Does anyone know how to upload images using PHP and calling the UploadHandler.php?
I'm not sure what information needs to be passed and in what format.
Here's what I have so far:
$prop="test";
session_id($prop);
@session_start();
$url = 'http://christinewilson.ca/wp-content/uploads/2013/02/port_rhdefence.png';
$file_name[] = file_get_contents($url);
error_reporting(E_ALL | E_STRICT);
require('UploadHandler.php');
$upload_handler = new UploadHandler(array(
'user_dirs' => true
));
The response is contained within the UploadHandler class object and can be retrieved like shown below.
$upload_handler = new UploadHandler();
$response = $upload_handler->response;
$files = $response['files'];
$file_count = count($files);
for ($c = 0; $c < $file_count; $c++) {
if (isset($files[$c]->error))
continue;
$type = $files[$c]->type;
$name = $files[$c]->name;
$url = $files[$c]->url;
}
I could not find a way to get the file name via php so I had to do it myself.
First you need to add a public variable under UploadHandler.php
class UploadHandler
{
public $file_name;
protected $options;
and then add that to the function that creates the name
protected function get_file_name($name,
$type = null, $index = null, $content_range = null) {
$this->file_name = $this->get_unique_filename(
$this->trim_file_name($name, $type, $index, $content_range),
$type,
$index,
$content_range
);
return $this->file_name;
}
then under index.php you could do something like this
$upload_handler = new UploadHandler();
echo "\r\n [" . $upload_handler->fileName . "]\r\n";
I hope this help you or save someone some time :)
you can use the basic plugin:
<!DOCTYPE HTML>
<html>
<head>
<meta charset="utf-8">
<title>jQuery File Upload Example</title>
</head>
<body>
<input id="fileupload" type="file" name="files[]" data-url="server/php/" multiple>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="js/vendor/jquery.ui.widget.js"></script>
<script src="js/jquery.iframe-transport.js"></script>
<script src="js/jquery.fileupload.js"></script>
<script>
$(function () {
$('#fileupload').fileupload({
dataType: 'json',
done: function (e, data) {
$.each(data.result.files, function (index, file) {
$('<p/>').text(file.name).appendTo(document.body);
});
}
});
});
</script>
</body>
I ran into the same problem, where in the PHP I wanted to write all the URLS that UploadHandler.php had created to a mySQL database. If you look through the code, you'll see that
public function post($print_response = true)
actually returns the data structure from generate_response (which is a array with all the processed image metadata like image size, sanitized url, etc), but the call to $this->post() never does anything which it. So I add a variable
protected $upload_content = [];
to the class definition and changed the logic in function
protected function initialize()
to
case 'POST':
$this->upload_content = $this->post(false);
break;
to update this variable after the images have been processed (you would need to do something similar with the GET case if you are using that). Then, I add a public function to the class to get this variable
public function get_upload_content() {
return $this->upload_content;
}
and now the UploadHandler class can be called like this
$upload_handler = new UploadHandler();
$images = $upload_handler->get_upload_content();
// Call a function that writes the urls in $images array to a database
Hope this helps!
First of all you should create protected variable:
protected $options;
protected $uploaded_files = [];
then you should assign to this variable the response value in post() method:
$this->uploaded_files = $response;
return $this->generate_response($response, $print_response);
then you should create public method which would return that response:
public function get_uploaded_files() {
return $this->uploaded_files;
}
and finally you should initiate the class and call the method:
$uploadPicture = new UploadHandler();
$images = $uploadPicture->get_uploaded_files();
Hope this Helps !
'param_name' => 'files'in the class construct, so i guess it expects<input type="file" name="files" />$_FILESarray with the path to the file and info. You access that in the same way as any other array e.g:$_FILES['files']. Check out php.net/manual/en/features.file-upload.post-method.php