RxJS-如何将新道具传递给观察者

Question

使用RxJS如何在观察者上传递新属性？所以基本上我希望道具“customProp”可以被观察到

const { Observable } = require('rxjs');

const observable = Observable.create(function (observer) {
    console.log(observer.customProp); //How to get this working?
    observer.next(1);
    observer.next(2);
    observer.next(3);
    setTimeout(() => {
        observer.next(4);
        observer.complete();
    }, 1000);
});

console.log('just before subscribe');
observable.subscribe({
    next: x => console.log('got value ' + x),
    error: err => console.error('something wrong occurred: ' + err),
    complete: () => console.log('done'),
    customProp: 'Hello World RxJS',
});
console.log('just after subscribe');

---->>> Output
just before subscribe
undefined //How can I get this working, please?
got value 1
got value 2
got value 3
just after subscribe
=> undefined
got value 4
done

添加更多的信息-所以基本上，我试图创建一个冷可观察的，其中制片人需要一个道具，应该来自订户

//Cold observable
var coldObservable = new Observable((observer) => {
    var myProducerObj = new MyProducer();
    myProducerObj.executeQuery(observer.customProp);
   // How could we get customProp over here?
});

添加用法信息--coldObservable是DB连接函数，customProp是需要在DB上执行的查询

Answer 1

如果不扩展可观察的类，这是不可能的，这是不需要的。它可以是工厂功能：

const createObservableWithCustomProp = (customProp, observer) => {
  return new Observable((observer) => {
    var myProducerObj = new MyProducer(customProp);
    ...
  });
};

通常很少需要手动构造观测值，因为RXJSAPI提供了丰富的特性集来创建、转换和组合观测值。