jQuery:将匿名回调重写为命名函数 [英] jQuery: Rewriting Anonymous Callback to a Named Function

问题描述

如果我这样做:

$('h1').slideUp('slow', function() { $('div:first').fadeOut(); });

h1 会向上滑动，然后第一个 div 会淡出.

h1 will slide up, then the first div will fade out.

但是，如果我这样做:

function last() { $('div:first').fadeOut(); }

$('h1').slideUp('slow', last());

h1 会向上滑动，div 会同时淡出！

h1 will slide up and div will fade out at the same time!

如何使我的第二个示例与第一个示例一样工作，其中fadeOut() 在slideUp() 之后被调用?

How can I make my second example work the same as the first one, where fadeOut() is called AFTER slideUp()?

推荐答案

你不需要使用函数返回值(你通过调用函数得到的)，而是函数体:

You don't need to use the function return value (which you get by calling the function), but the function body:

$('h1').slideUp('slow', last);

<小时>

你所做的和这个一样:

---

What you did is the same as this:

var returned = last();             // call to last returns undefined
                                   // so returned has the value undefined
$('h1').slideUp('slow', returned); // simply sending undefined as a callback

所以您只是内联执行 last 函数，然后将返回值(它是 undefined，因为它不返回任何内容)作为参数传递给 slideUp 的回调函数.

So you were just executing the last function inline, and then passing the return value (which is undefined since it returns nothing) as a parameter to the slideUp's callback function.

希望这个例子能帮助你理解:

Hope this example will help you understand:

function outer() {
  function inner() {};
  return inner;
}

alert(outer);    // returns the outer function body
alert(outer());  // returns the outer function's return value, which is the inner function

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