设置超时不使用功能 [英] set time out not working with function

问题描述

我使用以下内容暂停javascript几秒钟：

I am using the following to pause the javascript for a few seconds:

 setTimeout(start_countdown(),3000);

它不起作用，无论秒数如何都会调用该函数。但是，以下函数可以正常工作，它不使用函数。

It does not work, the function is called regardless of the seconds. The following function does however work, which doesnt use a function.

setTimeout(alert('hi'),3000);

我该如何解决这个问题？

How can I solve this?

推荐答案

您需要传递函数引用。你正在传递函数的返回值。

You need to pass a function reference. You are passing a function's return value.

区别在于：一个是你想要发生的函数的蓝图，另一个是你立即执行函数的蓝图。将其返回值传递给 setTimeout 。

The difference is this: one is a blueprint of the function you want to happen, the other means you are executing the function immediately and passing its return value to setTimeout.

setTimeout(start_countdown, 3000);

如果你想做一些比简单地调用命名函数更复杂的事情，或者你想传递一个对于命名函数的参数，您需要将匿名函数传递给超时并在其中调用您的函数：

If you want to do something more complex than simply call a named function, OR you want to pass a param to the named function, you'll need to instead pass an anonymous function to the timeout and call your function within that:

setTimeout(function() {
    start_countdown(/* possible params */);
    /* other code here as required */
}, 3000);

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