這不是一張HTML表格,但我使用jqGrid以及Craig Stuntz的helper functions將任何IQueryable<T>
「導出」爲JSON。助手功能ToJqGridData
發送JSON恰好由jqGrid的要求的格式,所以你的代碼是好的,整齊:
MyObjectRepository rep = new MyObjectRepository();
var myObjects = from o in rep.SelectAll()
select new
{
Prop1 = o.Prop1,
Prop2 = o.Prop2
...
}
return Json(apps.ToJqGridData(page, rows, sidx, null, null), JsonRequestBehavior.AllowGet);
要知道,你還需要更新您jqGrids全局設置,以使它們與兼容通過ToJqGridData
使用的命名約定(我只是包括在我的母版頁這個腳本):
$(document).ready(function() {
GridDemo.SiteMaster.setDefaults();
});
var GridDemo = {
Home: {
GridDemo: {}
},
SiteMaster: {
setDefaults: function() {
$.jgrid.defaults = $.extend($.jgrid.defaults, {
datatype: 'json',
height: 'auto',
imgpath: '/Scripts/jqGrid/themes/lightness/images',
jsonReader: {
root: "Rows",
page: "Page",
total: "Total",
records: "Records",
repeatitems: false,
userdata: "UserData",
id: "Id"
},
loadui: "block",
mtype: 'GET',
multiboxonly: true,
rowNum: 20,
rowList: [10, 20, 50],
viewrecords: true
});
}
}
};
謝謝扎克瑞,這真是太棒了! – DaveDev 2010-04-02 09:39:12