JqG​​rid使用jquery和jsonresult [英] JqGrid using jquery and jsonresult

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本文介绍了JqG​​rid使用jquery和jsonresult的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

iam使用jqgrid使用json结果但是网格没有填充plz帮助我

这里是我的代码

控制器:

  public  JsonResult GetEmployeeData()
{
List< models.empdetails> emp = new List< models.empdetails>();
con.Open();
string query = 从EmpDetail中选择*其中StatusTypeId = EmpId DESC的1个订单;
SqlCommand cmd = new SqlCommand(query,con);
SqlDataAdapter da = new SqlDataAdapter(cmd);
cmd.CommandType = CommandType.Text;
DataSet ds = new DataSet();
da.Fill(ds);
foreach (DataRow dr in ds.Tables [ 0 ]。行)
{
emp.Add( new Models.EmpDetails()
{
Name = dr [ 0 ]。ToString(),
EmpId = Convert.ToInt32(dr [ 1 ]。ToString()),
Salary = Convert.ToInt32(dr [ 2 ]。ToString()),
地址= dr [ 3 ]。ToString(),
ZipCode = Convert.ToInt32(dr [ 4 ]。ToString()),
PhoneNumber = Convert.ToInt32(dr [ 5 ]。ToString()),
Permanent = Boolean .Parse(dr [ 7 ]。ToString() ),
Birthday = Convert.ToDateTime(dr [ 8 ]。ToString()),
Country = dr [ 9 ]。ToString()
});
}
cmd.ExecuteNonQuery();
return Json(emp,JsonRequestBehavior.AllowGet);
}



jquery:

 $(document).ready(function(){
$(' #grd')。jqGrid(
{
dataType:' json'
height: 250
mtype:' 获取'
url: ' / Home / GetEmployeeData'
colNames:[' 名称'' 地址'' 薪水'' PhoneNumber'' Zipcode'' Permanent'' 生日'' 国家'],
colModel:[{name:' 名称',索引: ' 名称',宽度: 80 },
{name:' 地址',索引:' 地址',宽度: 100 },
{name: ' 薪水',索引:' 薪水',width: 60 },
{name:' PhoneNumber',index:' PhoneNumber',width: 75 },
{name:' ZipCode',index:' ZipCode',width: 75 },
{name:' Permanent',index:' Permanent',width: 60 },
{name:' Birthday',index:' 生日',宽度: 75 },
{name:'
Country',index:' Country',width: 100 }

],
cation:' EmplyoeeDetails'
pager:$(' #pager'),
sortdesc:' desc'
viewrecords: true

});
});

解决方案

(document).ready(function(){


< blockquote>(' #grd')。jqGrid(
{
dataType: ' json'
height: 250
mtype:' 获取'
url:' / Home / GetEmployeeData'
colNames:[ ' 名称'' 地址'' 薪水'' PhoneNumber'' Zi pcode'' Permanent'' 生日'' 国家'],
colModel:[{name:' 名称',索引:' 名称',宽度: 80 },
{名称:' 地址',索引: ' 地址',宽度: 100 },
{name:' 薪水',索引:' 薪水',宽度: 60 },
{n ame:' PhoneNumber',index:' PhoneNumber',width: 75 },
{name:' ZipCode',index:' ZipCode',width: 75 },
{name:' 永久',索引:' Permanent',width: 60 },
{name:' Birthday',index:' 生日',宽度: 75 },
{name:' Country',index:' Country',width: 100 }

],
cation:' EmplyoeeDetails'
pager:


' #pager'),
sortdesc:' desc'
viewrecords: true

});
});


Hi, iam using jqgrid using json result but grid is not populating plz help me
here is my code
for controller:

public JsonResult GetEmployeeData()
        {
            List<models.empdetails> emp = new List<models.empdetails>();
            con.Open();
            string query = "select * from EmpDetail where StatusTypeId=1  order by EmpId DESC";
            SqlCommand cmd = new SqlCommand(query, con);
            SqlDataAdapter da = new SqlDataAdapter(cmd);
            cmd.CommandType = CommandType.Text;
            DataSet ds = new DataSet();
            da.Fill(ds);
            foreach (DataRow dr in ds.Tables[0].Rows)
            {
                emp.Add(new Models.EmpDetails()
                {
                    Name = dr[0].ToString(),
                    EmpId = Convert.ToInt32(dr[1].ToString()),
                    Salary = Convert.ToInt32(dr[2].ToString()),
                    Address = dr[3].ToString(),
                    ZipCode = Convert.ToInt32(dr[4].ToString()),
                    PhoneNumber = Convert.ToInt32(dr[5].ToString()),
                    Permanent = Boolean.Parse(dr[7].ToString()),
                    Birthday = Convert.ToDateTime(dr[8].ToString()),
                    Country = dr[9].ToString()
                });
            }
            cmd.ExecuteNonQuery();
            return Json(emp, JsonRequestBehavior.AllowGet);
        }


jquery:

$(document).ready(function () {
        $('#grd').jqGrid(
        {
            dataType: 'json',
            height: 250,
            mtype: 'Get',
            url: '/Home/GetEmployeeData',
            colNames: ['Name', 'Address', 'Salary', 'PhoneNumber', 'Zipcode', 'Permanent', 'Birthday', 'Country'],
            colModel: [{ name: 'Name', index: 'Name', width: 80 },
                       { name: 'Address', index: 'Address', width: 100 },
                       { name: 'Salary', index: 'Salary', width: 60 },
                       { name: 'PhoneNumber', index: 'PhoneNumber', width: 75 },
                       { name: 'ZipCode', index: 'ZipCode', width: 75 },
                       { name: 'Permanent', index: 'Permanent', width: 60 },
                       { name: 'Birthday', index: 'Birthday', width: 75 },
                       { name: 'Country', index: 'Country', width: 100 }

            ],
            cation: 'EmplyoeeDetails',
            pager: $('#pager'),
            sortdesc: 'desc',
            viewrecords: true

        });
    });

解决方案

(document).ready(function () {


('#grd').jqGrid( { dataType: 'json', height: 250, mtype: 'Get', url: '/Home/GetEmployeeData', colNames: ['Name', 'Address', 'Salary', 'PhoneNumber', 'Zipcode', 'Permanent', 'Birthday', 'Country'], colModel: [{ name: 'Name', index: 'Name', width: 80 }, { name: 'Address', index: 'Address', width: 100 }, { name: 'Salary', index: 'Salary', width: 60 }, { name: 'PhoneNumber', index: 'PhoneNumber', width: 75 }, { name: 'ZipCode', index: 'ZipCode', width: 75 }, { name: 'Permanent', index: 'Permanent', width: 60 }, { name: 'Birthday', index: 'Birthday', width: 75 }, { name: 'Country', index: 'Country', width: 100 } ], cation: 'EmplyoeeDetails', pager:


('#pager'), sortdesc: 'desc', viewrecords: true }); });


这篇关于JqG​​rid使用jquery和jsonresult的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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