ajax调用的alert-ing返回值

发布于 2024-12-14 14:27:03 字数 679 浏览 2 评论 0原文

（1. 你觉得拥有一个可以调用 ajax(url,data,async) 的全局 jQuery ajax “助手”有用吗？）

function ajax(ajax_url, ajax_data, ajax_async){
  ajax_async = typeof(ajax_async) != 'undefined' ? ajax_async : true;

  $.ajax({
    type: 'GET',
    url: ajax_url,
    data: ajax_data,
    async: ajax_async,
    dataType: 'text',
    success: function(response) {
        return response;
    },
    error: function(){

    }
  });
}

如果我调用一个放置 GET 和返回成功响应。

alert(ajax('localhost/hello','',false);

如果我执行alert(get_ajax())，我会收到“未定义”的警报。即使有返回值，

我也会看到它的事件循环问题，但我async:false< /code> 在 ajax 调用中。想法？

原文

(1. Do you find it useful to have a global jQuery ajax 'helper' that you can call ajax(url,data,async) ? )

function ajax(ajax_url, ajax_data, ajax_async){
  ajax_async = typeof(ajax_async) != 'undefined' ? ajax_async : true;

  $.ajax({
    type: 'GET',
    url: ajax_url,
    data: ajax_data,
    async: ajax_async,
    dataType: 'text',
    success: function(response) {
        return response;
    },
    error: function(){

    }
  });
}

If I'm calling a function that places a GET and returns the success response.

alert(ajax('localhost/hello','',false);

If I do alert(get_ajax()) I get an alert of 'undefined'.. even if there was a return value

I see its an event loop issue, but I async:false in the ajax call. thoughts?

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甜嗑 2024-12-21 14:27:04

对于同步调用，我相信你会这样做

var data;
$.ajax({
    ...
    async: false,
    success: function(response) {
        data = response;
    }
});
return data;

For an synchronous call, I believe you would do

var data;
$.ajax({
    ...
    async: false,
    success: function(response) {
        data = response;
    }
});
return data;

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要走就滚别墨迹 2024-12-21 14:27:04

您的 ajax 函数包装了 $.ajax 调用，但不返回任何内容：这就是 undefined 的来源。要在 alert 示例中获取一些值，您应该在函数底部添加 return 语句。

要获得带有响应的警报，应执行以下操作：

success: function(response) {
    alert(response);
},

Your ajax function wraps the $.ajax call, does doesn't return anything: that's where the undefined comes from. To get some value in your alert example, you should add a return statement at the bottom of your function.

To get an alert with the response, the following should do:

success: function(response) {
    alert(response);
},

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唐婉 2024-12-21 14:27:03

这是已经被问过一百万次的问题之一，但几乎不可能搜索到。不管怎样，问题是当你从这里返回时......

success: function(response) {
    return response;
}

你不是从ajax()函数返回，而是从匿名success函数返回。要从主函数返回值，您必须将字符串保存到变量中，以便可以从正确的函数返回。

var val = '';

$.ajax({
    //       ....
    success: function(response) {
       val = response;
    },
    //       ....
});

return val;

请注意，这仅在您进行同步调用时才有效。对于异步调用，最好提供回调函数作为参数并将其用作成功函数。

This is one of those questions that has been asked a million times, but it's nearly impossible to search. Anyway, the problem is that when you return from here...

success: function(response) {
    return response;
}

...you're not returning from the ajax() function but the anonymous success function. To return the value from the main function you have to save the string to a variable so that you can return from the correct function.

var val = '';

$.ajax({
    //       ....
    success: function(response) {
       val = response;
    },
    //       ....
});

return val;

Note that this only works when you're making a synchronous call. For asynchronous calls it's better to give a callback function as an argument and use it as the success function.

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