douqiao8032 2017-02-20 03:38
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AJAX从PHP返回变量

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);
?>
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3条回答 默认 最新

  • dpwh11290 2017-02-20 03:41
    关注

    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.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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