Rx 合并 + 组合最新
本文关键字:最新 组合 合并 Rx | 更新日期: 2023-09-27 17:55:56
当两个可观察量都启动时,CombineLatest
开始。
A 1----------2---------------
B -----a----------b---c------
C -----1a----2a---2b--2c----- C = A.CombineLatest(B)
Merge
运算符在 A 或 B 启动时启动。但是,它不能组合 A 和 B 最新值。
A 1----------2---------------
B -----a----------b---c------
C 1----a-----2----b---c------ C = A.Merge(B)
我需要一个行为类似于Merge
的运算符,除了当两个可观察值都开始时,它允许我组合 A 和 B 的最新值:
A 1----------2---------------
B -----a----------b---c------
C 1----1a----2a---2b--2c----- C = A.MergeOrCombineLatest(B)
它的签名可能如下所示:
Observable<C> MergeOrCombineLatest<A, B, C>(
this IObservable<A> a,
IObservable<B> b,
Func<A, C> aResultSelector, // When A starts before B
Func<B, C> bResultSelector, // When B starts before A
Func<A, B, C> bothResultSelector) // When both A and B have started
如何实现此运算符?
这对
我有用:
public static IObservable<C> MergeOrCombineLatest<A, B, C>(
this IObservable<A> a,
IObservable<B> b,
Func<A, C> aResultSelector, // When A starts before B
Func<B, C> bResultSelector, // When B starts before A
Func<A, B, C> bothResultSelector) // When both A and B have started
{
return
a.Publish(aa =>
b.Publish(bb =>
aa.CombineLatest(bb, bothResultSelector).Publish(xs =>
aa
.Select(aResultSelector)
.Merge(bb.Select(bResultSelector))
.TakeUntil(xs)
.SkipLast(1)
.Merge(xs))));
}
然后这个:
var a = new Subject<int>();
var b = new Subject<string>();
var C = a.MergeOrCombineLatest(b, x => $"{x}!!", y => $"{y}!!", (x, y) => $"{x}{y}");
C.Subscribe(x => Console.WriteLine(x));
b.OnNext("x");
b.OnNext("y");
b.OnNext("z");
a.OnNext(1);
a.OnNext(5);
a.OnNext(6);
b.OnNext("a");
a.OnNext(2);
b.OnNext("b");
b.OnNext("c");
。给出这个:
x!!y!!z!!1z5z6Z6一2一2b2c
首先,你为 A、B 选择你的可观察量永远不会发出的特殊值,让它为 null:
A specialA = null;
B specialB = null;
然后
Observable<C> MergeOrCombineLatest<A, B, C>(
this IObservable<A> a,
IObservable<B> b,
Func<A, C> aResultSelector, // When A starts before B
Func<B, C> bResultSelector, // When B starts before A
Func<A, B, C> bothResultSelector) // When both A and B have started
{
return a.StartWith(specialA).CombineLatest(b.StartWith(specialB),
(aval, bval) => {
if (aval == specialA) return bval == specialB ? default(C) : bResultSelector(bval);
if (bval == specialB) return aResultSelector(bval);
return bothResultSelector(aval, bval);
}
).skip(1); // skip the first emission where both are special values
}