如何只另一个函数完成后执行一个JavaScript / jQuery的功能？

问题描述

我知道这是一个简单的问题，但我真的不知道该怎么做。我执行两个功能中的我的code，'luxboxEngine'和'fitToScreen'，后者需要前者的完成。我如何告诉fitToScreen'到'luxboxEngine后，只执行已完成？现在我有这个，它得到了预期的效果：

I know this is a simple question, but I don't really know how to do it. I'm executing two functions in my code, 'luxboxEngine' and 'fitToScreen', the latter of which requires the completion of the former. How do I tell 'fitToScreen' to only execute after 'luxboxEngine' has completed? Right now I have this, which gets the desired results:

luxboxEngine(self);
setTimeout(fitToScreen, 1000);

...但我知道，这不是一个很好的解决方案。我读过有关回调函数，但我怕我真的不明白他们是什么/如何将它们。谢谢阅读。

...But I know that's not a good solution. I've read about callback functions, but I'm afraid I don't really get what they are / how to incorporate them. Thanks for reading.

编辑：那么作为Sudharsan斯里兰卡和Brian McGinity下面回答，回调是解决方案。这里的code我用得到什么，我需要在我的程序：

So as Sudharsan Sri and Brian McGinity answered below, a callback is the solution. Here's the code I used to get what I need in my program:

    luxboxEngine(self, function () {
       fitToScreen();
    });

这luxboxEngine完成后触发fitToScreen功能。

That fires the fitToScreen function after the completion of luxboxEngine.

推荐答案

使用的回调：

luxboxEngine(self , function() { fitToScreen(); });

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