使用.done的嵌套AJAX调用 [英] Nested AJAX Calls using .done

问题描述

我对AJAX还是很陌生，但是从事的项目需要ajax调用来验证特定值，然后如果第一个返回期望值，则再进行一次ajax调用.我正在尝试实现.done/.fail模型，但是找不到阻止两个调用同时发生的方法，而不是一旦第一个调用完成并成功就阻止了这两种方法.

I'm fairly new to AJAX, but working on a project that requires an ajax call to validate a specific value, then make another ajax call if the first returns the expected value. I am trying to implement the .done/.fail model, but can't find a way to prevent both calls from happening simultaneously, rather than once the first call is done and successful.

以下代码将两次调用ajaxCall函数，但同时调用而不是连续调用.我研究了很多代码，包括嵌套AJAX调用，jQuery摆脱了嵌套的Ajax函数和$.when.done回调不起作用，但似乎没有一个适合我的实际情况，或者也许我只是不理解代码.无论哪种方式，我都无法找到解决方案，任何帮助将不胜感激！

The following code will call the ajaxCall function twice, but concurrently rather than consecutively. I have researched a ton of code, including Nested AJAX Calls, jQuery getting rid of nested ajax functions, and $.when.done callback isn't working, but none seem to fit my exact scenario, or maybe I just don't understand the code. Either way, I haven't been able to find a solution, and any help will be much appreciated!

var xReturn = ajaxCall("1");    

xReturn.done(function(msg){
    console.log("callback "+msg+" successful");

    // if successful, place second call
    if(parseInt(msg)==1)
        xReturn = ajaxCall("2");    
});


function ajaxCall(mop){
    return $.ajax({
         url: "getItem.php",
         type: "POST",
         data: {code: '<?php echo $code; ?> ', op:mop}
    });
}

看来 promise 可能是可行的方法，但是在这种情况下，我无法确定如何使用它们.在此先感谢您指出正确方向的任何指针.

It seems like promises may be the way to go, but I can't wrap my head around how to use them in this scenario. Thanks in advance for any pointers in the right direction.

更新:

我进行了一系列测试，结果各不相同.为了进行昨晚的最终测试，我每次在ajaxCall("2");之后直接放置一个console.log(msg);，每次生成的味精始终为"1"时，使我相信呼叫无法正常进行.这个结果告诉我xReturn.done(function(msg)...仅被调用一次，但是我认为每次ajax调用都会被调用.

I ran through a battery of tests with different results. For my final test last night, I placed another console.log(msg); directly after ajaxCall("2"); Each time the resulting msg was always "1", leading me to believe the calls were not happening properly. This result tells me that xReturn.done(function(msg)... is only being called once, but I thought it would be called with each ajax call.

有了这些新信息，我将在今晚进行其他测试并报告.
谢谢

With this new information, I will perform additional testing tonight and report back.
Thanks

推荐答案

您需要将.done()方法绑定到每个承诺. xReturn.done()将功能绑定到该诺言.

You need to bind a .done() method to each promise. xReturn.done() binds a function to that promise.

当您执行xReturn = ajaxCall("2");时，您将 替换 使用不同的对象.此对象没有绑定的.done()方法.

When you do xReturn = ajaxCall("2");, you are replacing xReturn with a different object. This object does not have a .done() method bound to it.

您需要将.done()绑定到每个诺言，但这不会自动发生.

You need to bind .done() to each promise, that doesn't happen automatically.

var xReturn = ajaxCall("1");
// This binds the callback to this *specific* promise    
xReturn.done(ajaxDone);    

function ajaxCall(mop){
    return $.ajax({
         url: "getItem.php",
         type: "POST",
         data: {code: '<?php echo $code; ?> ', op:mop}
    });
}

function ajaxDone(msg){
    console.log("callback "+msg+" successful");

    // if successful, place second call
    if(parseInt(msg)==1){
        xReturn = ajaxCall("2");

        // Bind a callback to this *new* object
        xReturn.done(ajaxDone);
    }
}

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