Question

看起来很简单但是我从jqGrid代码中得到了这个神秘错误，说“元素不是表”。这是代码：

var grid_data = [ {id:"1",invdate:"2007-10-01",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"1",invdate:"2010-05-24",name:"test",note:"note",tax:"10.00",total:"2111.00"} , {id:"2",invdate:"2010-05-25",name:"test2",note:"note2",tax:"20.00",total:"320.00"}, {id:"3",invdate:"2007-09-01",name:"test3",note:"note3",tax:"30.00",total:"430.00"}, {id:"4",invdate:"2007-10-04",name:"test",note:"note",tax:"10.00",total:"210.00"}, {id:"5",invdate:"2007-10-05",name:"test2",note:"note2",tax:"20.00",total:"320.00"}, {id:"6",invdate:"2007-09-06",name:"test3",note:"note3",tax:"30.00",total:"430.00"}, {id:"7",invdate:"2007-10-04",name:"test",note:"note",tax:"10.00",total:"210.00"}, {id:"8",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"21.00",total:"320.00"}, {id:"9",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"11",invdate:"2007-10-01",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"12",invdate:"2007-10-02",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"13",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"14",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"15",invdate:"2007-10-05",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"16",invdate:"2007-09-06",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"17",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"18",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"19",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"21",invdate:"2007-10-01",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"22",invdate:"2007-10-02",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"23",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"24",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"25",invdate:"2007-10-05",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"26",invdate:"2007-09-06",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, {id:"27",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, {id:"28",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, {id:"29",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}  ]; 
var default_grid = {
            datatype: "local", 
            height: 80, 
            rowNum: 10, 
            rowList: [10,20,30], 
            multiselect: true, 
            rowList:[10,20,30], 
            sortname: 'id', viewrecords: true, sortorder: "asc",
            footerrow: true,
            loadui: "block"
};
var grid_detail = $.extend({ 
    data: grid_data, 
    colNames:['Inv No','Date', 'Client', 'Amount','Tax','Total','Notes'], 
    colModel:[ 
        {name:'id',index:'id', width:60, sorttype:"int"}, {name:'invdate',index:'invdate', width:90, sorttype:"date", formatter:"date"},
        {name:'name',index:'name', width:100}, {name:'amount',index:'amount', width:80, align:"right",sorttype:"float", formatter:"number"}, 
        {name:'tax',index:'tax', width:80, align:"right",sorttype:"float"}, {name:'total',index:'total', width:80,align:"right",sorttype:"float"},
        {name:'note',index:'note', width:150, sortable:false}
    ],
    caption: "Test Grid" ,
    pager: '#plist2'
},default_grid);

$('#abutton').click(function(e){
    e.preventDefault();
    $('<div></div>').dialog().jqGrid(grid_detail);
});

Answer 1

这根本不是一个神秘的信息。元素不是表格，这意味着您尝试变成网格的元素不是表格。这很有道理：你正在使用div：

$('<div></div>')

所以不要在javascript中完成所有操作，只需制作以下HTML：

<div id="myDialog"><table id="myGrid"></table></div>

然后制作对话框和网格：

$('#myDialog').dialog();
$('#myGrid').jqGrid ...

编辑：完全动态地执行此操作：

$('<div><table></table></div>').dialog().find('table:first').jqGrid(...);

jqGrid错误 - “元素不是表”

1 个答案: