使用jquery

时间:2017-02-25 22:13:22

标签: jquery html html-table

我使用jquery动态构建了一个表。在我的代码下面:

<div class="scroll-pane">
   <table class="table support-table">
    <thead>
        <th id="support-field-1"></th>
        <th id="support-field-3"></th>
        <th id="support-field-4"></th>
        <th id="support-field-5"></th>
        <th id="support-field-6"></th>
    </thead>
    <tbody>
    </tbody>
   </table>
</div>

<script>
var container = $('.support-table tbody');
var ticket;
var j;
for(var i = 0; i < support.length; i++) {
    var id=support[i].identification!=null ? support[i].identification : '';
    var sdateTime = support[i].sdateTime!=null ? support[i].sdateTime : '';
    var type = support[i].type!=null ? support[i].type : '';
    var subtype = support[i].subtype!=null ? support[i].subtype : '';
    var status = support[i].status!=null ? support[i].status : '';
    var feedback = support[i].feedback!=null ? support[i].feedback : '';
    ticket = '<tr class="ticketRow" id="ticket'+i+'"><td>'+sdateTime+'</td>';
    ticket += '<td>'+type+'</td><td>'+subtype+'</td><td>'+status+'</td>   <td>'+feedback+'</td></tr>';
    container.append(ticket);
}
</script>

我根据JSON对象的长度动态构建了表,并且#34;支持。&#34; 现在我想更改表格第三行中的子类型文本。我怎么能用jquery做到这一点?我正在尝试:

$('table support-table').children('tbody').children('tr').eq(3).eq(4).html("POSITIVE");

但我没有成功。我做错了什么?

0 个答案:

没有答案