如何在TypeScript中声明一个可为空的类型？

Question

我在TypeScript中有一个接口。

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

我想提出“工资”作为一个可空场（就像我们可以用C＃做​​的）。这可能在TypeScript中做到吗？

Answer 1

在JavaScript（和在打字稿）所有字段可以具有值null或undefined。

您可以使字段可选这是不同于可空。

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; 
} 

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

只需在可选字段中添加一个问号?即可。

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

Answer 3

我有同样的问题而回。在TS所有的类型都是空的，因为空虚是所有类型的亚型（不像，例如，斯卡拉） 。

看看该流程图帮助 - https://github.com/bcherny/language-types-comparison#typescript

Answer 4

联盟类型是在这种情况下，我心中最好的选择：

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 };