Question

Suppose I have this class:

export class DoSomethingClass<TEntity> {
    constructor(entity: TEntity) {
    }
}

I can call it like this from other code:

new DoSomethingClass<Person>(myPersonObject);

I would like to achieve the following shorthand syntax:

DoSomething<Person>(myPersonObject);

Which I can do with something like:

export function DoSomething<TEntity>(entity: TEntity): DoSomethingClass<TEntity>{
    return new DoSomethingClass<TEntity>(entity);
}

But of course this involves a lot of duplication. Ideally I would like:

export DoSomething = DoSomethingClass.prototype.constructor;

But this doesn't work, otherwise I wouldn't be asking :)

Is there a non-duplication way of exporting the constructor of a class as an individual, named, short-hand function?

Answer 1

首先，是的，你可以使用下面的东西来做到这一点：

export const DoSomething = DoSomethingClass.prototype.constructor;

这将避免您可能面临的编译错误。

但是，等等......

我个人的意见是避免这种构造。

问题是你的用例到底是做什么的。如果您只想执行在构造函数中编写的代码，那么它可能会满足您的需求。在这种情况下，一个明智的选择是改为导出function。

但是，当使用构造函数时，通常的情况是实例化给定类的对象。这样做是使用new operator。

请参阅以下示例，其中var1未定义，其中new未使用，var2不是这种情况。请注意，虽然示例是在JavaScript中，但在编译之后，TypeScript也将转换为非常相似的东西。

＆＃13;

var myFunction = function(prop1) {
    this.prop1 = prop1;
    console.log("myFunction called. prop1:" + this.prop1);

    this.method1 = function() {
        console.log("Method1")
    }
}

var var1 = myFunction("called as normal func");
var var2 = new myFunction("called with new");

console.log(var1); //undefined
console.log(var2);

// var1.method1(); //error
var2.method1();

console.log(var2.__proto__ === myFunction.prototype); //true

＆＃13;

似乎这不是你要找的答案，但我希望这会有所帮助。

其他好的读物：

Exporting a class constructor as a named function using TypeScript

1 个答案:

但是，等等......