RxJS6:完成后重置流

Question

我一直在玩RxJS 6。我正在尝试实现dragstart, dragmove and dragend。这是我到目前为止掌握的密码。

import { fromEvent } from 'rxjs';
import { concatMap } from 'rxjs/internal/operators/concatMap';
import { map } from 'rxjs/internal/operators/map';
import { takeUntil } from 'rxjs/internal/operators/takeUntil';
import { repeat } from 'rxjs/internal/operators/repeat';
import { first } from 'rxjs/internal/operators/first';

const mousedown = fromEvent<MouseEvent>(window, 'mousedown');
const mousemove = fromEvent<MouseEvent>(window, 'mousemove');
const mouseup = fromEvent<MouseEvent>(window, 'mouseup');

const dragstart = mousedown.pipe(
  first()
);

const dragmove = mousedown.pipe(
  concatMap((dragStartEvent) => mousemove.pipe(
    takeUntil(mouseup))
  )
);

const dragend = mousedown.pipe(
  (dragEvent) => mouseup.pipe(first())
);

const log = (prefix: string) => (data: MouseEvent) => console.log(`${prefix}: x: ${data.clientX}, y:${data.clientY}`);

dragstart.subscribe(log('dragstart'));
dragmove.subscribe(log('dragmove'));
dragend.subscribe(log('dragend'));

问题是运算符takeUntil和first将流标记为已完成。这意味着dragstart和dragend只会触发一次。在此事件发生后，是否有办法以某种方式重置流？例如，当dragstart流接收到事件时，通过重置mouseup流。

附加解释

在当前实现中，dragstart和dragend将在加载页面(&拖动)后准确地记录到控制台一次。在此之后，流就完成了，并且不会通过它发送更多的事件。我想要重置流，每次拖动操作，以便dragstart & dragend将工作的每一个拖动操作，而不是仅仅是在加载页面后的第一次。

当前行为([]表示发生的鼠标操作，斜体文本是控制台输出)：

首次拖动操作

• 滑鼠
• 拖起
• 摩丝
• 拖曳
• 鼠嘴
• 拖尾

二次拖动操作

• 滑鼠
• (无论拖放启动记录到控制台，什么都不会发生)
• 摩丝
• 拖曳
• 鼠嘴
• (任何事情都不会发生，但是应该将拖尾记录到控制台)

Answer 1

请参阅这里的工作示例https://codepen.io/anon/pen/bMWjEV在这里输入链接描述

​

const { fromEvent } = Rx.Observable;
const target = document.querySelector('.box');

const mouseup = fromEvent(target, 'mouseup');
const mousemove = fromEvent(document, 'mousemove');
const mousedown = fromEvent(target, 'mousedown');
let log = (prefix: string, x:number, y:number) => console.log(`${prefix}: x: ${x}, y:${y}`);
const mousedrag = mousedown.selectMany((md) => {

  const startX = md.clientX + window.scrollX,
        startY = md.clientY + window.scrollY,
        startLeft = parseInt(md.target.style.left, 10) || 0,
        startTop = parseInt(md.target.style.top, 10) || 0;
  
  return mousemove.map((mm) => {
    mm.preventDefault();
   log('mousemove',mm.clientX, mm.clientY);
    return {
      left: startLeft + mm.clientX - startX,
      top: startTop + mm.clientY - startY
    };
  }).takeUntil(mouseup);
});

subscription = mousedrag.subscribe((pos) => {
  log('dragstart',pos.top, pos.left)
  target.style.top = pos.top + 'px';
  target.style.left = pos.left + 'px';
});

.box {
  position: relative;
  width: 100px;
  height: 100px;
  background: red;
  cursor: pointer;
  border:solid 10px green;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/rxjs/2.5.2/rx.all.js"></script>
<div class="box"></div>

​

Answer 2

但你告诉我这样做:)

如果您从dragstart流中移除管道和"first“运算符

const dragstart = mousedown

这条小溪不会流完。拖移不会完成，因为它是从鼠标向下映射的，而不是dragstart流。

如果您告诉结果流只选择第一个流，则该流将在发出一次后完成。

这一守则应适用于：

const mousedown = fromEvent<MouseEvent>(window, 'mousedown');
const mousemove = fromEvent<MouseEvent>(window, 'mousemove');
const mouseup = fromEvent<MouseEvent>(window, 'mouseup');

// I dont think it's necessary to duplicate
const dragstart = mousedown

// No need to map to something else
const dragend = mouseup

const dragmove = dragstart.pipe(
  concatMapTo( 
    mousemove.pipe(
        takeUntil( dragend )
    )
  )
);



const log = (prefix: string) => (data: MouseEvent) => console.log(`${prefix}: x: ${data.clientX}, y:${data.clientY}`);

dragstart.subscribe(log('dragstart'));
dragmove.subscribe(log('dragmove'));
dragend.subscribe(log('dragend'));