在每个表上运行jQuery [英] Run jquery on each table

问题描述

下面，我有一些jQuery，它们通过类名= sizetable遍历每个表

Below I have some jQuery that runs through each table with the classname = sizetable

我要执行此操作:

$("#frontshade :radio").click(function() {});

但是我不确定如何用数组中的ID替换#frontshade

But I'm not sure how to replace the #frontshade with the id from the array

  var table_ids = new Array();
  $('.sizetable')
  .each(function(e){
    table_ids[] = $(this).attr('id');
  // JQUERY TO EXECUTE ON EACH TABLE        

  $("#frontshade :radio").click(function() {};

  //

  });

推荐答案

您可以编写$(this).find('input:radio').
.find()方法查找与选择器匹配的所有后代.

You can write $(this).find('input:radio').
The .find() method finds all descendants that match a selector.

请注意， input:radio比:radio 快.
如文档所述，

Note that input:radio is faster than :radio.
As the documentation states,

$(':radio')等效于 $('[type=radio]').和其他一样 伪类选择器(那些 建议以:"开头 在其前面加上标签名称或一些 其他选择器否则， 暗含通用选择器("*"). 换句话说，裸$(':radio') 等价于$('*:radio')，所以 应使用$('input:radio') 代替.

$(':radio') is equivalent to $('[type=radio]'). As with other pseudo-class selectors (those that begin with a ":") it is recommended to precede it with a tag name or some other selector; otherwise, the universal selector ("*") is implied. In other words, the bare $(':radio') is equivalent to $('*:radio'), so $('input:radio') should be used instead.

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