我正在尝试使用json在extjs中创建gridview。出于某种原因,我的gridview没有显示任何数据。我试着用firebug调试它。我可以在“响应”部分看到结果。 这就是我在“Reponse”中所拥有的。
{“ContentEncoding”:null,“ContentType”:null,“Data”:“{\ r \ n \”myTable \“:[\ r \ n {\ r \ n \”\ Q1 \“:\” 1 \“,\ r \ n \”Q2 \“:\”1 \“,\ r \ n \”Q3 \“:\”1 \“,\ r \ n \”Q4 \“:\”1 \ “,\ r \ n \”改进\“:\”\“,\ r \ n \”评论\“:\”1 \“\ r \ n},\ r \ n {\ r \ n \”Q1 \“:\”1 \“,\ r \ n \”Q2 \“:\”2 \“,\ r \ n \”Q3 \“:\”3 \“,\ r \ n \”Q4 \“ :\“4 \”,\ r \ n \“改进\”:\“Iphone5 \”,\ r \ n \“评论\”:\“Iphone14 \”\ r \ n},\ r \ n \ n \ r \ n \“Q1 \”:\“1 \”,\ r \ n \“Q2 \”:\“1 \”,\ r \ n \“Q3 \”:\“3 \”,\ r \ n \“Q4 \”:\“3 \”,\ r \ n \“改进\”:\“这是Comment1-3 \”,\ r \ n \“评论\”:\“这是评论2-3 \“\ r \ n} \ r \ n] \ r \ n}”,“JsonRequestBehavior”:0}
的更新 的 其实我现在在Json看到这个,但我的gridview仍然是空的 Please click here to see my JSON
/GridViewApp.js
Ext.define('GridViewApp.view.GridViewApp', {
alias: 'widget.gridviewapp',
width: 800,
title: 'My Grid Panel',
grid: null,
store: null,
layout: {
type: 'anchor'
},
constructor: function () {
this.callParent(arguments);
var store = Ext.create('Ext.data.Store', {
storeId: 'myData',
scope: this,
fields: [
{ name: 'Q1', type: 'int' },
{ name: 'Q2', type: 'int' },
{ name: 'Q3', type: 'int' },
{ name: 'Q4', type: 'int' },
{ name: 'Q5', type: 'int' },
{ name: 'Improvements', type: 'string' },
{ name: 'Comments', type: 'string' }
],
sorters: [
{
//property: 'myData',
direct: 'ASC'
}
],
proxy: {
type: 'ajax',
scope: this,
url: 'GridView/writeRecord',
reader: {
type: 'json',
root: 'myTable',
idProperty: 'ID'
}
}
});
store.load();
this.grid = Ext.create('Ext.grid.Panel', {
title: 'GridView App',
store: this.store,
columns: [
{header: 'Q1', width: 100,
sortable: true, dataIndex: 'Q1'
},
{ header: 'Q2', width: 100,
sortable: true, dataIndex: 'Q2'
},
{ header: 'Q3', width: 100,
sortable: true, dataIndex: 'Q3'
},
{ header: 'Q4', width: 100,
sortable: true, dataIndex: 'Q4'
},
{ header: 'Improvements', width: 200,
sortable: true, dataIndex: 'Improvements'
},
{ header: 'Comments', width: 200,
sortable: true, dataIndex: 'Comments'
}
],
stripeRows: true,
width: 800,
renderTo: Ext.getBody()
});
this.add(this.grid);
}
});
和/GridViewController.cs
namespace GridViewApp.Controllers
{
public class GridViewController : Controller
{
public ActionResult Index()
{
return View();
}
public JsonResult writeRecord()
{
SqlConnection conn = DBTools.GetDBConnection("ApplicationServices2");
string sqlquery = "SELECT Q1, Q2, Q3, Q4, Improvements, Comments FROM myTable";
SqlDataAdapter cmd = new SqlDataAdapter(sqlquery, conn);
DataSet myData = new DataSet();
cmd.Fill(myData, "myTable");
conn.Open();
conn.Close();
string myData1 = JsonConvert.SerializeObject(myData, Formatting.Indented,
new JsonSerializerSettings
{
ReferenceLoopHandling = ReferenceLoopHandling.Ignore
});
return Json(new { data = myData1 }, JsonRequestBehavior.AllowGet);
}
}
}
任何形式的输入都将是一个很大的帮助......谢谢
答案 0 :(得分:0)
它返回一个字符串是非常合乎逻辑的。
您使用json.net序列化为字符串。然后用Json()序列化该字符串。您应该只使用json.net序列化。
public string writeRecord()
{
SqlConnection conn = DBTools.GetDBConnection("ApplicationServices2");
string sqlquery = "SELECT Q1, Q2, Q3, Q4, Improvements, Comments FROM myTable";
SqlDataAdapter cmd = new SqlDataAdapter(sqlquery, conn);
DataSet myData = new DataSet();
cmd.Fill(myData, "myTable");
conn.Open();
conn.Close();
return myData1 = JsonConvert.SerializeObject(myData, Formatting.Indented,
new JsonSerializerSettings
{
ReferenceLoopHandling = ReferenceLoopHandling.Ignore
});
}