我有下表
FILETIME
我正在尝试为每个struct.unpack
添加不同的属性。理想情况下使用jquery .index和.attr
如果没有ID或类,请不确定如何选择<table class="results">
<thead>
<tr>
<th>Title1</th>
<th>Title2</th>
</tr>
</thead>
<tbody>
<tr>
<td>a</td>
<td>e</td>
</tr>
<tr>
<td>b</td>
<td>f</td>
</tr>
<tr>
<td>c</td>
<td>g</td>
</tr>
<tr>
<td>d</td>
<td>h</td>
</tr>
</tbody>
我猜我们可以使用.results类吗?
答案 0 :(得分:1)
我想为第2个添加不同的属性,例如th = =#34;一个&#34; th = =#34;两个&#34;
要实现这一点,您可以使用映射到表中th
元素的索引的类名数组。试试这个:
var classes = ['one', 'two'];
$('.results th').addClass(function(i) {
return classes[i];
});
&#13;
.one {
color: red;
}
.two {
color: blue;
}
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table class="results">
<thead>
<tr>
<th>Title1</th>
<th>Title2</th>
</tr>
</thead>
<tbody>
<tr>
<td>a</td>
<td>e</td>
</tr>
<tr>
<td>b</td>
<td>f</td>
</tr>
<tr>
<td>c</td>
<td>g</td>
</tr>
<tr>
<td>d</td>
<td>h</td>
</tr>
</tbody>
</table>
&#13;
答案 1 :(得分:1)
试试这个:
$(document).ready(function(){
$("th").each(function(index,ele){
var cla = (index===0) ? "one" : "two";
$(this).addClass(cla);
})
})
最终代码:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
<style>
.one {
color: red;
}
.two {
color: blue;
}
</style>
</head>
<body>
<table class="results">
<thead>
<tr>
<th>Title1</th>
<th>Title2</th>
</tr>
</thead>
<tbody>
<tr>
<td>a</td>
<td>e</td>
</tr>
<tr>
<td>b</td>
<td>f</td>
</tr>
<tr>
<td>c</td>
<td>g</td>
</tr>
<tr>
<td>d</td>
<td>h</td>
</tr>
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("th").each(function(index,ele){
var cla = (index===0) ? "one" : "two";
$(this).addClass(cla);
})
})
</script>
</body>
</html>
&#13;