从Ajax / jQuery的呼叫使用变量的成功的功能外 [英] Use variable outside the success function from an ajax/jquery call

问题描述

我有以下的code

var test;
     $.ajax({
        type: "GET",
        url: "../views/person/controller.php?actor=person&action=checkAge",
        data: "age=" + value,
        success: function(msg){
            console.log(msg);
            test = msg; 
        },
    });
    Validate.fail(test);

现在测试VAR应该给假真像控制台不会说。 但测试VAR给了我不确定为什么？

Now the test var should give true of false like the console does say. But test var gives me undefined why?

推荐答案

大概是因为Validate.fail（测试）的异步调用之后立即发生。请记住这是异步的，这意味着它执行平行于JavaScript的网页上运行。

Probably because Validate.fail(test) occurs immediately after the asynchronous call. Remember it is ASYNCHRONOUS, meaning it executes parallel to javascript running on your page.

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