[英]Raphael .mouseup() function only firing once
我正在尝试将可拖动的元素拖动到Rapheal中另一个元素的位置。 我遇到的问题是.mouseup()
函数仅在其中执行一次功能。 再次拖动或移动该元素后,它将不再执行我在其中具有的定位功能。
我的最终目标是:
这是我尝试使用的代码,但似乎无法正常运行:
JSfiddle: http : //jsfiddle.net/4GWEU/3/
Javascript:
//Makes elements Draggable.
Raphael.st.draggable = function() {
var me = this,
lx = 0,
ly = 0,
ox = 0,
oy = 0,
moveFnc = function(dx, dy) {
lx = dx + ox;
ly = dy + oy;
me.transform('t' + lx + ',' + ly);
},
startFnc = function() {
//window.draggedElement = this;
},
endFnc = function() {
ox = lx;
oy = ly;
};
this.drag(moveFnc, startFnc, endFnc);
};
var container = document.getElementById('container');
var paper = Raphael(container, '539', '537');
var shape1 = paper.rect(50,50, 50,50);
shape1.attr({x: '50',y: '50',fill: 'red','stroke-width': '0','stroke-opacity': '1'});
shape1Set = paper.set(shape1);
shape1Set.draggable();
var shape2 = paper.rect(50,50, 50,50);
shape2.attr({x: '150',y: '50',fill: 'blue','stroke-width': '0','stroke-opacity': '1'});
shape1Set.mousedown(function(event) {
console.log('mousedown');
});
shape1Set.mouseup(function(event) {
console.log('mouseup');
positionElementToElement(shape1, shape2);
});
$('#runPosition').click(function () {
positionElementToElement(shape1, shape2);
});
$('#runPosition2').click(function () {
positionElementToElement2(shape1, shape2);
});
function positionElementToElement(element, positionTargetElement)
{
var parentBBox = positionTargetElement.getBBox();
parent_x = parentBBox.x;
parent_y = parentBBox.y;
parent_width = parentBBox.width;
parent_height = parentBBox.height;
var elementBBox = element.getBBox();
element_width = elementBBox.width;
element_height = elementBBox.height;
var x_pos = parent_x + (parent_width / 2) - (element_width / 2) + 100;
var y_pos = parent_y + (parent_height / 2) - (element_height / 2) + 100;
console.log('Positioning element to: '+x_pos+' '+y_pos);
element.animate({'x' : x_pos, 'y' : y_pos}, 100);
}
function positionElementToElement2(element, positionTargetElement)
{
var parentBBox = positionTargetElement.getBBox();
parent_x = parentBBox.x;
parent_y = parentBBox.y;
parent_width = parentBBox.width;
parent_height = parentBBox.height;
var elementBBox = element.getBBox();
element_width = elementBBox.width;
element_height = elementBBox.height;
var x_pos = parent_x + (parent_width / 2) - (element_width / 2);
var y_pos = parent_y + (parent_height / 2) - (element_height / 2);
console.log('Positioning element to: '+x_pos+' '+y_pos);
element.animate({'x' : x_pos, 'y' : y_pos}, 100);
}
HTML:
<a href="#" id="runPosition">Run Position</a>
<a href="#" id="runPosition2">Run Position2</a>
<div id="container"></div>
笔记:
positionElementToElement()
函数,并使用偏移量设置了其中之一。 我已经将两个功能绑定到“ Run Position 1
和Run Position 2
链接。 Run Position 1
链接不再将正方形设置回应该去的位置(即使该函数记录的x / y坐标与工作时相同)。 我已经知道如何正确地做到这一点。
您必须直接修改元素的x
和y
属性。
同样重要的是要注意,当使用element.attr('x');
从元素中检索x和y属性时element.attr('x');
或element.attr('y');
它返回一个字符串值,而不是整数。 因此,您必须对这些返回值使用parseInt()
,以正确地将移动x和y值相加,以在元素移动时应用于元素。
当移动红色方块时,以下代码会将红色方块对齐到蓝色方块。
工作示例: http : //jsfiddle.net/naQQ2/2/
window.onload = function () {
var R = Raphael(0, 0, "100%", "100%"),
shape1 = R.rect(50,50, 50,50);
shape1.attr({x:'50',y:'50',fill: 'red','stroke-width': '0','stroke-opacity': '1'});
shape2 = R.rect(50,50, 50,50);
shape2.attr({x:'150',y:'50',fill: 'blue','stroke-width': '0','stroke-opacity': '1'});
var start = function () {
console.log(this);
this.ox = parseInt(this.attr('x'));
this.oy = parseInt(this.attr('y'));
this.animate({opacity: .25}, 500, ">");
},
move = function (dx, dy) {
this.attr({x: this.ox + dx, y: this.oy + dy});
},
up = function () {
//Snap to shape2 on mouseup.
var snapx = parseInt(shape2.attr("x"));
snapy = parseInt(shape2.attr("y"));
this.animate({x: snapx, y: snapy}, 100);
this.animate({opacity: 1}, 500, ">");
};
R.set(shape1, shape2).drag(move, start, up);
};
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.