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
简单的例子
interface Person {
name: string;
age: number;
}
abstract class Trackable {
static TrackInstances: number;
}
class Pablo extends Trackable implements Person {
constructor(public name: string, public age: number) { Pablo.TrackInstances+=1; }
}
console.log(Pablo.TrackInstances);
其他回答
遵循@Duncan的@Bartvds的回答,在这里提供了一个可行的方法。
在Typescript 1.5发布后(@Jun 15 '15),你的有用界面
interface MyType {
instanceMethod();
}
interface MyTypeStatic {
new():MyType;
staticMethod();
}
可以在decorator的帮助下以这种方式实现。
/* class decorator */
function staticImplements<T>() {
return <U extends T>(constructor: U) => {constructor};
}
@staticImplements<MyTypeStatic>() /* this statement implements both normal interface & static interface */
class MyTypeClass { /* implements MyType { */ /* so this become optional not required */
public static staticMethod() {}
instanceMethod() {}
}
参考我在github issue 13462的评论。
视觉效果: 编译错误,提示缺少静态方法。
静态方法实现后,提示方法缺失。
在静态接口和正常接口完成后进行编译。
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
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();
可以使用相同的名称将接口和命名空间合并:
interface myInterface { }
namespace myInterface {
Name:string;
}
但是这个接口只有知道它的属性Name才有用。你不能实现它。
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。