Rx js 理解提升方法 [英] Rx js understanding the lift method

问题描述

我想创建一个新的运算符，我在 文档，其中一种方法是做这样的事情:

I want to create a new operator and I find in the documentation that one of the ways is to do something like this:

class MyObservable extends Observable {
  lift(operator) {
    const observable = new MyObservable()
    observable.source = this;
    observable.operator = operator;
    return observable;
  }

  // put it here .. or ..
  customOperator() {
    /* do things and return an Observable */
  }
}

// ... put it here...
MyObservable.prototype.mySimpleOperator = mySimpleOperator;

我不明白什么是 lift 方法以及这里发生了什么，有人可以帮忙吗?

I don't understand what is the lift method and what is going on here, can someone help, please?

推荐答案

lift 在 RxJS 5 内部一直使用.lift 的原理是你准备一个新的 Observable，在订阅时以操作员定义的方式转发事件.Paul Taylor 有一个关于它的好视频 (https://youtu.be/QhjALubBQPg?t=19m).Lift 是一个非常基础的构建块.

lift is used all the time internally in RxJS 5. The principle of lift is that you prepare a new Observable that upon subscribe will forward the events in the way the operator defines. There is a good video about it by Paul Taylor (https://youtu.be/QhjALubBQPg?t=19m). Lift is a very fundamental building block.

不是创建一个新类——扩展 Observable——你也可以只创建 Operator 本身.操作员的用户然后可以通过编写来调用它:

Instead of creating a new class - extending Observable - you could also just create the Operator itself. Users of the operator can then call it by writing:

Observable.of(1, 2, 3)
  .lift(new MyCustomOperator)
  .subscribe()

这意味着没有人必须知道 Observable API 中还有另一个操作符可用，而是看到它是在别处定义的东西.

This means no-one has to learn that yet another operator is available in the Observable API, but instead sees that it is something defined elsewhere.

理想情况下你可以写

someObservable::myCustomOperator();

但不幸的是，绑定运算符可能会很远/永远不会发生，所以 .lift(operator) 似乎是最明确/最干净的方式.

but unfortunately the bind-operator might be long away / never gonna happen, so the .lift(operator) seems like the most explicit / clean way.

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