我使用以下方法将数据加载到jqgrid中。我无法将json数据加载到jqgrid中。因此,我将json解析为数组,如mydata=json.parse(jsondata)。然后使用datatype: "local"将此数组(mydata)绑定到jqgrid上。我的问题是如何将json数据绑定到jqgrid?
$("#datagrid").jqGrid({
datatype: "local",
data: mydata,
colNames:['companyid','company', 'price', 'Change','perchange','LastUpdated'],
colModel:[
{name:'companyid',index:'companyid', width:100,editable:true,editoptions:{size:10}},
{name:'company',index:'company', width:100,editable:true},
{name:'price',index:'price', width:100,editable:true,editoptions:{size:10}},
{name:'Change',index:'Change', width:100,editable:true,editoptions:{size:25}},
{name:'perchange',index:'perchange', width:100, align:"right",editable:true,editoptions:{size:10}},
{name:'LastUpdated',index:'LastUpdated', width:200, align:"right",editable:true,editoptions:{size:10}}
],
rowNum:10,
rowList:[3,6],
loadonce: true,
pager: '#navGrid',
sortname: 'companyid',
sortorder: "asc",
height: 210,
width:600,
onSelectRow: function(id)
{
getID = jQuery("#datagrid").jqGrid('getCell', id, 'companyid')
},
viewrecords: true,
caption:"JQ GRID"
});
JSON格式:
[
{
"company": "test",
"price": 98,
"Change": 8,
"perchange": 8,
"LastUpdated": "2",
"companyid": 2
},
{
"company": "test123",
"price": 1,
"Change": 1,
"perchange": 1,
"LastUpdated": "1",
"companyid": 3
},
{
"company": "abc",
"price": 1234,
"Change": 123,
"perchange": 1,
"LastUpdated": "1",
"companyid": 1
}
]