Angular2 rxjs 按可观察字段对对象列表进行排序(可观察)

Question

我想按可观察字段对事物列表进行排序，但无法全神贯注于可观察对象以使其正常工作。有人知道如何实现这一目标吗？

初始情况是这样的:

Thing[] things;

interface Thing {
  name: Observable<string>
}

<ul>
  <li *ngFor="const thing for things">
    {{thing.name | async}}
  </li>
</ul>

因为我显然没有正确描述我的问题:我想要对事物列表进行排序的字段是一个 Observable，而不是一个纯字符串。我想通过 websockets 保持字段更新，以便正确检测更改我必须使用我可以订阅的 Observable 字段。

Answer 1

感谢您澄清问题，Phosphoros。 :)

这里是你如何做你所要求的:

// Function to compare two objects by comparing their `unwrappedName` property.
const compareFn = (a, b) => {
  if (a.unwrappedName < b.unwrappedName)
    return -1;
  if (a.unwrappedName > b.unwrappedName)
    return 1;
  return 0;
};

// Array of Thing objects wrapped in an observable.
// NB. The `thing.name` property is itself an observable.
const thingsObs = Observable.from([
  { id: 1, name: Observable.of('foo') },
  { id: 2, name: Observable.of('bar') },
  { id: 3, name: Observable.of('jazz') }
]);

// Now transform and subscribe to the observable.
thingsObs

  // Unwrap `thing.name` for each object and store it under `thing.unwrappedName`.
  .mergeMap(thing =>
    thing.name.map(unwrappedName => Object.assign(thing, {unwrappedName: unwrappedName}))
  )

  // Gather all things in a SINGLE array to sort them.
  .toArray()

  // Sort the array of things by `unwrappedName`.
  .map(things => things.sort(compareFn))

  .subscribe();

将发出的值记录到控制台将显示一组按其 unwrappedName 属性排序的 Thing 对象:

[
  { id: 2, name: ScalarObservable, unwrappedName: "bar" },
  { id: 1, name: ScalarObservable, unwrappedName: "foo" },
  { id: 3, name: ScalarObservable, unwrappedName: "jazz" }
]

如果您对此代码有任何疑问，请告诉我。