Rag*_*azs 20 jquery jquery-selectors
我有这个东西:
<table>
<tr>
<td>nonono</td> <!-- FIND THIS -->
</td>foobar</td>
</tr>
<tr>
<td>nonono2</td> <!-- FIND THIS -->
</td>foobar2</td>
</tr>
<tr>
<td>nonono3</td> <!-- FIND THIS -->
</td>foobar2</td>
</tr>
</table>
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我试过$('td:first')没有运气;
预期收益:<td>nonono</td>,<td>nonon2</td>和<td>nonon3</td>
提前致谢!
Jam*_*lly 45
您应该使用:first-child而不是:first:
听起来你想要迭代它们.你可以使用.each().
例:
$('td:first-child').each(function() {
console.log($(this).text());
});
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结果:
nonono
nonono2
nonono3
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如果你不想迭代,那就太过分了:
$('td:first-child').css('background', '#000');
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Moh*_*dil 11
试试这个选择器 -
$("tr").find("td:first")
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演示--> http://jsfiddle.net/66HbV/
要么
$("tr td:first-child")
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演示--> http://jsfiddle.net/66HbV/1/
</td>foobar</td> 应该 <td>foobar</td>
你可以这样做
$(function(){
$("tr").find("td:eq(0)").css("color","red");
})Run Code Online (Sandbox Code Playgroud)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
<tr>
<td>col_1</td>
<td>col_2</td>
</tr>
<tr>
<td>col_1</td>
<td>col_2</td>
</tr>
<tr>
<td>col_1</td>
<td>col_2</td>
</tr>
<tr>
<td>col_1</td>
<td>col_2</td>
</tr>
</table>Run Code Online (Sandbox Code Playgroud)