本文介绍了有效检测兄弟元素重叠的时间的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
示例:
<div id="big"> </div> <div class="small"> </div> <div class="small"> </div> <div class="small"> </div> <div class="small"> </div> <div class="small"> </div> <!-- ...and so on -->#big"绝对位于".small"部分的后面,但是 不是父元素.
"#big" is positioned absolutely behind a portion of the ".small"s, but is not a parent element.
我一直在这样做:
var smallArray = []; var $big = $('#big'); var $bigPos = $big.offset(); $('div.small').each(function() { var $this = $(this); var $thisPos = $this.offset(); if( $thisPos.left >= $bigPos.left && $thisPos.left <= $bigPos.left+$big.outerWidth() && $thisPos.top >= $bigPos.top && $thisPos.top <= $bigPos.top+$big.outerHeight() ) smallArray.push($this); });...但是这似乎很糊涂.我是否错过了jQuery的某些方法 或香草JavaScript,让我可以更优雅地完成此操作 &有效的方式?
...but this seems kludgy. Am I missing out on some methods of jQuery or vanilla JavaScript that will allow me to do this in a more elegant & efficient manner?
感谢您提供的任何帮助.
Thanks ahead for any help you can provide.
推荐答案此公式将检测是否有任何指定的元素与目标元素重叠:
This formula will detect if any of the specified elements is overlapping a target element:
function findIntersectors(targetSelector, intersectorsSelector) { var intersectors = []; var $target = $(targetSelector); var tAxis = $target.offset(); var t_x = [tAxis.left, tAxis.left + $target.outerWidth()]; var t_y = [tAxis.top, tAxis.top + $target.outerHeight()]; $(intersectorsSelector).each(function() { var $this = $(this); var thisPos = $this.offset(); var i_x = [thisPos.left, thisPos.left + $this.outerWidth()] var i_y = [thisPos.top, thisPos.top + $this.outerHeight()]; if ( t_x[0] < i_x[1] && t_x[1] > i_x[0] && t_y[0] < i_y[1] && t_y[1] > i_y[0]) { intersectors.push($this); } }); return intersectors; }这是一个POC.
此SO问题对于解决此问题非常有帮助.
This SO question was very helpful in solving this problem.
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有效检测兄弟元素重叠的时间
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