如何迭代表并获取每个复选框值

Question

我有这张桌子:

<table ="myTable">
<tr>
<td id="username_57">lillie</td>
<td id="text_57"/>
<td>1969-12-31</td>
<td>24.xx.84.xx</td>
<td>
<button id="edit_text_57" name="id_57">Edit</button>
<input type="checkbox" id="delit_57"/>
</td>
</tr>
<tr>
<td id="username_39">test</td>
<td id="text_39">asdasdasdsdasdasd</td>
<td>2012-06-04</td>
<td>217.xx.237.6</td>
<td>
<button id="edit_text_39" name="id_39">Edit</button>
<input type="checkbox" id="delit_39"/>
</td>
</tr>
<tr>
<td id="username_45">admin</td>
<td id="text_45">sadasdasdad</td>
<td>2012-09-04</td>
<td>217.xx.237.6</td>
<td>
<button id="edit_text_45" name="id_45">Edit</button>
<input type="checkbox" id="delit_45"/>
</td>
</tr>
</table>

我喜欢用id delit_*遍历所有复选框并检查它的值是否被选中？我怎么能以jquery风格做到这一点？

Answer 1

选择器实际上可以返回被检查的选择器.不需要迭代.

$('input[type="checkbox"][id^="delit_"]:checked').each(function()
{
// Do something.
});

如果要检查和取消选中,只需删除:checked选择器的一部分,然后if (this.checked) { }在.each循环中进行比较.

jsFiddle:http://jsfiddle.net/jimp/rKjtr/