jQuery的：$就成功回调不会让我分配到外部私有变量 [英] jQuery: $.ajax success callback doesn't let me assign to outside private variable

问题描述

var doCheck = function() {
    var data;

    $.ajax({
        url: 'check.php',
        data: 'ch=check',
        success: function(resp) {
            data = resp;
        }
    });
console.log(data);
    return data == 1;
};

以上code，不管数据永远只能返回 0 或 1 。内的成功的回调的范围，这是事实。参数 RESP 的 0 或 1 取决于值在输入。

The above code, regardless of the data only ever returns a 0 or a 1. Within the scope of the success callback, this is true. The argument resp has a value of 0 or 1 depending on input.

不过，每当我尝试访问私有变量（应该不会受到影响范围），没有任何反应;当的console.log（数据）; 被称为所有写入控制台是未定义。

However, whenever I try to access a private variable (should not be affected by scope), nothing happens; and when console.log(data); is called all that is written to the console is undefined.

这功能没有父母，所以不用担心其他某种范围的干扰。

This function has no parent, so don't worry about some other kind of scope interference.

推荐答案

阿贾克斯是异步的。这就是为什么你要组织你的逻辑和回调：

Ajax is asynchronous. Which is why you have to organize your logic with callbacks:

var doCheck = function(callback) {
    var data;

    $.ajax({
        url: 'check.php',
        data: 'ch=check',
        success: function(resp) {
            callback(data == 1);
        }
    });
};

doCheck(function(result) {
    // result is true or false
});

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