[英]AJAX return variable from PHP
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 $zip) from php as a response but instead it is alerting [object Object]. 当用户单击下载时,它将成功在服务器上创建包含文件的zip,然后它应从php警报zip位置(变量$ zip)作为响应,但它会向[object Object]发出警报。 Everything else is working how it should.
其他一切都按预期进行。 What am I doing wrong?
我究竟做错了什么?
JQuery: JQuery的:
$('.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:
<?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);
?>
The return type is a JavaScript object, which will result in what you see. 返回类型是一个JavaScript对象,它将产生您所看到的内容。
First, you should console.log(data)
, to get the structure. 首先,您应该
console.log(data)
,以获取结构。 You can also do this by looking at the Network Tab in Chrome. 您也可以通过查看Chrome中的“网络”标签来执行此操作。
After you know the structure of data
, you can use the value. 知道
data
的结构后,就可以使用该值。
For example, then alert(data.location)
, to alert the actual value. 例如,然后
alert(data.location)
来警告实际值。
remove your dataType
from the ajax, it alert [Object Object] because your result becomes json object if you specify dataType: 'json',
, 从ajax中删除
dataType
,它会警告[Object Object],因为如果您指定dataType: 'json',
,
and in php- 和在PHP-
// to echo the location of the zipfile
echo 'kits/'.$numbercode.'.zip';
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 $zip was defined as a zip archive not a string. 多亏@ jake2389,我才知道我在做什么错,我基本上只需要在PHP中创建一个新变量,我将其想要发送回AJAX的数据称为$ link,因为$ zip被定义为zip归档文件而不是字符串。 。 Here is what I changed and now it is working.
这是我更改的内容,现在可以正常工作了。
PHP: 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);
?>
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