jqgrid后台传参数

JScript code // 获得当前postData选项的值

var postData = $("#proList").jqGrid("getGridParam", "postData"); // 将查询参数融入postData选项对象

$.extend(postData, [color=#FF0000]$("#searchVoForm").serializeArray()[/color]);

$("#proList").jqGrid('setGridParam',{search: true}).trigger("reloadGrid", [{page:1}]);

 

HTML code

<s:form id="searchVoForm">

<table id="searchVoInfo" class="form_table" width="100%" border="1" cellpadding="0" cellspacing="0">

<tr>

<td class="td_title_right" width="10%"><s:label value="项目名称 "/></td>

<td width="40%" align="left"><s:textfield id="proName" name="searchVo.proName" size="30" maxlength="50" cssStyle="width:155px"/></td>

<td class="td_title_right" width="10%"><s:label value="建设地区"/></td>

<td width="40%" align="left"><s:textfield id="buildArea" name="searchVo.buildArea" size="30" maxlength="50" cssStyle="width:155px"/></td>

</tr>

</table>

</s:form>

<div style="padding: 4px 0px 4px 0px; text-align: center;">

<div id="searchId" onclick="searchProList();">查询</div>

<div id="clearId" onclick="clearParam();">清空</div></div>

<div id="proGrid" class="ui-layout-west ui-widget ui-widget-content"><table id="proList" width="100%"></table><div id="pager" style="width: 100%;"></div></div>

 

$("#list").setGridParam({postData:{"param1": "value1","param2": param2}}).trigger("reloadGrid");
假设jqgrid的id是list主要是改变postData的数据就行 更简单的是对象化form中所以要提交的控件
$.fn.serializeObject = function(){ 
 var o = {}; var a = this.serializeArray(); 
 $.each(a, function() {  
  if (o[this.name] !== undefined)
   {   
   if (!o[this.name].push)
    {    
    o[this.name] = [o[this.name]];   
    }   
    o[this.name].push(this.value || '');  
    } else {   
     o[this.name] = this.value || '';  
    } 
   }); 
    return o;};
然后var serialformdate = $("form").serializeObject()$("#list").setGridParam({page:1, postData:serialformdate}).trigger("reloadGrid");

提交所以form的数据导control了

 

http://www.oschina.net/question/69055_29415 ------jqGrid如何传递id修改URL