java脚本代码
$("#category select").change(function(){
var category = $('select option:selected').html();
$.ajax({
type:"Post",
url:"collectiepost.php",
data:"category="+category,
cache:"false",
success:function(html){
$("select").html($(html).find("option.category"));
$("#test").html($(html).find("#testdata"));
}
});
});
}); collectiepost.php页面上的html输出
<select id="ontwerper">
<option class="desinger">vikas</option>
</select>
<select id="category">
<option class="category">cloth1</option>
<option class="category">cloth2</option>
</select>
<div id="testdata">test data</div>需要输出
<select>
<option class="category">cloth1</option>
<option class="category">cloth2</option>
</select>
<div id="test">
<div id="testdata">test data</div>
</div>问题
但是我的输出没有div标签
<select>
<option class="category">cloth1</option>
<option class="category">cloth2</option>
</select>发布于 2012-07-19 12:22:25
删除以下行中"div #test“中的空格:
$("div #test").html($(html).find("div #testdata"));发布于 2012-07-19 12:26:30
我想你可以得到这个:
<div id="test">
<div id="testdata">test data</div>
</div>因为在html输出中没有"test“div:
<select id="ontwerper">
<option class="desinger">vikas</option>
</select>
<select id="category">
<option class="category">cloth1</option>
<option class="category">cloth2</option>
</select>
<div id="testdata">test data</div>
No test div here而$("div #test").html($(html).find("div #testdata"));失败了
把它包装成这样:
$('#testdata').wrap('<div id="test" />');https://stackoverflow.com/questions/11560416
复制相似问题