javascript - RxJs:在不保持外部状态的情况下，通过 distinctUntilChanged() 过滤获取哪些对象键按顺序更改

Question

令 orders$ 是如下所示的 RxJs 对象序列。我们使用 distinctUntilChanged() 监听更改，并根据更改的键采取不同的操作。我们能否在 tap() 内部知道哪个对象键实际发生了变化，而无需通过外部 previousItem 变量 ?

例子:

let orders$ = of([
  { main: 'pizza', drink: 'cola', id: 0},
  { main: 'pizza', drink: 'cola', id: 1},
    ...
  { main: 'pizza', drink: 'cola', id: 3332},
  { main: 'pizza', drink: 'fanta', id: 3333},    // <-- drink changes
]); 

let previousItem = null;

orders$.pipe(
  distinctUntilChanged((o0, o1) => o0.main === o1.main || o0.drink === o1.drink),
  tap(item => {

      /* Which value actually changed -'main' or 'drink' ?! */

      if (previousItem && previousItem.main !== item.main) {
        doSthForMainChanged();
      }

      doSthForDrinkChanged();
      previousItem = item;        // <-- ... so that not to tweak via external state

  })).subscribe();

Answer 1

我建议用 scan 来完成:

source$.pipe(
  scan((last: any, item: any) => {
    let changes: any = {}
    last.main !== item.main ? changes.main = true : false
    last.drink !== item.drink ? changes.drink = true : false
    return {item, changes}
  }, {}),
  tap(({item, changes}) => {
    if (changes.main) {
      doSthForMainChanged();
    }
    if (changes.drink) {
      doSthForDrinkChanged();
    }    
  })
)

如果需要，您还可以在末尾添加 distinctUntilChanged