如何将回调作为参数传递给另一个函数 [英] How to pass a callback as a parameter into another function

问题描述

我是ajax和回调函数的新手，如果我的概念都错了，请原谅我。

I'm new to ajax and callback functions, please forgive me if i get the concepts all wrong.

问题：我可以吗？将 回调函数 作为参数发送给另一个执行回调的函数？

Problem: Could i send a callbackfunction as a parameter to another function that will execute the callback?

function firstFunction(){
    //some code

    //a callback function is written for $.post() to execute
    secondFunction("var1","var2",callbackfunction);
}

function secondFunction(var1, var2, callbackfunction) {
    params={}
    if (event != null) params = event + '&' + $(form).serialize();

    // $.post() will execute the callback function
    $.post(form.action,params, callbackfunction);
}

推荐答案

是的。函数引用就像任何其他对象引用一样，你可以将它们传递给你的内容。

Yup. Function references are just like any other object reference, you can pass them around to your heart's content.

这是一个更具体的例子：

Here's a more concrete example:

function foo() {
    console.log("Hello from foo!");
}

function caller(f) {
    // Call the given function
    f();
}

function indirectCaller(f) {
    // Call `caller`, who will in turn call `f`
    caller(f);
}

// Do it
indirectCaller(foo); // alerts "Hello from foo!"

你也可以通过在 foo 的参数中：

You can also pass in arguments for foo:

function foo(a, b) {
    console.log(a + " + " + b + " = " + (a + b));
}

function caller(f, v1, v2) {
    // Call the given function
    f(v1, v2);
}

function indirectCaller(f, v1, v2) {
    // Call `caller`, who will in turn call `f`
    caller(f, v1, v2);
}

// Do it
indirectCaller(foo, 1, 2); // alerts "1 + 2 = 3"

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