Question

我想知道是否可以从我创建的php页面返回错误回调到我的jquery，然后会根据我的php页面中发生的操作显示警告。我尝试创建一个404错误的标题，但似乎没有用。

示例JQuery代码：

$(document).ready(function()
    {
        var messageid= '12233122abdbc';
        var url = 'https://mail.google.com/mail/u/0/#all/' + messageid;
        var encodedurl = encodeURIComponent(url);
        var emailSubject = 'Testing123';
        var fromName = 'email@emailtest.com';

        var dataValues = "subject=" + emailSubject + "&url=" + encodedurl + "&from=" + fromName + "&messageID=" + messageid;
        $('#myForm').submit(function(){
            $.ajax({
                type: 'GET',
                data: dataValues,
                url: 'http://somepage.php',
                success: function(){
                     alert('It Was Sent')
                }
                error: function() {
                    alert('ERROR - MessageID Duplicate')
                }
            });
            return false;
        });
    });

示例PHP代码又名somepage.php：

<?php
include_once('test1.php');
include_once('test2.php');


if(isset($_GET['subject']))
{
   $subject=$_GET['subject'];
}
else
{
   $subject="";
}

if(isset($_GET['url']))
{
   $url=$_GET['url'];
}
else
{
   $url="";
}
if(isset($_GET['from']))
{
    $from=$_GET['from'];
}
else
{
    $from="";
}

if(isset($_GET['messageID']))
{
    $messageID = $_GET['messageID'];
}
else
{
    $messageID="";
}



    $stoqbq = new test2($from, $messageID);
    $msgID = new stoqbq->getMessageID();
    if($msgID = $messageID)
    {
        header("HTTP/1.0 404 Not Found");
        exit;
    }
    else
    {
        $userID = $stoqbq->getUser();
        $stoqb = new test1($subject,$url,$messageID,$userID);
        $stoqb->sendtoquickbase();
    }

     ?>

- 编辑 -

如果在使用json时收到无效的标签消息，这就是我解决此问题的方法：

服务器端PHP代码部分 -

if($msgID == $messageID)
{
    $response["success"] = "Error: Message Already In Quickbase";
    echo $_GET['callback'].'('.json_encode($response).')';
}
else
{
    $userID = $stoqbq->getUser();
    $stoqb = new SendToQuickbase($subject,$url,$messageID,$userID);
    $stoqb->sendtoquickbase();
    $response["success"] = "Success: Sent To Quickbase";
    echo $_GET['callback'].'('.json_encode($response).')';
}

客户端JQuery部分 -

$('#myForm').submit(function(){
            $.ajax({
                type: 'GET',
                data: dataValues,
                cache: false,
                contentType: "application/json",
                dataType: "json",
                url: "http://somepage.php?&callback=?",
                success: function(response){
                    alert(response.success);
                }
            });
            return false;
        });

Answer 1

您可以使用成功布尔值从PHP返回JSON响应。

if($msgID = $messageID)
{
    echo json_encode(array('success' => false));
}
else
{
    $userID = $stoqbq->getUser();
    $stoqb = new test1($subject,$url,$messageID,$userID);
    $stoqb->sendtoquickbase();
    echo json_encode(array('success' => true));
}

并在您的Javascript中：

 $.ajax({
            type: 'GET',
            dataType: 'json',
            data: dataValues,
            url: 'http://somepage.php',
            success: function(response){
                 if(response.success) {
                   alert('Success');
                 }
                 else {
                   alert('Failure');
                 }
            }
        });

Answer 2

有一个接受的q / a具有相同的内容：How to get the jQuery $.ajax error response text?

基本上你需要从错误回调函数中获取响应消息：

$('#myForm').submit(function(){
    $.ajax({
        type: 'GET',
        data: dataValues,
        url: 'http://somepage.php',
        success: function(){
             alert('It Was Sent')
        }
        error: function(xhr, status, error) {
      alert(xhr.responseText);
    }
    });
    return false;
});

如何在php中返回错误回调？

2 个答案: