Java中jqGrid-jQuery的外部分页

Question

我已经开发了如下代码。

jQuery(document).ready(function(){
    jQuery("#records").jqGrid({
        height:350,
        datatype: 'local',
        colNames:['Policy Name','Policy Type', 'Time allowed (HH:mm)','Expiration Duration (days)','Session Pulse(minutes)','Description'],
        colModel :[
            {name:'pName', index:'pName', editable:true,sorttype:'text',width:150,editoptions:{size:10},formatter:'showlink',formatoptions:{baseLinkUrl:'javascript:' , showAction: "GetAndShowUserData(jQuery('#list2'),'",addParam: "');"}},
            {name:'pType', index:'pType', sorttype:'text',editable:true,width:150,editoptions:{size:10}},
            {name:'timeAllowed', index:'timeAllowed', sorttype:'text',editable:true,width:200, align:"right",editoptions:{size:10}},
            {name:'expDuration', index:'expDuration',  sorttype:'text',editable:true,width:200, align:"right",editoptions:{size:10}},
            {name:'sessionPulse', index:'sessionPulse',sorttype:'int',editable:true,width:200, align:"right",editoptions:{size:10}},
            {name:'description', index:'description', sortable:false,editable:true,width:200,editoptions:{size:10}}],
        pager:jQuery('#pager'),
        rowNum:10,
        sortname: 'pName',
        autowidth:true,
        altRows:true,
        drag:true,
        sortorder: "asc",
        rowList:[2,5,10,20],
        viewrecords: true,
        loadonce:false,
        multiselect: true,
        /*
            onSelectRow: function(id){
                var gr = jQuery("#records").jqGrid('getGridParam','selrow');
                if( gr != null ) jQuery("#records").jqGrid('editGridRow',gr,{height:280,reloadAfterSubmit:false});
                else alert("Please Select Row");
            },
            editurl: "server.php",
        */
        caption:'Manage Policy'
    });
});

现在，当用户按下＆gt;＆gt;时，我想向servlet发出Ajax请求以获取下一条记录。 jqGrid的（下一个）按钮。我在互联网上搜索了很多，但我发现了很多PHP代码，但我无法理解PHP;我想用Java开发那个东西。我怎么能这样做？