ajax提交数据返回后,gridview中__doPostBack回发失效的问题 20C

我用ajax提交数据后,gridview中模板列的控件__doPostBack回发就失效了,
前台代码:

 $.ajax(
                     {
                         type: 'POST',
                         url: 'xinghao.aspx',
                         //async: false,//异步的话则注释
                         data: { 'strAction': 'dlg', 'bj_cid': DataKeyName, 'Sel': value },
                         success: function (res) {
                             var dvGrid = $(res).find("#dvGrid").html();
                             $("#dvGrid").html(dvGrid);

                         }
                     })

通过html中返回代码

 <table cellspacing="0" cellpadding="4" rules="cols" id="gridXh" style="color:Black;background-color:White;border-color:#DEDFDE;border-width:1px;border-style:None;width:100%;border-collapse:collapse;">
            <tbody><tr style="color:White;background-color:#5D7B9D;border-style:Solid;font-weight:bold;height:20px;">
                <th scope="col">&nbsp;</th><th scope="col">型号</th>
            </tr><tr style="color:#333333;background-color:#F7F6F3;">
                <td class="grid_style" align="center" style="width:2%;">

                                                                        <input id="gridXh_chk_0" type="checkbox" name="gridXh$ctl02$chk">
                                                                    </td><td class="grid_style" ondblclick="javascript:__doPostBack('gridXh$ctl02$ctl00','2')" style="width:10%;cursor:pointer;cursor:hand;">
                                                                        <span id="gridXh_lb_2_0">1×1</span>


                                                                    </td>
            </tr>
        </tbody></table>

1个回答

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