Question

function make(callback) {
    //some other manipulation to get data to pass it into $.post()
    $.post(data, function(response) {
        // do something
        callback()
    });
}

function two() {
    make(function() {
        console.log('hello');
    });
}

尽管我使用了回调，

console.log('hello')仍将首先触发。如何让make（）运行直到一切都完成然后触发console.log('hello')？

Answer 1

您的代码应在您致电two()后立即生效。

这可能会更好地显示您的情况：

function requestSomething(callback) {
    $.post(data, function(response) {
        // do something
        callback();
    });
}

function callBackFunction() {
    console.log('done!');
}

// Pass callBackFunction, which gets called after request.
requestSomething(callBackFunction);

javascript回调函数

1 个答案: