这些评论应该可以很好地说明问题，但是如果您有任何疑问，请告诉我。
这是玩的小提琴：http ://jsfiddle.net/Af9Jt/2/
var Draggable = React.createClass({
getDefaultProps: function () {
return {
// allow the initial position to be passed in as a prop
initialPos: {x: 0, y: 0}
}
},
getInitialState: function () {
return {
pos: this.props.initialPos,
dragging: false,
rel: null // position relative to the cursor
}
},
// we Could get away with not having this (and just having the listeners on
// our div), but then the experience would be possibly be janky. If there's
// anything w/ a higher z-index that gets in the way, then you're toast,
// etc.
componentDidUpdate: function (props, state) {
if (this.state.dragging && !state.dragging) {
document.addEventListener('mousemove', this.onMouseMove)
document.addEventListener('mouseup', this.onMouseUp)
} else if (!this.state.dragging && state.dragging) {
document.removeEventListener('mousemove', this.onMouseMove)
document.removeEventListener('mouseup', this.onMouseUp)
}
},
// calculate relative position to the mouse and set dragging=true
onMouseDown: function (e) {
// only left mouse button
if (e.button !== 0) return
var pos = $(this.getDOMNode()).offset()
this.setState({
dragging: true,
rel: {
x: e.pageX - pos.left,
y: e.pageY - pos.top
}
})
e.stopPropagation()
e.preventDefault()
},
onMouseUp: function (e) {
this.setState({dragging: false})
e.stopPropagation()
e.preventDefault()
},
onMouseMove: function (e) {
if (!this.state.dragging) return
this.setState({
pos: {
x: e.pageX - this.state.rel.x,
y: e.pageY - this.state.rel.y
}
})
e.stopPropagation()
e.preventDefault()
},
render: function () {
// transferPropsTo will merge style & other props passed into our
// component to also be on the child DIV.
return this.transferPropsTo(React.DOM.div({
onMouseDown: this.onMouseDown,
style: {
left: this.state.pos.x + 'px',
top: this.state.pos.y + 'px'
}
}, this.props.children))
}
})
从一开始，“谁应该拥有什么状态”是一个重要的问题。在“可拖动”组件的情况下，我可以看到几种不同的情况。
602392714
清零编程群