> <i>(size=0)</i>
<i><font color='#888a85'>empty</font></i>
</pre>
I can't see what i am doing wrong, I am getting a result so it is getting to the right place, can anyone point me in the right direction?
EDIT: added #logo_upload in form
var file_data = $('#logo_location')[0].files[0];
EDIT: replaced data with formData variable
EDIT: added attribute: enctype="multipart/form-data"
New Result:
<pre class='xdebug-var-dump' dir='ltr'>
<b>array</b> <i>(size=1)</i>
'file' <font color='#888a85'>=></font> <small>string</small> <font color='#cc0000'>'undefined'</font> <i>(length=9)</i>
</pre>
Upvotes: 0
Views: 203
Reputation: 97672
You're appending file_data[0]
to the formdata object, file_data
is the file not an array, use file_data
.
$('#logo_upload').submit(function(e) {
e.preventDefault();
var formData = new FormData();
var file_data = $('#logo_location')[0].files[0];
formData.append("file", file_data);
$.ajax({
url: "../../../controllers/ajax_controller.php?action=image_upload",
type: 'POST',
data: formData ,
contentType: false,
processData: false,
success: function(data){
console.log(data);
}
});
});
also you can instantiate the form data object with the form in question instead of doing the append.
$('#logo_upload').submit(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
...
Upvotes: 1
Reputation: 616
Hope this would help you
Upvotes: 1
Reputation: 5792
You are passing wrong variable here and you are not defining success in your ajax request:
$('#logo_upload').submit(function(e) {
e.preventDefault();
var formData = new FormData($('#your_form_id')[0]);
$.ajax({
url: "../../../controllers/ajax_controller.php?action=image_upload",
type: 'POST',
data: formData,
success: function(result){
alert(result);
}
cache: false,
contentType: false,
processData: false
});
});
Upvotes: 0