您可以使用三元运算符和新的空合并运算符轻松做到这一点。

首先:使用三元来检查它是否为真。如果是，则返回false，因此If语句不会运行。

第二:因为现在知道值是假的，所以如果值为空，可以使用空合并运算符返回true。由于它将为任何其他值返回自身，如果它不为null，则将使if语句正确失败。

let x = true; console.log("starting tests") if (x?false:x ?? true){ console.log(x,"is nullish") } x = false if (x?false:x ?? true){ console.log(x,"is nullish") } x = 0; if (x?false:x ?? true){ console.log(x,"is nullish") } x=1; if (x?false:x ?? true){ console.log(x,"is nullish") } x=""; if (x?false:x ?? true){ console.log(x,"is nullish") } x="hello world"; if (x?false:x ?? true){ console.log(x,"is nullish") } x=null; if (x?false:x ?? true){ console.log(x,"is nullish") } x=undefined; if (x?false:x ?? true){ console.log(x,"is nullish") }

我有这个问题，一些答案工作只是很好的JS，但不是TS这里的原因。

//JS
let couldBeNullOrUndefined;
if(couldBeNullOrUndefined == null) {
  console.log('null OR undefined', couldBeNullOrUndefined);
} else {
  console.log('Has some value', couldBeNullOrUndefined);
}

这很好，因为JS没有类型

//TS
let couldBeNullOrUndefined?: string | null; // THIS NEEDS TO BE TYPED AS undefined || null || Type(string)

if(couldBeNullOrUndefined === null) { // TS should always use strict-check
  console.log('null OR undefined', couldBeNullOrUndefined);
} else {
  console.log('Has some value', couldBeNullOrUndefined);
}

在TS中，如果变量未定义为null，当您试图检查该null时，tslint |编译器将报错。

//tslint.json
...
"triple-equals":[true],
...

 let couldBeNullOrUndefined?: string; // to fix it add | null

 Types of property 'couldBeNullOrUndefined' are incompatible.
      Type 'string | null' is not assignable to type 'string | undefined'.
        Type 'null' is not assignable to type 'string | undefined'.

我在typescript操场上做了不同的测试:

http://www.typescriptlang.org/play/

let a;
let b = null;
let c = "";
var output = "";

if (a == null) output += "a is null or undefined\n";
if (b == null) output += "b is null or undefined\n";
if (c == null) output += "c is null or undefined\n";
if (a != null) output += "a is defined\n";
if (b != null) output += "b is defined\n";
if (c != null) output += "c is defined\n";
if (a) output += "a is defined (2nd method)\n";
if (b) output += "b is defined (2nd method)\n";
if (c) output += "c is defined (2nd method)\n";

console.log(output);

给:

a is null or undefined
b is null or undefined
c is defined

so:

检查(a == null)是否正确，以知道a是否为空或未定义 检查(a != null)是否正确，以知道是否定义了a 检查(a)是否错误，以知道a是否被定义

我们使用一个helper hasValue来检查null /undefined，并通过TypeScript确保不执行不必要的检查。(后者类似于TS如何抱怨if ("a" === undefined)，因为它总是假的)。

始终使用这个始终是安全的，不像!val匹配空字符串，零等。它还避免了模糊==匹配的使用，这几乎总是一个坏的做法-没有必要引入异常。

type NullPart<T> = T & (null | undefined);

// Ensures unnecessary checks aren't performed - only a valid call if 
// value could be nullable *and* could be non-nullable
type MustBeAmbiguouslyNullable<T> = NullPart<T> extends never
  ? never
  : NonNullable<T> extends never
  ? never
  : T;

export function hasValue<T>(
  value: MustBeAmbiguouslyNullable<T>,
): value is NonNullable<MustBeAmbiguouslyNullable<T>> {
  return (value as unknown) !== undefined && (value as unknown) !== null;
}

export function hasValueFn<T, A>(
  value: MustBeAmbiguouslyNullable<T>,
  thenFn: (value: NonNullable<T>) => A,
): A | undefined {
  // Undefined matches .? syntax result
  return hasValue(value) ? thenFn(value) : undefined;
}

你可以用:

if (!!variable) {}

它等于写作

it (variable != null && variable != undefined) {}

如果你想传递tslint而不设置严格布尔表达式为allow-null-union或allow-undefined-union，你需要从节点的util模块中使用isNullOrUndefined或滚动你自己的:

// tslint:disable:no-null-keyword
export const isNullOrUndefined =
  <T>(obj: T | null | undefined): obj is null | undefined => {
    return typeof obj === "undefined" || obj === null;
  };
// tslint:enable:no-null-keyword

不完全是语法糖，但当你的tslint规则很严格时很有用。