Java 数据表-在datatable中加载JSON响应

使用RESTAPI，我从下面加载datatable的Ajax中获得JSON响应

我收到错误：

未捕获类型错误：无法读取未定义的属性“length”。

$('#example').dataTable({
    "bSort" : false,
    "bDestroy" : true,
    "sPaginationType" : "full_numbers",
    "bProcessing" : false,
    "bFilter" : true,
    "bServerSide" : true,
    "bAutoWidth" : false,
    "sAjaxSource" : 'exampledatagrid.json?token=' + sessionId,
    "aoColumns" : [{
        "mData" : "0",
        "mData" : "1",
        "mData" : "2",
        "mData" : "3"
    }],
    "fnServerParams" : function(aoData) {
        alert(aoData);
        aoData.push({
            "name" : "colNameMode",
            "value" : "userStatusList"
        }, {
            "name" : "colName_0",
            "value" : "userIDList"
        }, {
            "name" : "colName_1",
            "value" : "userNameList"
        }, {
            "name" : "colName_2",
            "value" : "userRemarksList"
        });
    },
    "fnRowCallback" : function(nRow, aData, iDisplayIndex) {
        alert(aData);
    },
    "fnDrawCallback" : function(oSettings) {
        if ( oSettings.bSorted || oSettings.bFiltered ) {
            for ( var i=0, iLen=oSettings.aiDisplay.length ; i<iLen ; i++ ) {
                $('td:eq(0)', oSettings.aoData[ oSettings.aiDisplay[i] ].nTr ).html( i+1 );
            }
        }
    }
});

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我想知道我是否必须更改JSON格式，或者是否可以从上面的JSON响应本身检索？

我通过使用hibernate“org.hibernate.query”查询结果，在后端进行了更改，从而得到了答案。从中，我能够获得记录列表（一个接一个），并将其分别加载到网格中

{"userRemarksList":["TECHNOLOGY","GUEST","HOUSE","SECURITY"],"userIDList":[23,45,23,54],"userNameList":["Michael","Danny","Rahil","Ramesh"],"userStatusList":[0,0,1,1]}