jQuery选择问题

Question

好的，所以我需要选择一个TD并让它改变里面的图像，这很好。但是，我需要能够同时显示所选td的ID，但是我的功能必须选择图像才能更改它。有没有办法做到这一点，所以它会选择图像，改变它，然后显示带有td id的警报？下面的代码

的jQuery：

$(document).ready(function(){ 
    $("td img").click(function() { 
    $(this).attr('src', 'images/d.gif'); 
    alert($(this).attr("td id")); 
    return false; 
    }); 
}); 

HTML

<div id='plan'> 
<table> 
<tr> 
    <td class='n' id='a1'><img src='images/a.gif'/></td> 
    <td class='n' id='a2'><img src='images/a.gif'/></td> 
    <td class='n' id='a3'><img src='images/a.gif'/></td> 
    <td class='n' id='a4'><img src='images/a.gif'/></td> 
    <td></td> 
    <td class='n' id='a6'><img src='images/a.gif'/></td> 
    <td class='n' id='a7'><img src='images/a.gif'/></td> 
    <td class='n' id='a8'><img src='images/a.gif'/></td> 
    <td class='n' id='a9'><img src='images/a.gif'/></td> 
</tr> 
<tr> 
    <td class='n' id='b1'><img src='images/a.gif'/></td> 
    <td class='n' id='b2'><img src='images/a.gif'/></td> 
    <td class='n' id='b3'><img src='images/1.gif'/></td> 
    <td class='n' id='b4'><img src='images/1.gif'/></td> 
    <td></td> 
    <td class='n' id='b6'><img src='images/1.gif'/></td> 
    <td class='n' id='b7'><img src='images/a.gif'/></td> 
    <td class='n' id='b8'><img src='images/a.gif'/></td> 
    <td class='n' id='b9'><img src='images/a.gif'/></td> 
</tr> 
<tr> 
    <td class='n' id='c1'><img src='images/1.gif'/></td> 
    <td class='n' id='c2'><img src='images/1.gif'/></td> 
    <td class='n' id='c3'><img src='images/a.gif'/></td> 
    <td class='n' id='c4'><img src='images/a.gif'/></td> 
    <td></td> 
    <td class='n' id='c6'><img src='images/1.gif'/></td> 
    <td class='n' id='c7'><img src='images/1.gif'/></td> 
    <td class='n' id='C8'><img src='images/a.gif'/></td> 
    <td class='n' id='C9'><img src='images/a.gif'/></td> 
</tr> 
<tr> 
    <td class='n' id='d1'><img src='images/1.gif'/></td> 
    <td class='n' id='d2'><img src='images/1.gif'/></td> 
    <td class='n' id='d3'><img src='images/1.gif'/></td> 
    <td class='n' id='d4'><img src='images/1.gif'/></td> 
    <td></td> 
    <td class='n' id='d6'><img src='images/1.gif'/></td> 
    <td class='n' id='d7'><img src='images/1.gif'/></td> 
    <td class='n' id='d8'><img src='images/1.gif'/></td> 
    <td class='n' id='d9'><img src='images/1.gif'/></td> 
</tr> 
<tr> 
    <td class='p' id='e1'><img src='images/1.gif'/></td> 
    <td class='p' id='e2'><img src='images/1.gif'/></td> 
    <td class='p' id='e3'><img src='images/1.gif'/></td> 
    <td class='p' id='e4'><img src='images/1.gif'/></td> 
    <td></td> 
    <td class='p' id='e6'><img src='images/a.gif'/></td> 
    <td class='p' id='e7'><img src='images/a.gif'/></td> 
    <td class='p' id='e8'><img src='images/1.gif'/></td> 
    <td class='p' id='e9'><img src='images/1.gif'/></td> 
</tr> 
</table> 
</div> 

Answer 1

我想你的意思$(this).parent().attr("id");

您的代码：

alert($(this).attr("td id")); 

实际上是寻找一个名为属性td id的img标签。