使用jquery ajax返回true时没有回调 [英] use jquery ajax don't have callback when return true

问题描述

我尝试使用查询ajax，当我返回false"时，我也会得到回调

I try use query ajax and when I have "return false" I get callback as well

$(document).ready(function(){

    $("form#Form").submit(function(){
        $.ajax({
            url: "ajax.php",
            type: "POST", 
            data: $(this).serialize(), 
            success: function(callback){ 
                alert(callback);            
            } //success
        }); //ajax

        return false;

    }); //submit
}); //ready

在php文件中，我只是回显"dennis"，它起作用并且我收到内容"dennis"的警报. 当我尝试这样的事情时:

in the php file I just echo "dennis", it works and I get alert with content "dennis". when I try something like this:

$(document).ready(function(){

    $("form#Form").submit(function(){
        $.ajax({
            url: "ajax.php",
            type: "POST", 
            data: $(this).serialize(), 
            success: function(callback){ 
                alert(callback);            
            } //success
        }); //ajax

    if (callback == "dennis"){
        return false;
    }else{
        return true;
    }   

    }); //submit
}); //ready

我得到空警报，并且表单的行为就像他得到返回真"一样. 我也尝试把变量放回去，但还是一样..

I get empty alert and the form act like he get "return true". I also try put variable and return him, but still same problem..

为什么会这样?如何解决?

why it happened? how fix it?

推荐答案

这是因为ajax-scope将异步执行，因此当您检查if (callback == "dennis")时，您的ajax调用 may 甚至都没有达到您的php，因此回调将为 undefined .

This is because the ajax-scope will execute asynchronous, hence when you check if (callback == "dennis"), you ajax call may not even reached your php yet, therefore callback will be undefined.

您需要在成功范围内检查callback(它实际上是一个错误的变量名，因为它不是回调，而是ajax调用的结果).

You need to check callback (it's actually a bad variable name as it is not a callback, but a result from your ajax-call) in the success scope.

$.ajax({
    url: "ajax.php",
    type: "POST", 
    data: $(this).serialize(), 
    success: function(callback){ 
       alert(callback);
       if (callback == 'dennis') {
           return true;
       } else return false;
    } //success
}); //ajax

现在，您的下一个问题是您的表单可能已经返回true，因此与在此处返回true | false相比，用结果调用新函数更好.

Now, your next problem is that your form probably already returns true, so better calling a new function with your result, than returning true|false here.

 // ajax metod success function
    success: function(result){ 
       alert(callback);
       onFormReturn(result);
    } //success

并定义处理结果的功能

// elsewhere, outside ajax-scope - even outside document.ready
function onFormReturn(result) {
   if (result == "dennis") {
      // do your stuff
   } else {
      // do something else
   }
}

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