我可以在 JavaScript 的 switch 语句中处理“未定义”的情况吗？
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Can I handle an "undefined" case in a switch statement in JavaScript?
提问by egucciar
If I'm passing an object to a case statement, and there is a case where it is undefined, can I handle that case? If so then how? If its not possible, then what is the best practice for handling an undefined case for a switch?
如果我将一个对象传递给 case 语句，并且存在未定义的情况，我可以处理这种情况吗？如果是，那么如何？如果不可能，那么处理开关未定义情况的最佳实践是什么？
回答by Jason McCreary
Add a casefor undefined.
添加一个casefor undefined。
case undefined:
// code
break;
Or, if all other options are exhausted, use the default.
或者，如果所有其他选项都用尽，请使用default.
default:
// code
break;
Note:To avoid errors, the variable supplied to switchhas to be declaredbut can have an undefinedvalue. Reference this fiddleand read more about defined and undefined variables in JavaScript.
注意：为避免错误，switch必须声明提供给的变量，但可以有一个undefined值。参考此小提琴并阅读有关JavaScript 中已定义和未定义变量的更多信息。
回答by ircmaxell
Well, the most portable way would be to define a new variable undefinedin your closure, that way you can completely avoid the case when someone does undefined = 1;somewhere in the code base (as a global var), which would completely bork most of the implementations here.
好吧，最可移植的方法是undefined在你的闭包中定义一个新变量，这样你就可以完全避免有人undefined = 1;在代码库中的某个地方（作为全局变量）做的情况，这会使这里的大多数实现完全无聊。
(function() {
var foo;
var undefined;
switch (foo) {
case 1:
//something
break;
case 2:
//something
break;
case undefined:
// Something else!
break;
default:
// Default condition
}
})();
By explicitly declaring the variable, you prevent integration issues where you depend upon the global state of the undefinedvariable...
通过显式声明变量，可以防止依赖于undefined变量全局状态的集成问题......
回答by I Hate Lazy
If you're comparing object references, but the variable may not be assigned a value, it'll work like any other case to simply use undefined.
如果您正在比较对象引用，但变量可能没有被赋值，那么它会像任何其他情况一样简单地使用undefined.
var obs = [
{},
{}
];
var ob = obs[~~(Math.random() * (obs.length + 1))];
switch(ob) {
case obs[0]:
alert(0);
break;
case obs[1]:
alert(1);
break;
case undefined:
alert("Undefined");
break;
default: alert("some unknown value");
}
回答by jAndy
Since undefinedreally is just another value('undefined' in window === true), you can check for that.
由于undefined真的只是另一个值( 'undefined' in window === true)，您可以检查它。
var foo;
switch( foo ) {
case 1:
console.log('1');
break;
case 2:
console.log('2');
break;
case 3:
console.log('3');
break;
case undefined:
console.log('undefined');
break;
}
works just about right.
工作得恰到好处。
