我如何退出JavaScript脚本,就像PHP的退出或死亡?我知道这不是最好的编程实践,但我需要这样做。
当前回答
如果你只是想停止执行更多的代码而不“抛出”任何错误,你可以暂时覆盖window。如cross-exit所示:
function exit(code) {
const prevOnError = window.onerror
window.onerror = () => {
window.onerror = prevOnError
return true
}
throw new Error(`Script termination with code ${code || 0}.`)
}
console.log("This message is logged.");
exit();
console.log("This message isn't logged.");
其他回答
如果你不关心它是一个错误,可以这样写:
fail;
这将阻止您的主(全局)代码继续进行。 对调试/测试的某些方面很有用。
我知道这是旧的,但如果你想要一个类似的PHP die()函数,你可以这样做:
function die(reason) {
throw new Error(reason);
}
用法:
console.log("Hello");
die("Exiting script..."); // Kills script right here
console.log("World!");
上面的例子只打印“Hello”。
我使用这段代码来停止执行:
throw new FatalError("!! Stop JS !!");
虽然你会得到一个控制台错误,但这对我来说很好。
"exit"函数通常退出程序或脚本,并以错误消息作为参数。例如php中的die(…)
die("sorry my fault, didn't mean to but now I am in byte nirvana")
在JS中等效的是用throw关键字发出错误信号,如下所示:
throw new Error();
你可以很容易地测试这个:
var m = 100;
throw '';
var x = 100;
x
>>>undefined
m
>>>100
If you're looking for a way to forcibly terminate execution of all Javascript on a page, I'm not sure there is an officially sanctioned way to do that - it seems like the kind of thing that might be a security risk (although to be honest, I can't think of how it would be off the top of my head). Normally in Javascript when you want your code to stop running, you just return from whatever function is executing. (The return statement is optional if it's the last thing in the function and the function shouldn't return a value) If there's some reason returning isn't good enough for you, you should probably edit more detail into the question as to why you think you need it and perhaps someone can offer an alternate solution.
注意,在实践中,大多数浏览器的Javascript解释器在遇到错误时会简单地停止运行当前脚本。所以你可以做一些事情,比如访问一个未设置变量的属性:
function exit() {
p.blah();
}
它可能会中止脚本。但你不应该指望它,因为它根本不是标准的,而且它看起来真的是一个糟糕的做法。
编辑:好吧,也许这不是一个很好的答案Ólafur的光。尽管他链接到的die()函数基本上实现了我的第二段,即它只是抛出一个错误。