由于TypeScript是强类型的,简单地使用if(){}来检查null和undefined听起来并不正确。

TypeScript有专门的函数或语法吗?


当前回答

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") }

其他回答

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

可能已经晚了!但是你可以用??typescript中的运算符。 参见https://mariusschulz.com/blog/nullish-coalescing-the-operator-in-typescript

在TypeScript 3.7中,我们现在有可选的链接和Nullish Coalescing来同时检查null和undefined,例如:

let x = foo?.bar.baz();

这段代码将检查foo是否有定义,否则它将返回undefined

旧方法:

if(foo != null && foo != undefined) {
   x = foo.bar.baz();
} 

这样的:

let x = (foo === null || foo === undefined) ? undefined : foo.bar();

if (foo && foo.bar && foo.bar.baz) { // ... }

与可选的链接将:

let x = foo?.bar();

if (foo?.bar?.baz) { // ... }

另一个新特性是Nullish Coalescing,例如:

let x = foo ?? bar(); // return foo if it's not null or undefined otherwise calculate bar

老方法:

let x = (foo !== null && foo !== undefined) ?
foo :
bar();

奖金

通常我做杂耍检查,芬顿已经说过了。 为了让它更具可读性,你可以使用ramda中的isNil。

import * as isNil from 'ramda/src/isNil';

totalAmount = isNil(totalAmount ) ? 0 : totalAmount ;

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;
}