首页 \ 问答 \ jQuery按名称选择元素(jQuery selecting elements by name)

jQuery按名称选择元素(jQuery selecting elements by name)

我有以下的html:

<td valign="top" align="left"> <input style="width:250px;" name="men_url" type="text" /></td>


<td valign="top" align="left"> 
<select name="men_page">
 <option value="">Wybierz stronę</option>
 <option value="index.php?page=8">O firmie</option>
 <option value="index.php?page=9">Referencje</option>

</select>
</td>

和两个相应的jQuery选择器:

$("select[name='men_page']")

$("input[name='men_url']")

第一个很好,第二个不会返回。 这里可能有什么问题?

尤其是alert($(“input [name ='men_url'”)。name); 显示“未定义”


I have the following html:

<td valign="top" align="left"> <input style="width:250px;" name="men_url" type="text" /></td>


<td valign="top" align="left"> 
<select name="men_page">
 <option value="">Wybierz stronę</option>
 <option value="index.php?page=8">O firmie</option>
 <option value="index.php?page=9">Referencje</option>

</select>
</td>

and two corresponding jQuery selectors:

$("select[name='men_page']")

and

$("input[name='men_url']")

The first one works great, the second one returns nothing. What might be wrong here?

Especially alert($("input[name='men_url'").name); displays "undefined"


原文:https://stackoverflow.com/questions/3128705
更新时间:2023-07-22 14:07

最满意答案

应该是if (shape.next().contains对象没有contains方法, shape.next()返回的Shape对象具有它。在这种情况下,你应该移除shape.next();因为你将移动到if的下一个对象或相应地修改整个代码。


Should be if (shape.next().contains. The Iterator object don't have the contains method, the Shape object returned by shape.next() has it. In that case you should remove the line shape.next(); since you'll move to the next object inside the if or modify the entire code accordingly.

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