AJAX return variable from PHP

Question

When a user clicks download it will successfully create a zip on server with the files, then it should alert the the zips location (variable $zip) from php as a response but instead it is alerting [object Object]. Everything else is working how it should. What am I doing wrong?

JQuery:

$('.download').click(function() { 
window.keys = [];
$('.pad').each(function(i, obj) {
    var key = $(this).attr('key');
        keys.push(key)
});
var jsonString = JSON.stringify(keys);
$.ajax({
      type:'post',
    url:'download.php',
    data: {data : jsonString}, 
        cache: false,
   dataType: 'json',
    success: function(data){

       alert(data);

      }
 });
});

PHP:

<?php


$data = json_decode(stripslashes($_POST['data']));

$numbercode = md5(microtime());
$zip = new ZipArchive();
$zip->open('kits/'.$numbercode.'.zip', ZipArchive::CREATE);

foreach($data as $d) {

$zip->addFile($d);  

}

$zip->close();



echo json_encode($zip);
?>

$zip is an object... you are declaring it as new ZipArchive(), therefore it cannot be a string (the file's location) — jake2389
– jake2389, Commented Feb 20, 2017 at 3:40

l3utterfly · Accepted Answer · 2017-02-20 03:41:25Z

2

The return type is a JavaScript object, which will result in what you see.

First, you should console.log(data), to get the structure. You can also do this by looking at the Network Tab in Chrome.

After you know the structure of data, you can use the value.

For example, then alert(data.location), to alert the actual value.

answered Feb 20, 2017 at 3:41

l3utterfly

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Comments

Veshraj Joshi · Accepted Answer · 2017-02-20 03:42:53Z

1

remove your dataType from the ajax, it alert [Object Object] because your result becomes json object if you specify dataType: 'json',,

and in php-

// to echo the location of the zipfile
echo 'kits/'.$numbercode.'.zip';

answered Feb 20, 2017 at 3:42

Veshraj Joshi

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Comments

Jeff · Accepted Answer · 2017-02-20 03:44:42Z

0

Thanks to @jake2389 I see what I was doing wrong, I basically just had to create a new variable within PHP which I called $link with the data I wanted to send back to AJAX because $zip was defined as a zip archive not a string. Here is what I changed and now it is working.

PHP:

<?php


$data = json_decode(stripslashes($_POST['data']));

$numbercode = md5(microtime());
$zip = new ZipArchive();
$zip->open('kits/'.$numbercode.'.zip', ZipArchive::CREATE);

foreach($data as $d) {

$zip->addFile($d);  

}

$zip->close();

$link = 'kits/'.$numbercode.'.zip';

echo json_encode($link);
?>

answered Feb 20, 2017 at 3:44

Jeff

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1 Comment

Veshraj Joshi Over a year ago

I think, there is no need of json_encode in string, if you did change your ajax call, alert problem will not be solved

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