Ajax request returns nothing. why?

Question

Below is the ajax request.

$.post(\'delete.php\', {\'deletearray\':deletearray, \'dir\':dir}, function(deleted, undeleted){
    if(undeleted == 0) {
        aler         


        

Answer 1

You should use json_encode like following:

json_encode(array('deleted' => $deleted, 'undeleted' => $undeleted));

And you have to get vars with data.undeleted and data.deleted

$.post('delete.php', {'deletearray':deletearray, 'dir':dir}, function(data) {
    if(data.undeleted == 0) {
        alert('All ' + data.deleted + ' files delted from the server');
    } else {
        alert(data.deleted + ' files deleted and ' + data.undeleted + ' files could not be deleted');
    }
}, 'json');

Answer 2

The best way to do jquery + ajax + php is as next:

jquery:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript">

function do_ajax() { 
        //set data
        var myData  = new Array();
        myData.push({name:'deletearray',value:'deletearray'});
        myData.push({name:'dir',value:'dir'});
        //ajax post
        $.ajax({
            dataType: 'json',
            url: 'delete.php',
            type: 'post',
            data: myData,
            success: function(returnData) {
                if(returnData.undeleted == 0) {
                    alert('All ' + returnData.deleted + ' files delted from the server');
                } else {
                    alert(returnData.deleted + ' files deleted and ' + returnData.undeleted + ' files could not be deleted');
                }
            }
        });
}            
</script>

PHP:

<?php
    $myData = $_POST;
    if(isset($myData['deletearray']) AND isset($myData['dir'])) {
        $files = $myData['deletearray'];
        $dir = $myData['dir'];
        $deleted = 0;
        $undeleted = 0;

        foreach($files as $file) {
            if(unlink($dir.$file) && unlink($dir.'thumb/'.$file)) {
                $deleted ++;
            } else {
                $undeleted ++;
            }
        }
        print(json_encode(array('deleted' => $deleted, 'undeleted' => $undeleted)));
        exit();
    }
?>

Answer 3

First thing-

Use $_POST['deletearray'] instead of $_POST[deletearray]

Second thing-

You cannot return different variables from the PHP scrtipt, every thing you print there is returned in the ajax callback, so just write this-

PHP

json_encode(array('totalDeleted' => $deleted, 'totalUndeleted' => $undeleted));

AJAX

...
function(response){
     response=JSON.parse(response);
     console.log(response);
}