我有一个接口在TypeScript。
interface Employee{
id: number;
name: string;
salary: number;
}
我想把工资作为一个可空字段(就像我们可以在c#中做的那样)。这可能在TypeScript中实现吗?
我有一个接口在TypeScript。
interface Employee{
id: number;
name: string;
salary: number;
}
我想把工资作为一个可空字段(就像我们可以在c#中做的那样)。这可能在TypeScript中实现吗?
当前回答
JavaScript(和TypeScript)中的所有字段的值都可以是null或undefined。
您可以将字段设置为可选的,而不是可空的。
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;
}
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';
}
其他回答
就加个问号?到可选字段。
interface Employee{
id: number;
name: string;
salary?: number;
}
你可以像下面这样实现一个用户定义的类型:
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];
type Nullable<T> = {
[P in keyof T]: T[P] | null;
};
然后你就可以用它了
Nullable<Employee>
这样,您仍然可以像在其他地方一样使用Employee界面
我通过编辑tsconfig解决了这个问题。json文件。
下面:"strict": true, 加上这两行:
"noImplicitAny": false,
"strictNullChecks": false,
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
}
或者你可以关闭strictNullChecks;)
反过来说:
type WithNonNullableFields<T, Fields> = {
[K in keyof T]: K extends Fields
? NonNullable<T[K]>
: T[K]
}