如何在TypeScript中将类型声明为可为空?

时间:2013-06-20 17:33:23

标签: javascript typescript

我在TypeScript中有一个界面。

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中完成吗?

9 个答案:

答案 0 :(得分:206)

JavaScript(和TypeScript)中的所有字段都可以包含值nullundefined

您可以将可选字段设为与可为空的

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: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';
}

答案 1 :(得分:80)

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:要使其按预期工作,您应在strictNullChecks中启用tsconfig

答案 2 :(得分:27)

只需在可选字段中添加问号?

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

答案 3 :(得分:17)

要更像C#,请定义Nullable类型,如下所示:

type Nullable<T> = T | null;

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

答案 4 :(得分:6)

您可以仅实现如下所示的用户定义类型:

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];

答案 5 :(得分:4)

我有一段时间也有同样的问题.. ts中的所有类型都可以为空,因为void是所有类型的子类型(例如,scala不同)。

查看此流程图是否有用 - https://github.com/bcherny/language-types-comparison#typescript

答案 6 :(得分:2)

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

答案 7 :(得分:0)

输入您的号码的值未定义

var user: Employee = { name: null, salary: undefined }; 

答案 8 :(得分:0)

可空类型可以调用运行时错误。 因此,我认为最好使用编译器选项--strictNullChecks并将number | null声明为类型。同样在嵌套函数的情况下,尽管输入类型为null,但是编译器不知道它会中断什么,因此我建议使用!(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