如何在三元运算符中检查未定义的变量? [英] How to check undefined variable in a ternary operator?

问题描述

三元运算存在问题:

let a = undefined ? "Defined!" : "Definitely Undefined",
    b = abc ? "Defined!" : "Definitely Undefined", // ReferenceError
    c = (abc !== undefined) ? "Defined!" : "Definitely Undefined", // ReferenceError
    d = (typeof abc !== "undefined") ? "Defined!" : "Definitely Undefined"

// results: a = d = "Definitely Undefined", 
// while b and c throw ReferenceError when abc is undefined

在访问abc属性并分配空白对象之前，检查abc是否为 undefined 的 最佳和简短方法 是什么?{} 是否为 undefined ?

What is the best and short way to check if abc is undefined before accessing its properties as well as assign blank object {} if undefined?

let a = [[best way to check abc]] ? {[abc.label1]: 2, [abc.label2]: 1} : {}

PS:我目前正在使用(typeof abc！=="undefined")代替 [[检查abc的最佳方法]]

PS: I am currently using (typeof abc !== "undefined") in place of [[best way to check abc]]

推荐答案

b和c时，当abc未定义时抛出ReferenceError

while b and c throw ReferenceError when abc is undefined

所以 abc 不仅是未定义的，它是未声明的.那里有很大的不同.

So abc isn't just undefined, it's undeclared. There's a big difference there.

如果您需要处理未声明的 abc ，唯一安全的方法(无需 try / catch )是使用 typeof :

If you need to handle abc being undeclared, the only safe way to do that (without try/catch) is with typeof:

typeof abc === "undefined"

如果 abc 是未声明的标识符，那将是正确的，没有错误.如果声明了 abc 并包含值 undefined .

That will be true, without error, if abc is an undeclared identifier. It will also be true if abc is declared and contains the value undefined.

在访问其属性之前检查 abc 是否未定义以及分配空白对象 {} (如果未定义)的最佳和简短方法是什么?

What is the best and short way to check if abc is undefined before accessing its properties as well as assign blank object {} if undefined?

可能使用 var 来确保已声明:

Probably using var to ensure it's declared:

var abc = abc || {};

重复的 var 声明不是错误(重复的 let 声明是).因此，使用上面的代码，如果未声明 abc ，则会使用初始值 undefined 对其进行声明，然后将其分配为 {} .如果已声明，则在错误的情况下将其值替换为 {} .但是，如果可能会或可能不会使用 let 或 const 进行声明，则上述内容也会引发错误.

Duplicate var declarations are not errors (duplicate let declarations are). So with the above, if abc is undeclared, it gets declared with the initial value undefined and we assign it {}. If it's declared, we replace its value with {} if it's falsy. But, if it may or may not be declared with let or const, then the above will throw an error as well.

因此，要处理可能会或可能不会使用 let 或 const 声明的情况，我们完全需要一个不同的变量:

So to handle the case where it may or may not be declared with let or const, we need a different variable entirely:

let ourabc = typeof abc === "undefined" || !abc ? {} : abc;

如果未声明 abc 或包含虚假值，则将 ourabc 设置为 {} .由于所有非 null 对象引用都是真实的，并且您已经说过要访问对象属性，所以这可能是最短的方法.

That sets ourabc to {} if abc is undeclared or if it contains a falsy value. Since all non-null object references are truthy, and you've said you want to access object properties, that's probably the shortest way.

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