RxJS - 从 observable 中取出最后 n 个元素 [英] RxJS - take n last elements from an observable

问题描述

我想从 observable 中取出 3 个最后的元素.假设我的时间线如下所示:

I want to take 3 last elements from an observable. Let's say that my timeline looks like this:

--a---b-c---d---e---f-g-h-i------j->

其中:a、b、c、d、e、f、g、h、i、j 是发射值

每当发出新值时，我想立即获取它，因此它看起来像这样:

Whenever a new value is emitted I want to get it immediately, so it can look like this:

[a]
[a, b]
[a, b, c]
[b, c, d]
[c, d, e]
[d, e, f]
[e, f, g]
[f, g, h]
... and so on

我认为这非常有用.想象一下，建立一个聊天室，您希望显示 10 条最后的消息.每当收到新消息时，您都希望更新您的视图.

I think that this is super useful. Imagine building a chat where you want to display 10 last messages. Whenever a new message comes you want to update your view.

我的尝试:演示

推荐答案

您可以使用 scan 来解决这个问题:

You can use scan for this:

from(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u'])
  .pipe(
    scan((acc, val) => {
      acc.push(val);
      return acc.slice(-3);
    }, []),
  )
  .subscribe(console.log);

这将打印:

[ 'a' ]
[ 'a', 'b' ]
[ 'a', 'b', 'c' ]
[ 'b', 'c', 'd' ]
[ 'c', 'd', 'e' ]
...
[ 's', 't', 'u' ]

bufferCount 不会做你想做的事.仅当每个缓冲区恰好是 === 3 时它才会发出，这意味着您在发布至少 3 条消息之前不会得到任何发射.

The bufferCount won't do what you want. It'll emit only when each buffer is exactly === 3 which means you won't get any emission until you post at least 3 messages.

2019 年 1 月:针对 RxJS 6 更新

Jan 2019: Updated for RxJS 6

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