in database table is thier . in that table fields are
photoid , photoname,photo... i want display image in the
gridview
Answers were Sorted based on User's Feedback
Answer / .....
initially while inserting image into database take the
database field in oracle as BLOB in sql as image
during insertion convert the image file into bytes and
insert it into the database
during the display of image in datagrid take a image
template and display the converted values
sample code to get the image:
public byte[] spic(string cm,int id)
{
con = new OracleConnection(cm);
OracleCommand cmd = new OracleCommand("select
photo from kir_image where eid= "+id +" ", con);
con.Open();
byte[] img = (byte[])cmd.ExecuteScalar();
con.Close();
return img;
}
use Response.BinaryWrite to display the image
Is This Answer Correct ? | 3 Yes | 0 No |
Answer / nagaraj
pls post a mail to below mail id answer if any knows
Is This Answer Correct ? | 2 Yes | 0 No |
Answer / krishnasamy2008
Using C#(Asp.net)
-------------
<asp:TemplateColumn HeaderText ="Photo" >
<ItemTemplate>
<asp:image
</ItemTemplate>
<asp:Image ID="udphoto" ImageUrl='<%#Eval("Photo")%>'
runat="server" />
<HeaderStyle ForeColor ="blue" />
</asp:TemplateColumn>
Using Vb.Net(Asp.net)
-------------
<asp:TemplateColumn HeaderText ="Photo" >
<ItemTemplate>
<asp:image
</ItemTemplate>
<asp:Image ID="udphoto"
ImageUrl='<%#DataBinder.Eval(Container.DataItem,"Photo")%>'
runat="server" />
<HeaderStyle ForeColor ="blue" />
</asp:TemplateColumn>
Is This Answer Correct ? | 2 Yes | 0 No |
Answer / deepika
Use template column and put image as control in the
itemTemplate
Is This Answer Correct ? | 2 Yes | 1 No |
Answer / appu
<asp:GridView ID="GridView1" runat="server"
AutoGenerateColumns="False" DataKeyNames="ID"
DataSourceID="SqlDataSource1">
<Columns>
<asp:BoundField DataField="ID" HeaderText="ID"
InsertVisible="False" ReadOnly="True"
SortExpression="ID" />
<asp:BoundField DataField="ImageName" HeaderText="ImageName"
SortExpression="ImageName" />
<asp:TemplateField HeaderText="Image">
<ItemTemplate>
<asp:Image ID="Image1" runat="server"
ImageUrl='<%# "Handler.ashx?ID=" + Eval("ID")%>'/>
</ItemTemplate>
</asp:TemplateField>
</Columns>
</asp:GridView>
<asp:SqlDataSource ID="SqlDataSource1" runat="server"
ConnectionString="<%$ ConnectionStrings:ConnectionString %>"
SelectCommand="SELECT [ID], [ImageName], [Image]
FROM [Images]"></asp:SqlDataSource>
Handler.ashx
<%@ WebHandler Language="C#" Class="Handler" %>
using System;
using System.Web;
using System.Configuration;
using System.Data.SqlClient;
public class Handler : IHttpHandler {
public void ProcessRequest (HttpContext context)
{
SqlConnection con = new SqlConnection();
con.ConnectionString = ConfigurationManager.ConnectionStrings
["ConnectionString"].ConnectionString;
// Create SQL Command
SqlCommand cmd = new SqlCommand();
cmd.CommandText = "Select ImageName,Image from Images" +
" where ID =@ID";
cmd.CommandType = System.Data.CommandType.Text;
cmd.Connection = con;
SqlParameter ImageID = new SqlParameter
("@ID", System.Data.SqlDbType.Int);
ImageID.Value = context.Request.QueryString["ID"];
cmd.Parameters.Add(ImageID);
con.Open();
SqlDataReader dReader = cmd.ExecuteReader();
dReader.Read();
context.Response.BinaryWrite((byte[])dReader["Image"]);
dReader.Close();
con.Close();
}
Is This Answer Correct ? | 1 Yes | 0 No |
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