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This is not like that, that you are thinking.
You have to use AJAX for that. Beause you dont know how much time will it take for load the report.
Use AJAX UpdatePanel and Update ProgressBar.
These will resolve your problem.
For your Ref.
The UpdatePanel and UpdateProgress Controls[^]
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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Hi Abhijit,
Thanks for your reply..Actually i am using VS 2003,So any javascripts can be used?
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On Page Init Event On your Server side Code You Can Do Like
txtSearchSelect.Attributes.Add("onClick", "if (this.value == 'Search Keyword') this.value = 'Please wait Report is being Generated';");
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Hi Dear, I've a serious problem, that is:
I've three asp:Contents in my content page. Content1 for head, Content2 is activated when user is logged in; By contrast, Content3 is for anonymous user.
In my program asp:UpdatePanel works properly in both content(Content2 & Content3). asp:UpdateProgress works properly in Content3(used for anonymous user) but doesn't work for Content2(used for logged in user)
<asp:Content ID="Content1" ContentPlaceHolderID="head" runat="server">
</asp:Content>
<asp:Content ID="Content2" ContentPlaceHolderID="ContentPlaceHolder1" runat="server">
<asp:UpdatePanel ID="UpdatePanel2" runat="server">
<ContentTemplate>
<%=DateTime.Now.ToLongTimeString() %>
<asp:Button ID="Button1" runat="server" Text="Button" OnClick="UpdateButton_Click"/>
</ContentTemplate>
</asp:UpdatePanel>
<asp:UpdateProgress ID="UpdateProgress1" runat="server" AssociatedUpdatePanelID="UpdatePanel2">
<ProgressTemplate>
<img alt="Loading..." src="../Images/ajax-loader.gif" />
Please wait...
</ProgressTemplate>
</asp:UpdateProgress>
</asp:Content>
<asp:Content ID="Content3" ContentPlaceHolderID="ContentPlaceHolder2" runat="server">
<asp:UpdatePanel ID="UpdatePanel1" runat="server">
<ContentTemplate>
<%=DateTime.Now.ToLongTimeString() %>
<asp:Button ID="UpdateButton" runat="server" Text="Button" OnClick="UpdateButton_Click" />
</ContentTemplate>
</asp:UpdatePanel>
<asp:UpdateProgress ID="UpdateProgress2" runat="server" AssociatedUpdatePanelID="UpdatePanel1">
<ProgressTemplate>
<img alt="Loading..." src="../Images/ajax-loader.gif" />
Please wait...
</ProgressTemplate>
</asp:UpdateProgress>
</asp:Content>
I don't understand which is my fault. Could you help me?
shahdat
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Plz avoid double posting.
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OK boss, I am highly worried about my problem. But I don't get any answer yet within 9 hours ago.
Plz help me. If it's not beyond your capacity.
shahdat
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Cross Post will not help you to get answer faster.
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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Yes I will remain later.
Plz answer me
shahdat
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I have answered. Please check.
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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I am answering you over here, because, It is better readable than the last one.
MD. SHAHDAT HOSAIN wrote: but doesn't work for Content2(used for logged in user)
How did you think, its not working ? Are you getting any specific error ? Yor code seems like good. So, give us little bit details on your problem/error .
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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Thanks a lot
No specific error show from compiler. When I click update button, page is partially updated. But progressing message (with Image) never show for a logged in user. Interest that their is no problem for anonymous user.
I tall you again
In my program the Content2 is activated only for logged in user. Is it problem?
shahdat
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i m trying to display the Images in a grid view with refrence to the following article
it is the article if Abhijit Jana on code project
but i m getting the following errors
error: Both DataSource and DataSourceID are defined on 'GridView1'. Remove one definition.
in the following lines
GridView1.DataSource = ds; // give data to gridview
GridView1.DataBind();
can anyone help me ??
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swtlibra wrote: it is the article if Abhijit Jana on code project
Yes I am on Code Project . UP and Running 
swtlibra wrote: error: Both DataSource and DataSourceID are defined on 'GridView1'. Remove one definition.
in the following lines
GridView1.DataSource = ds; // give data to gridview
GridView1.DataBind();
Because, I guess, You are using Two Data Source for GridView1. One From Code Behind. and Another is on ASPX PAge.
As per my article goes, Displaying Image in Gridview from Database[^] There is only one datasource that is created only on CodeBehind.
Remove DataSourceID of GridView From aspx page.
