如何结合运营商RXJS？

Question

我有以下四个听众：

    this.reportPropertiesChanges$.pipe(filter(() => this.loaded)).subscribe((v) => {
        this.registryReportSettings.reportProperties = v;
        this.registryReportSettings$.next(this.registryReportSettings);
    });

    this.objectsProperties$.pipe(filter(() => this.loaded)).subscribe((v) => {
        this.registryReportSettings.reportObjectsProperties = v;
        this.registryReportSettings$.next(this.registryReportSettings);
    });

    this.textPropertiesChanges$.pipe(filter(() => this.loaded)).subscribe((v) => {
        this.registryReportSettings.textProperties = v;
        this.registryReportSettings$.next(this.registryReportSettings);
    });

    this.toggleLabels$.pipe(filter(() => this.loaded)).subscribe((v) => {
        this.registryReportSettings.visibleText = v;
        this.registryReportSettings$.next(this.registryReportSettings);
    });

如何将它们合并为一个运算符？我无法使用 combineLatest，因为并非所有流都可以启动。

Answer 1

您可以使用 merge，如果任何 Observable 将发出一个值，它将触发

this.reportPropertiesChanges$.pipe(map(reportProperties => ({reportProperties})));
this.objectsProperties$.pipe(map(reportObjectsProperties => ({reportObjectsProperties})));
this.textPropertiesChanges$.pipe(map(textProperties => ({textProperties})))
this.toggleLabels$.pipe(map(visibleText => ({visibleText})));

merge(
    this.reportPropertiesChanges$,
    this.objectsProperties$,
    this.textPropertiesChanges$,
    this.toggleLabels$
)
.pipe(
    filter(() => this.loaded)
).subscribe((p) => {
    this.registryReportSettings = {...this.registryReportSettings, ...p };
    this.registryReportSettings$.next(this.registryReportSettings);
});

Answer 2

我会将其重构为：

import { merge } from 'rxjs';
import { map } from 'rxjs/operators';
...

const forwardProp = prop => value => ({prop, value});

const obs = [
    this.reportPropertiesChanges$.pipe(map(forwardProp('reportProperties'))),
    this.objectsProperties$.pipe(map(forwardProp('reportObjectsProperties'))),
    this.textPropertiesChanges$.pipe(map(forwardProp('textProperties'))),
    this.toggleLabels$.pipe(map(forwardProp('visibleText')))
];

merge(...obs).pipe(filter(() => this.loaded)).subscribe(({prop, value}) => {
  this.registryReportSettings[prop] = value;
  this.registryReportSettings$.next(this.registryReportSettings);
});

Answer 3

您也可以使用 rxjs 中的 combineLatest 运算符。它将使您能够访问一个操作员中的所有 4 个可订阅对象。像这样：

const reportProps = this.reportPropertiesChanges$;
const objectsProperties = this. objectsProperties$;
const textPropertiesChanges = this.textPropertiesChanges$;
const toggleLabels = this. toggleLabels$;

combineLatest([
   reportProps,
   objectsProperties,
   textPropertiesChanges,
   toggleLabels
]).pipe(filter(() => this.loaded)).subscribe(([
     reportPropsValue, 
     objPropsValue,
     textChangesValue,
     toggleLabelsValue
   ]) => {
      this.registryReportSettings.reportProperties = reportPropsValue;
      this.registryReportSettings.reportObjectsProperties = objPropsValue;
      this.registryReportSettings.textProperties = textChangesValue;
      this.registryReportSettings.visibleText = toggleLabelsValue;

      // Once you set all the values, you can update the Behavior Subject
      this.registryReportSettings$.next(this.registryReportSettings);
   })

如何结合运营商RXJS？

How to combine operators RXJS?

rxjs

angular