如何在TypeScript中将类型声明为可为空？

Question

我在TypeScript中有一个接口.

interface Employee{
   id: number;
   name: string;
   salary: number;
}

我想把'薪水'作为一个可以为空的领域(就像我们在C#中所做的那样).这可以在TypeScript中做到吗？

Answer 1

JavaScript(和TypeScript)中的所有字段都可以具有值null或undefined.

您可以将字段设为可选字段,该字段与可空字段不同.

interface Employee1 {
    name: string;
    salary: number;
}

var a: Employee1 = { name: 'Bob', salary: 40000 }; // OK
var b: Employee1 = { name: 'Bob' }; // Not OK, you must have 'salary'
var c: Employee1 = { name: 'Bob', salary: undefined }; // OK
var d: Employee1 = { name: null, salary: undefined }; // OK

// OK
class SomeEmployeeA implements Employee1 {
    public name = 'Bob';
    public salary = 40000;
}

// Not OK: Must have 'salary'
class SomeEmployeeB implements Employee1 {
    public name: string;
}

与之比较:

interface Employee2 {
    name: string;
    salary?: number;
}

var a: Employee2 = { name: 'Bob', salary: 40000 }; // OK
var b: Employee2 = { name: 'Bob' }; // OK
var c: Employee2 = { name: 'Bob', salary: undefined }; // OK
var d: Employee2 = { name: null, salary: 'bob' }; // Not OK, salary must be a number

// OK, but doesn't make too much sense
class SomeEmployeeA implements Employee2 {
    public name = 'Bob';
}

Answer 2

在这种情况下,联盟类型是我心目中的最佳选择:

interface Employee{
   id: number;
   name: string;
   salary: number | null;
}

// Both cases are valid
let employe1: Employee = { id: 1, name: 'John', salary: 100 };
let employe2: Employee = { id: 1, name: 'John', salary: null };

编辑:为了使其按预期工作,您应该启用strictNullChecksin tsconfig.

Answer 3

要更像C#,请定义如下Nullable类型:

type Nullable<T> = T | null;

interface Employee{
   id: number;
   name: string;
   salary: Nullable<number>;
}

Answer 4

只需?在可选字段中添加问号即可.

interface Employee{
   id: number;
   name: string;
   salary?: number;
}

Answer 5

type Nullable<T> = {
  [P in keyof T]: T[P] | null;
};

然后你就可以使用它了

Nullable<Employee>

这样你仍然可以使用Employee接口，因为它在其他地方

Answer 6

type MyProps = {
  workoutType: string | null;
};

Answer 7

可空类型可能会引发运行时错误。所以我认为最好使用编译器选项--strictNullChecks并声明number | null为类型。同样在嵌套函数的情况下，虽然输入类型为空，但编译器无法知道它会破坏什么，所以我建议使用!（感叹号）。

function broken(name: string | null): string {
  function postfix(epithet: string) {
    return name.charAt(0) + '.  the ' + epithet; // error, 'name' is possibly null
  }
  name = name || "Bob";
  return postfix("great");
}

function fixed(name: string | null): string {
  function postfix(epithet: string) {
    return name!.charAt(0) + '.  the ' + epithet; // ok
  }
  name = name || "Bob";
  return postfix("great");
}

参考。 https://www.typescriptlang.org/docs/handbook/advanced-types.html#type-guards-and-type-assertions

Answer 8

type WithNullableFields<T, Fields> = {
  [K in keyof T]: K extends Fields 
    ? T[K] | null | undefined
    : T[K]
}

let employeeWithNullableSalary: WithNullableFields<Employee, "salary"> = {
  id: 1,
  name: "John",
  salary: null
}

或者您可以关闭 strictNullChecks ；）

以及相反的版本：

type WithNonNullableFields<T, Fields> = {
  [K in keyof T]: K extends Fields
    ? NonNullable<T[K]>
    : T[K]
}

Answer 9

您可以像下面这样实现用户定义的类型：

type Nullable<T> = T | undefined | null;

var foo: Nullable<number> = 10; // ok
var bar: Nullable<number> = true; // type 'true' is not assignable to type 'Nullable<number>'
var baz: Nullable<number> = null; // ok

var arr1: Nullable<Array<number>> = [1,2]; // ok
var obj: Nullable<Object> = {}; // ok

 // Type 'number[]' is not assignable to type 'string[]'. 
 // Type 'number' is not assignable to type 'string'
var arr2: Nullable<Array<string>> = [1,2];