求完整的代码：
用java将查询的列表封装成json传递到前台，前台获取json格式的数据
这边的url该怎么写
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Complex DataGrid - jQuery EasyUI Demo</title>
<link rel="stylesheet" type="text/css" href="../themes/default/easyui.css">
<link rel="stylesheet" type="text/css" href="../themes/icon.css">
<link rel="stylesheet" type="text/css" href="demo.css">
<script type="text/javascript" src="../jquery-1.6.min.js"></script>
<script type="text/javascript" src="../jquery.easyui.min.js"></script>
<script>
$(function(){
$('#test').datagrid({
title:'My DataGrid',
iconCls:'icon-save',
width:700,
height:350,
nowrap: false,
striped: true,
collapsible:true,
url:'datagrid_data.json', sortName: 'code',
sortOrder: 'desc',
remoteSort: false,
idField:'code',
frozenColumns:[[
{field:'ck',checkbox:true},
{title:'code',field:'code',width:80,sortable:true}
]],
columns:[[
{title:'Base Information',colspan:3},
{field:'opt',title:'Operation',width:100,align:'center', rowspan:2,
formatter:function(value,rec){
return '<span style="color:red">Edit Delete</span>';
}
}
],[
{field:'name',title:'Name',width:120},
{field:'addr',title:'Address',width:220,rowspan:2,sortable:true,
sorter:function(a,b){
return (a>b?1:-1);
}
},
{field:'col4',title:'Col41',width:150,rowspan:2}
]],
pagination:true,
rownumbers:true,
toolbar:[{
id:'btnadd',
text:'Add',
iconCls:'icon-add',
handler:function(){
$('#btnsave').linkbutton('enable');
alert('add')
}
},{
id:'btncut',
text:'Cut',
iconCls:'icon-cut',
handler:function(){
$('#btnsave').linkbutton('enable');
alert('cut')
}
},'-',{
id:'btnsave',
text:'Save',
disabled:true,
iconCls:'icon-save',
handler:function(){
$('#btnsave').linkbutton('disable');
alert('save')
}
}]
});
var p = $('#test').datagrid('getPager');
$(p).pagination({
onBeforeRefresh:function(){
alert('before refresh');
}
});
});
</script>
</head>
<body>
<table id="test"></table>
</body>
</html>
------解决方案--------------------
- Java code
$(function(){$('#test').datagrid({....url:'test.action',.....});});