0

当我发送请求时，警报为空白。文件上传正常，只是没有响应文本。我在这里做错了吗？ 

var request = new XMLHttpRequest();

request.open('POST','upload.php');
request.setRequestHeader('Cache-Control','no-cache');
request.send(data);
alert(request.responseText);

并上传.php

if(!empty($_FILES['file'])){
foreach  ($_FILES['file']['name'] as $key => $name) {
if($_FILES['file']['error'][$key] == 0 && move_uploaded_file($_FILES['file']['tmp_name'][$key],"video/$name"))
    {
       $x = "1";
    }
        else
    {
      $x = "2";
    }
  }
}
   if ($x == "1"){echo "success";}
   if ($x == "2"){echo "failed";}
4

1 回答 1

0

您已经处理了$x其值为 1 和 2 的输出。如果$x根本没有设置怎么办？

于 2013-02-03T17:48:57.123 回答