TECHNOLOGIES
FORUMS
JOBS
BOOKS
EVENTS
INTERVIEWS
Live
MORE
LEARN
Training
CAREER
MEMBERS
VIDEOS
NEWS
BLOGS
Sign Up
Login
No unread comment.
View All Comments
No unread message.
View All Messages
No unread notification.
View All Notifications
Answers
Post
An Article
A Blog
A News
A Video
An EBook
An Interview Question
Ask Question
Forums
Monthly Leaders
Forum guidelines
Deepak Ratan
NA
175
50.6k
Getting an ERROR to display the photo in gridview
May 20 2015 6:39 AM
ERROR :
DataBinding: 'System.Data.DataRowView' does not contain a property with the name 'EmpID'.
hi i tried to display image from sql table to gridview but its not displaying this s my code to bind gridview with sql table records..
When the user login the Login user photo should display
DataBase: TBL_PBLogin
EmpID PK (1,1),
UName varchar(50),
Password varchar(50),
photo image
PostBookChat.aspx
<body>
<form id="form1" runat="server">
<div>
<div id="header">
<img src="Image/book.png" height="60" width="140" style ="margin-left:0px;float:left;"/>
<div id="login">
<b><asp:Label ID="lblsession" runat="server" ForeColor="white" CssClass="label"></asp:Label>
<asp:GridView ID="gridviewphoto" runat="server" AutoGenerateColumns="false" BackColor="#CC3300" ForeColor="Black" ShowHeader="false" GridLines="None">
<Columns>
<asp:TemplateField ControlStyle-Width="100" ControlStyle-Height="100">
<ItemTemplate>
<img src="Handler.ashx?EmpID=<%#Eval("EmpID").ToString()%>" width="50" height="50"/>
</ItemTemplate>
</asp:TemplateField>
</Columns>
</asp:GridView>
<asp:Button ID="btnlogout" runat="server" Text="Sign Out" CssClass="myButton" OnClick="btnlogout_Click"/>
</div>
</div>
PostBookChat.aspx.cs
protected void Page_Load(object sender, EventArgs e)
{
if (!IsPostBack)
{
lblsession.Text = "Welcome" + Convert.ToString(Session["UName"]);
sessionimage();
}
}
private void sessionimage()
{
SqlConnection con = new SqlConnection(ConfigurationManager.ConnectionStrings["connect"].ConnectionString);
try
{
int EmpID = (int)Session["EmpID"];
SqlDataAdapter Adp = new SqlDataAdapter("select Photo from TBL_PBLogin where EmpID='" + EmpID + "'", con);
DataTable Dt = new DataTable();
con.Open();
Adp.Fill(Dt);
gridviewphoto.DataSource = Dt;
gridviewphoto.DataBind();
con.Close();
}
catch (Exception ex)
{
throw new Exception(ex.ToString());
}
}
Handler.ashx
<%@ WebHandler Language="C#" Class="Handler" %>
using System;
using System.Web;
using System.Data.SqlClient;
using System.Data;
using System.IO;
using System.Collections.Specialized;
using System.Configuration;
public class Handler : IHttpHandler
{
public void ProcessRequest(HttpContext context)
{
if (context.Request.QueryString["EmpID"] == null) return;
string connStr = ConfigurationManager.AppSettings["connect"].ToString();
string pictureId = context.Request.QueryString["EmpID"];
using (SqlConnection conn = new SqlConnection(connStr))
{
using (SqlCommand cmd = new SqlCommand("SELECT Photo FROm TBL_PBLogin WHERE EmpID = @EmpId", conn))
{
cmd.Parameters.Add(new SqlParameter("@EmpID", pictureId));
conn.Open();
using (SqlDataReader reader = cmd.ExecuteReader(CommandBehavior.CloseConnection))
{
reader.Read();
context.Response.BinaryWrite((Byte[])reader[reader.GetOrdinal("Photo")]);
reader.Close();
}
}
}
}
public bool IsReusable
{
get
{
return false;
}
}
}
Reply
Answers (
4
)
How to add column in Datatable
Take multiple dynamic checkbox value.