Ajax 数据结果到 jqgrid 显示答案

【问题标题】：Ajax Data Result to jqgrid displayAjax 数据结果到 jqgrid 显示
【发布时间】：2012-07-12 18:09:12
【问题描述】：

我从 AJAX 调用得到这个结果，我想将它传递给 jqgrid。格式与正常的 jqgrid 调用完全相同。我不知道为什么它不起作用，非常感谢任何帮助。

{"page":null,"total":2,"records":"16","rows":[{"id":"9910003","cell":["9910003","DEMO3" ,"Test3","Sat,Sun","08:00:00","17:00:00","17","0"]},{"id":"9910004","cell": ["9910004","DEMO4","Test4","Sat,Sun","08:00:00","17:00:00","19","0"]},{"id": "9910005","cell":["9910005","DEMO5","Test5","Sat,Sun","08:00:00","17:00:00","17","0" ]},{"id":"9910006","cell":["9910006","DEMO6","Test6","Sat,Sun","08:00:00","17:00:00" ,"17","0"]},{"id":"9910007","cell":["9910007","DEMO7","Test7","Sat,Sun","08:00:00" ,"17:00:00","14","0"]},{"id":"9910008","cell":["9910008","DEMO8","Test8","Sat,Sun" ,"21:00:00","06:00:00","19","0"]},{"id":"9910009","cell":["9910009","DEMO9"," Test9","Sat,Sun","21:00:00","06:00:00","14","0"]},{"id":"9910010","cell":[" 9910010","DEMO10","Test10","Sat,Sun","21:00:00","06:00:00","14","0"]},{"id":"9910011 ","cell":["9910011","DEMO11","Test11","Sat,Sun","21:00:00","06:00:00","14","0"]} ,{"id":"9910012","cell":["9910012","DEMO12","Test12","Sat,Sun","21:00:00","06:00:00"," 17","0"]},{"id":"9910013" ,"cell":["9910013","DEMO13","Test13","Sat,Sun","21:00:00","06:00:00","16","0"]}, {"id":"9910014","cell":["9910014","DEMO14","Test14","Sat,Sun","21:00:00","06:00:00","14 ","0"]},{"id":"9910015","cell":["9910015","DEMO15","Test15","Sat,Sun","21:00:00","06 :00:00","14","0"]},{"id":"9910016","cell":["9910016","DEMO16","Test16","Sat,Sun","21 :00:00","06:00:00","14","0"]}]}

下面是我的完整代码：

function report_grid() {
$.ajax({
    type: "POST",
    url: "filter_option.php?action=filter",
    data: $('#form1').serialize(),
    success: function(rdata) {

jQuery("#report-grid").jqGrid({
    data: rdata,
    datatype: "local",
    colNames:['Emp ID','firstName', 'surName', 'restDay','shift_In','shift_Out','late','undertime'],
    colModel:[
        {name:'emp_id',index:'emp_id', width:55},
        {name:'firstname',index:'firstname', width:90},
        {name:'surname',index:'surname', width:100},
        {name:'restday',index:'restday', width:80, align:"right"},
        {name:'shift_in',index:'shift_in', width:80, align:"right"},
        {name:'shift_out',index:'shift_out', width:80,align:"right"},
        {name:'late',index:'late', width:150, sortable:false},
        {name:'ut',index:'ut', width:150, sortable:false},
    ],
    height: 'auto',
    width: 'auto',
    rowNum:10,
    rowList:[10,15,20,30,50],
    pager: '#report-pager',
    sortname: 'emp_id',
    viewrecords: true,
    sortorder: "desc",
    caption:"Timesheet Summary"
});
jQuery("#report-grid").jqGrid('navGrid','#report-pager',{edit:false,add:false,del:false});

        }
    })
}

我在下面尝试但没有成功，我想我错过了一些东西

data:rdata.rows
datatype:"local" // shows only the id but with correct number of records

data:rdata.rows
datatype: "json" // shows nothing at all.

【问题讨论】：

标签： jqgrid jqgrid-php

【解决方案1】：

现在知道了，我的json格式不对，应该参考demo页面中的data数组。

【讨论】：