jQuery index（）与所有可见的兄弟姐妹相关 [英] jQuery index() in relation to all visible siblings

问题描述

我想获取与所有可见兄弟相关的元素索引。

I'd like to fetch the index of an element in relation to all visible siblings.

A td.index（）确实将td的索引提取给所有其他td兄弟。

A td.index() does fetch the index of the td to all other td siblings.

但当某些TD设置为时显示什么：无。我想在计算索引时排除它们。 td.index（'：visible'）似乎无效。

But what when some of those TDs are set to display:none. I want to exclude them when calculating the index. td.index(':visible') does not seem to work.

推荐答案

var $td = $("#theTD")

$td.siblings(":visible").andSelf().index($td);

以上内容应该按照您的要求进行。基本上获取要在其中搜索的元素集，然后获取元素的索引。

The above should do what you're asking. Basically get the set of elements you want to search within and then get the index of your element within them.

http://jsfiddle.net/3NYY9/

编辑：截至jquery 1.8 andSelf 已弃用， addBack 应在其位置使用：

As of jquery 1.8 andSelf has been deprecated and addBack should be used in its place:

var $td = $("#theTD")

$td.siblings(":visible").addBack().index($td);

http://jsfiddle.net/3NYY9/45/

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