我正在尝试使用YUI DataTable显示以下JSON对象。 我能够成功地在YUI DataTable中显示lastName,firstName,startDate,employeeCode,employeeStatus。 但是我无法显示内部对象中的值。 在列设置中,我尝试了user.userId,但在DataTable中显示为{value}。
[
{
"lastName": "MyLastName",
"firstName": "MyFirstName",
"startDate": "11-11-11",
"employeeCode": "124",
"employeeStatus": "Permanent",
"user": {
"key": {
"name": null,
"parent": {
"name": null,
"parent": null,
"id": 855,
"namespace": "",
"complete": true,
"kind": "Employee"
},
"id": 856,
"namespace": "",
"complete": true,
"kind": "Users"
},
"salt": null,
"userId": "myemail@example.com",
"status": true,
},
{
...
}
]
这里是 JavaScript 代码:
<script type="text/javascript">
YUI().use("jsonp", 'sortable', 'json-parse', 'datatable', "datatable-sort", "io", "node", function(Y) {
var nestedCols = [
{key : "employeeCode",label : "Employee Code", sortable:true},
{key : "firstName", label : "First Name",sortable: true},
{key : "lastName", label : "Last Name", sortable:true},
{key : "user.userId", label : "Email Id"},
];
Y.io('/Employee/AjaxList', {
on : {
success : function(tx, r) {
var data = Y.JSON.parse(r.responseText);
var table = new Y.DataTable.Base({
columnset : nestedCols,
recordset : data,
}).plug(Y.Plugin.DataTableSort);
table.render("#empTable");
}
}
});
});
</script>
这段代码有什么问题吗?我该如何在DataTable中显示user.userId的值?
注意:JSON是使用Jackson生成的,应用程序是在GAE/J中开发的。
更新:
我按照@Luke的建议使用了DataSource。这一次我只得到了带有标题的空DataTable。以下是代码片段。
YUI().use("datasource-get", "datatable-base", "datatable-datasource","datasource-arrayschema", function (Y) {
var url = "/Employee/AjaxList?";
var dataSource, table;
dataSource = new Y.DataSource.Get({ source: url });
dataSource.plug(Y.Plugin.DataSourceArraySchema, {
schema: {
resultFields: ["firstName", "lastName"]
}
});
var cols = ["firstName", "lastName"];
table = new Y.DataTable.Base({
columnset: cols,
});
table.plug(Y.Plugin.DataTableDataSource, { datasource: dataSource });
table.render("#empTable");
table.datasource.load();
});