我只是想在typescript接口中声明一个静态属性?我没有找到任何关于这方面的资料。
interface myInterface {
static Name:string;
}
这可能吗?
我只是想在typescript接口中声明一个静态属性?我没有找到任何关于这方面的资料。
interface myInterface {
static Name:string;
}
这可能吗?
当前回答
虽然静态关键字不支持接口在Typescript 但我们可以通过创建一个具有静态成员的函数接口来实现。
在下面的代码中,我创建了一个函数接口Factory,它有两个静态成员serialNumber和printSerial。
// factory is a function interface
interface Factory<T> {
(name: string, age: number): T;
//staic property
serialNumber: number;
//static method
printSrial: () => void;
}
class Dog {
constructor(public name: string, public age: number) { }
}
const dogFactory: Factory<Dog> = (name, age) => {
return new Dog(name, age);
}
// initialising static members
dogFactory.serialNumber = 1234;
dogFactory.printSrial = () => console.log(dogFactory.serialNumber);
//instance of Dog that DogFactory creates
const myDog = dogFactory("spike", 3);
//static property that returns 1234
console.log(dogFactory.serialNumber)
//static method that prints the serial 1234
dogFactory.printSrial();
其他回答
我为我的特定用例找到了这样做的方法(不需要装饰器)。
检查静态成员的重要部分是IObjectClass,并在createObject方法中使用cls: IObjectClass<T>:
//------------------------
// Library
//------------------------
interface IObject {
id: number;
}
interface IObjectClass<T> {
new(): T;
table_name: string;
}
function createObject<T extends IObject>(cls: IObjectClass<T>, data:Partial<T>):T {
let obj:T = (<any>Object).assign({},
data,
{
id: 1,
table_name: cls.table_name,
}
)
return obj;
}
//------------------------
// Implementation
//------------------------
export class User implements IObject {
static table_name: string = 'user';
id: number;
name: string;
}
//------------------------
// Application
//------------------------
let user = createObject(User, {name: 'Jimmy'});
console.log(user.name);
这里有一个相当简单的方法:
interface MyClass {
new (): MyClassInstance;
staticMethod(): string;
}
interface MyClassInstance {
instanceMethod(): string;
}
const Class: MyClass = class {
static staticMethod() {
return "This is a static method";
}
instanceMethod() {
return "This is an instance method";
}
}
Class.staticMethod();
// Has type MyClassInstance
const instance = new Class();
instance.instanceMethod();
请注意,这并不允许您像通常那样让类扩展接口,但对于许多情况来说,这已经足够好了。
解决方案
返回I的实例类型,并确保C扩展I:
type StaticImplements<I extends new (...args: any[]) => any, C extends I> = InstanceType<I>;
实例方法接口:
interface MyInstance {
instanceMethod();
}
接口采用静态方法:
interface MyClassStatic {
new (...args: any[]): MyInstance;
staticMethod();
}
类需要静态方法并使用自己的方法进行扩展:
class MyClass implements StaticImplements<MyClassStatic, typeof MyClass> {
static staticMethod();
static ownStaticMethod();
instanceMethod();
ownInstanceMethod();
}
推理
在接口中定义静态方法将在#33892中讨论,抽象静态方法将在#34516中讨论。
基于Val和Aleksey的回答(谢谢),这个解决方案:
不需要额外的运行时值 保留类自身的成员信息 允许构造函数约束
Test
原样-游乐场连结:
MyClass.staticMethod(); // OK
MyClass.ownStaticMethod(); // OK
new MyClass().instanceMethod(); // OK
new MyClass().ownInstanceMethod(); // OK
如果从MyClass - Playground中删除staticMethod:
class MyClass implements StaticImplements<MyClassStatic, typeof MyClass> {} // Type 'typeof MyClass' does not satisfy the constraint 'MyClassStatic'. Property 'staticMethod' is missing in type 'typeof MyClass' but required in type 'MyClassStatic'.
如果从MyClass - Playground中删除instanceMethod:
class MyClass implements StaticImplements<MyClassStatic, typeof MyClass> {} // Class 'MyClass' incorrectly implements interface 'MyInstance'. Property 'instanceMethod' is missing in type 'MyClass' but required in type 'MyInstance'.
你可以正常定义接口:
interface MyInterface {
Name:string;
}
但你不能这么做
class MyClass implements MyInterface {
static Name:string; // typescript won't care about this field
Name:string; // and demand this one instead
}
为了表示一个类应该遵循这个接口来获取它的静态属性,你需要一点技巧:
var MyClass: MyInterface;
MyClass = class {
static Name:string; // if the class doesn't have that field it won't compile
}
你甚至可以保留类名,TypeScript(2.0)不会介意:
var MyClass: MyInterface;
MyClass = class MyClass {
static Name:string; // if the class doesn't have that field it won't compile
}
如果你想静态地继承许多接口,你必须首先将它们合并到一个新的接口中:
interface NameInterface {
Name:string;
}
interface AddressInterface {
Address:string;
}
interface NameAndAddressInterface extends NameInterface, AddressInterface { }
var MyClass: NameAndAddressInterface;
MyClass = class MyClass {
static Name:string; // if the class doesn't have that static field code won't compile
static Address:string; // if the class doesn't have that static field code won't compile
}
或者如果你不想命名合并接口,你可以这样做:
interface NameInterface {
Name:string;
}
interface AddressInterface {
Address:string;
}
var MyClass: NameInterface & AddressInterface;
MyClass = class MyClass {
static Name:string; // if the class doesn't have that static field code won't compile
static Address:string; // if the class doesn't have that static field code won't compile
}
工作示例
这里没有提到的另一个选项是用表示静态接口的类型定义变量,并为其分配类表达式:
interface MyType {
instanceMethod(): void;
}
interface MyTypeStatic {
new(): MyType;
staticMethod(): void;
}
// ok
const MyTypeClass: MyTypeStatic = class MyTypeClass {
public static staticMethod() { }
instanceMethod() { }
}
// error: 'instanceMethod' is missing
const MyTypeClass1: MyTypeStatic = class MyTypeClass {
public static staticMethod() { }
}
// error: 'staticMethod' is missing
const MyTypeClass2: MyTypeStatic = class MyTypeClass {
instanceMethod() { }
}
效果与使用装饰器的应答相同,但没有装饰器的开销
操场上
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