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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Abhijit Jana wrote: Yes I am on Code Project . UP and Running Laugh
swtlibra found you.lol
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![Jig | [Dance]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/jig.gif)
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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swtlibra wrote: Both DataSource and DataSourceID are defined on 'GridView1'. Remove one definition.
Error is self describing.look at that. 
I think you defined DataSourceID in the html.Check your html for GridView and remove that.
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when i remove the datasource ID from gridview
it dosen't display the whole grid view 
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swtlibra wrote: it dosen't display the whole grid view
Show us the code of GriView from aspxpage. And DataBind code from Codebehind page.
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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<br />
<br />
<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False" <br />
DataKeyNames="Image_ID" <br />
AllowPaging="True" onrowdatabound="GridView1_RowDataBound" <br />
onselectedindexchanged="GridView1_SelectedIndexChanged"><br />
<Columns><br />
<asp:BoundField DataField="Image_ID" HeaderText="Image_ID" <br />
InsertVisible="False" ReadOnly="True" SortExpression="Image_ID" /><br />
<br />
<asp:TemplateField><br />
<ItemTemplate><br />
<asp:Image ID="img_1" ImageUrl='<%# Bind("Image_1") %>' runat="server" /><br />
</ItemTemplate><br />
</asp:TemplateField><br />
<br />
<asp:ImageField DataImageUrlField="Image_1"><br />
</asp:ImageField><br />
<asp:ImageField DataImageUrlField="Image_2"><br />
</asp:ImageField><br />
<asp:ImageField DataImageUrlField="Image_3"><br />
</asp:ImageField><br />
</Columns><br />
</asp:GridView><br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
using System;<br />
using System.Collections;<br />
using System.Configuration;<br />
using System.Data;<br />
using System.Linq;<br />
using System.Web;<br />
using System.Web.Security;<br />
using System.Web.UI;<br />
using System.Web.UI.HtmlControls;<br />
using System.Web.UI.WebControls;<br />
using System.Web.UI.WebControls.WebParts;<br />
using System.Xml.Linq;<br />
using System.Data.OleDb;<br />
using System.IO;<br />
<br />
public partial class samplepage : System.Web.UI.Page<br />
{<br />
OleDbConnection con;<br />
OleDbCommand cmd;<br />
OleDbDataReader rd;<br />
OleDbDataAdapter da;<br />
DataSet ds;<br />
protected void Page_Load(object sender, EventArgs e)<br />
{<br />
con = new OleDbConnection("Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" + HttpContext.Current.Server.MapPath("App_Data/testdb.mdb"));<br />
Load_GridData();<br />
}<br />
<br />
void Load_GridData()<br />
{<br />
con.Open();
OleDbDataAdapter da = new OleDbDataAdapter("select Image_ID, Image_1, Image_2, Image_3 from Images", con);<br />
DataSet ds=new DataSet();<br />
da.Fill(ds);
GridView1.DataSource = ds;
GridView1.DataBind();<br />
con.Close();<br />
}<br />
protected void GridView1_SelectedIndexChanged(object sender, EventArgs e)<br />
{<br />
<br />
}<br />
protected void GridView1_RowDataBound(object sender, GridViewRowEventArgs e)<br />
{<br />
if (e.Row.RowType == DataControlRowType.DataRow)<br />
{<br />
((Image)e.Row.FindControl("img_1")).ImageUrl = "'@~/Upload/'" ;<br />
<br />
}<br />
}<br />
<br />
<br />
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put a breakpoint at
GridView1.DataSource = ds;
and check if ds is filling correctly with data.
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ok if we use a simple code then how the image will be displayed using code
through code simple data is displaying in the grid
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swtlibra wrote: if we use a simple code then how the image will be displayed using code
Give the path of the image as relative url and check if that images existing in the solution.
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where to give the path
plz tell about the code
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Instead of
((Image)e.Row.FindControl("img_1")).ImageUrl = "'@~/Upload/'" ;
give something like,
((Image)e.Row.FindControl("img_1")).ImageUrl = @"~/Upload/imagename.jpg" ;
where imagename is the path from object source.
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swtlibra wrote: where to give the path
Image path should be the database column . And physically your image storage location would be hard drive.
When you will retrieve the path from database, image will be load from drive.
The Article, which you have mention above, clearly explained each and every steps.
Abhijit Jana | Codeproject MVP
Web Site : abhijitjana.net
Don't forget to click "Good Answer" on the post(s) that helped you.
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