RxJs：拖放示例：添加 mousedragstart

Question

如何使用 Drag and Drop RxJs's example.

实现 mousedragstart Observable

mousedragstart 应该在 mousedown 之后的第一个 mousedrag 之前发出，直到 mouseup。

我想我们必须玩 flatMap/take(1)/takeUntil(mouseup) 但我每次都失败..

更新

这里的难点是不能避免mousedrag在mousedragstart

之前发出

Answer 1

它应该像

一样简单

var mousedragstart = mousedown.flatMap(() => mousemove.takeUntil(mouseup).take(1));

但事实并非如此。 Chrome 在 mousedown 之后立即引发 mousemove 事件，这将导致上述逻辑在用户实际开始拖动之前错误地生成元素。所以你实际上需要这样的东西：

var mousedragstart = mousedown.flatMap(() =>
      mousemove
         .where(x => x.movementX !== 0 || x.movementY !== 0)
         .takeUntil(mouseup)
         .take(1)
);

Answer 2

基于我之前的非解决根本问题的答案以及您提供的信息，我们在概念上定义如下：

var dragTarget = document.getElementById('dragTarget');

var mouseup   = Rx.Observable.fromEvent(document,   'mouseup');
var mousemove = Rx.Observable.fromEvent(document,   'mousemove');
var mousedown = Rx.Observable.fromEvent(dragTarget, 'mousedown');

var dragstart = mousedown.flatMap(() =>
  mousemove
   .where(x => x.movementX !== 0 || x.movementY !== 0)
   .takeUntil(mouseup)
   .take(1)
);

var dragmove = mousedown.flatMap(() =>
  mousemove
    .where(x => x.movementX !== 0 || x.movementY !== 0)
    .takeUntil(mouseup)
);

这里的问题是事件之间的重叠；就与基础事件的关系而言，dragstart 的触发与第一个 dragmove 完全相同。在这种情况下，订阅顺序将决定执行顺序，正如您所说，这不是您想要依赖的。为了解决这个问题，我们必须控制底层事件。

这是一个简单的函数，它接受一个 observable 和 returns 一个包含两个 observable 的数组，这两个 observable 将被发出与原始 observable 相同的值，但其中事件将始终传递给第一个 observable 之前第二个 observable，不管第一个订阅的是哪个：

function prioritize(s$) {
  var first = new Rx.Subject();
  var second = s$.do(x => first.onNext(x)).share();

  return [
    Rx.Observable.using(
      () => second.subscribe(() => {}),
      () => first
    ),
    second
  ];
}

从那里，我们可以用这样的东西替换上面的适当部分：

var mousedowns = prioritize(mousedown);

var dragstart = mousedowns[0].flatMap(() =>
 mousemove
   .where(x => x.movementX !== 0 || x.movementY !== 0)
   .takeUntil(mouseup)
   .take(1)
);

var dragmove = mousedowns[1].flatMap(() =>
  mousemove
    .where(x => x.movementX !== 0 || x.movementY !== 0)
    .takeUntil(mouseup)
);

dragmove.subscribe(() => console.log('dragmove'));
dragstart.subscribe(() => console.log('dragstart'));

这是整个工作过程：

https://jsbin.com/qodilerofe/edit?js,console,output

Answer 3

另一种拖放解决方案：

let dragDiv = document.getElementById('drag');
let mouseisdown = false;
let startPos;

Observable.fromEvent(dragDiv, "mousedown").subscribe((e) => {
    mouseisdown = true;
    startPos = { x: e.offsetX, y: e.offsetY}
});
Observable.fromEvent(document, "mouseup").subscribe(e => mouseisdown = false); 


Observable
    .fromEvent(document, "mousemove")
    .filter(e => mouseisdown)
    .map((e) => {
        return   {
          left: e.clientX - startPos.x,
          top: e.clientY - startPos.y
        }
    })
    .subscribe( p => {
    dragDiv.style.top = p.top + "px";
    dragDiv.style.left = p.left + "px";
});

typescript版本：

let dragDiv = document.getElementById('drag');
let mouseisdown = false;
let startPos;

Observable.fromEvent(dragDiv, "mousedown").subscribe((e:MouseEvent) => {
    mouseisdown = true;
    startPos = { x: e.offsetX, y: e.offsetY}
});
Observable.fromEvent(document, "mouseup").subscribe(e => mouseisdown = false); 


Observable
    .fromEvent(document, "mousemove")
    .filter(e => mouseisdown)
    .map((e:MouseEvent) => {
        return   {
          left: e.clientX - startPos.x,
          top: e.clientY - startPos.y
        }
    })
    .subscribe( p => {
    dragDiv.style.top = p.top + "px";
    dragDiv.style.left = p.left + "px";
});

RxJs：拖放示例：添加 mousedragstart

RxJs: Drag and Drop example : add mousedragstart

javascript

system.reactive

rxjs