Question

我有一个PHP脚本响应正在上传的文件。我想“回应”这个回应，但我无法让它起作用

正如您在下面的脚本中看到的，我使用了'响应'，因为我无法想到任何事情。这是如何工作的？

success:function(response){
    if(response == 6563){
        var uploaddiv = document.getElementById('uploadattachment');
        uploaddiv.innerHTML = '<center><br><br><span style="font-size:18px; color:green;">File has successfully been uploaded.</span><br>You can now click this window away or upload another file</center>';
    } else {
        var uploaddiv = document.getElementById('uploadattachment');
        uploaddiv.innerHTML = '<center><br><br><span style="font-size:18px; color:red;">Oops, something went wrong</span><br>' response '</center>';
    }

Answer 1

您忘记了正确连接字符串（缺少'+'）：

uploaddiv.innerHTML = '<center><br><br><span style="font-size:18px; color:red;">Oops, something went wrong</span><br>'+response+'</center>';

Answer 2

如果要在不将其添加到DOM的情况下检查响应，请使用console.log(response); 它将被记录到您浏览器的开发控制台。

Answer 3

可能是指如何获取响应代码状态，代码示例：

   success: function(data, textStatus, xhr) {
       console.log(xhr.status);
       alert(xhr.status);
   },

因此，您可以根据状态代码处理响应代码并处理数据。

Answer 4

这是错误：您在+之前和之后错过了response。

var uploaddiv = document.getElementById('uploadattachment');
uploaddiv.innerHTML = '<center><br><br><span style="font-size:18px; color:red;">Oops, something went wrong</span><br>' + response + '</center>';

打印Javascript响应

4 个答案: