Question

我一直在玩 RxJS 6 。我正在尝试实施dragstart, dragmove and dragend。这是我到目前为止的代码。

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'));

问题是运营商takeUntil和first将流标记为已完成。这意味着dragstart和dragend只会触发一次。有没有办法在这个事件发生后以某种方式重置流？例如，当dragstart流收到事件时重置mouseup流。

补充说明

在当前实现中，dragstart和dragend将在加载页面后完全记录一次到控制台（＆amp;拖动）。之后，流完成，不再通过它发送任何事件。我想在每次拖动操作时重置流，所以dragstart＆amp; dragend将在每次拖动操作时工作，而不是仅在加载页面后的第一次操作。

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

[mousedown]
的dragstart
[鼠标移动]
dragmove
[鼠标松开]
dragend

第二次拖动操作

[鼠标按下]
（没有任何反应，但是应该将dragstart记录到控制台）
[鼠标移动]
dragmove
[鼠标松开]
（没有任何反应，但应将dragend记录到控制台）

Answer 1

请看这里的工作示例 https://codepen.io/anon/pen/bMWjEV enter link description here

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

但是您却告诉过这样做：）

如果您从dragstart流中删除管道和“ first”运算符

const dragstart = mousedown

该流不会完成。拖动动作无论如何都不会完成，因为它是从鼠标向下而不是从拖动开始流映射的。

如果您告诉结果流仅选择第一个流，则该流将在发出一次之后完成。

此代码应该有效：

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'));

RxJS6：完成后重置流

补充说明

当前行为（[]表示发生的鼠标操作，斜体文本是控制台输出）：

2 个答案: