元素相对于其父级的坐标

Question

该方法el.getBoundingClientRect()给出了结果相对于视口的左上角(一0,0),而不是相对于一个元素的父,而el.offsetTop,el.offsetLeft(等)得到的结果相对于父一个.

使元素的坐标相对于其父元素的最佳实践是什么？el.getBoundingClientRect()修改(如何？)使用父为(0,0)坐标,或静止el.offsetTop,el.offsetLeft等等？

Answer 1

您可以使用getBoundingClientRect(),只需减去父级的坐标:

var parentPos = document.getElementById('parent-id').getBoundingClientRect(),
    childPos = document.getElementById('child-id').getBoundingClientRect(),
    relativePos = {};

relativePos.top = childPos.top - parentPos.top,
relativePos.right = childPos.right - parentPos.right,
relativePos.bottom = childPos.bottom - parentPos.bottom,
relativePos.left = childPos.left - parentPos.left;

console.log(relativePos);
// something like: {top: 50, right: -100, bottom: -50, left: 100}

现在您拥有相对于其父级的子级坐标.

请注意,如果top或left坐标为负数,则表示子项在该方向上转义其父项.如果bottom或right坐标为正,则相同.

工作实例

var parentPos = document.getElementById('parent-id').getBoundingClientRect(),
    childPos = document.getElementById('child-id').getBoundingClientRect(),
    relativePos = {};

relativePos.top = childPos.top - parentPos.top,
relativePos.right = childPos.right - parentPos.right,
relativePos.bottom = childPos.bottom - parentPos.bottom,
relativePos.left = childPos.left - parentPos.left;

console.log(relativePos);

#parent-id {
    width: 300px;
    height: 300px;
    background: grey;
}

#child-id {
    position: relative;
    width: 100px;
    height: 200px;
    background: black;
    top: 50px;
    left: 100px;
}

<div id="parent-id">
    <div id="child-id"></div>
</div>