我创建了一个名为Circle
的对象,并且该对象的两个实例名为circle1
和circle2
我试图让它们交换位置,但只有其中一个正在移动那一刻。
使用Javascript:
var gap = 6, size= 60;
var Circle = function(t,l, color){
var elem = document.createElement('div');
t = t,
l = l;
elem.setAttribute("class", "circle");
elem.style.backgroundColor = color;
// init positions
elem.style.top = gap+t*(size+2*gap) + "px";
elem.style.left = gap+l*(size+2*gap) + "px";
document.body.appendChild(elem);
this.getPosition = function(){
return [t,l];
};
this.setPosition = function(arr){
t = arr[0];
l = arr[1];
elem.style.top = gap+t*(size+2*gap) + "px";
elem.style.left = gap+l*(size+2*gap) + "px";
};
}
// make two new circle objects
var circle1 = new Circle(0, 0, "blue");
var circle2 = new Circle(0, 1, "red");
// we need the circles to swap positions
setTimeout(function(){
circle1.setPosition(circle2.getPosition());
circle2.setPosition(circle1.getPosition()); // this is not working
}, 1000);
我已将此代码放在jsFiddle上以使其更容易: http://jsfiddle.net/rhL7671p/
答案 0 :(得分:4)
只需缓存第一个圆圈的位置:
setTimeout(function(){
var pos = circle1.getPosition();
circle1.setPosition( circle2.getPosition() );
circle2.setPosition( pos );
}, 1000);
在此行之后的代码中
circle1.setPosition(circle2.getPosition());
第一个圆圈的位置被第二个圆圈的位置覆盖。因此下一行没有效果。在JavaScript中没有并行执行代码的事情,因为只有一个线程(有一些例外......)。
规避这个问题:首先获得一个(或两个)位置,然后再设置它们。
答案 1 :(得分:3)
在更新位置之前存储左圆圈的位置。
setTimeout(function(){
var c = circle1.getPosition();
circle1.setPosition(circle2.getPosition());
circle2.setPosition(c);
}, 1000);
var gap = 6, size= 60;
var Circle = function(t,l, color){
var elem = document.createElement('div');
t = t,
l = l;
elem.setAttribute("class", "circle");
elem.style.backgroundColor = color;
elem.style.top = gap+t*(size+2*gap) + "px";
elem.style.left = gap+l*(size+2*gap) + "px";
document.body.appendChild(elem);
this.getPosition = function(){
return [t,l];
}
this.setPosition = function(arr){
t = arr[0];
l = arr[1];
elem.style.top = gap+t*(size+2*gap) + "px";
elem.style.left = gap+l*(size+2*gap) + "px";
}
}
var circle1 = new Circle(0,0, "blue");
var circle2 = new Circle(0,1, "red");
// we need the circles to swap positions
setTimeout(function(){
var c = circle1.getPosition();
circle1.setPosition(circle2.getPosition());
circle2.setPosition(c);
}, 1000);
console.log(circle2)
.circle{
width:60px;
height:60px;
border-radius: 50px;
background-color:red;
position:absolute;
-webkit-transition-property: top, left;
-webkit-transition-duration: 0.3s;
}
答案 2 :(得分:2)
getPosition
的结果不是您在移动第一个圆圈后所期望的,所以在移动之前将其缓存
setTimeout(function(){
var a = circle2.getPosition(),
b = circle1.getPosition();
circle1.setPosition(a);
circle2.setPosition(b);
}, 1000);
答案 3 :(得分:2)
这种情况正在发生,因为当您尝试获取元素的位置时会有新值。一种可能的解决方案是使用局部变量:
var gap = 6,
size = 60;
var Circle = function(t, l, color) {
var elem = document.createElement('div');
t = t,
l = l;
elem.setAttribute("class", "circle");
elem.style.backgroundColor = color;
elem.style.top = gap + t * (size + 2 * gap) + "px";
elem.style.left = gap + l * (size + 2 * gap) + "px";
document.body.appendChild(elem);
this.getPosition = function() {
return [t, l];
}
this.setPosition = function(arr) {
t = arr[0];
l = arr[1];
elem.style.top = gap + t * (size + 2 * gap) + "px";
elem.style.left = gap + l * (size + 2 * gap) + "px";
}
}
var circle1 = new Circle(0, 0, "blue");
var circle2 = new Circle(0, 1, "red");
// we need the circles to swap positions
setTimeout(function() {
//use local variables to keep circles position
var circle1Pos = circle1.getPosition();
var circle2Pos = circle2.getPosition()
circle1.setPosition(circle2Pos);
circle2.setPosition(circle1Pos);
}, 200);

.circle {
width: 60px;
height: 60px;
border-radius: 50px;
background-color: red;
position: absolute;
-webkit-transition-property: top, left;
-webkit-transition-duration: 0.3s;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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