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george.dt wrote: thanks. Good answer
You have to explain my friend ..!Thts what he means..!Thts not the answer..!
Did You Check again with different cases..!
LatestArticle :Log4Net
Why Do Some People Forget To Mark as Answer .If It Helps.
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Go to
Internet Explorer >> Tools >> Internet Options >> Advanced
Just Un-Check the Enable third-party browser extensions.
You are done.
Next time please try to explain what exactly you want.
Cheers.
![Rose | [Rose]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/rose.gif)
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Below is the Sample
<asp:content></asp:content>
Actually i wrote
<asp:content></asp:content>
in code block..!
LatestArticle :Log4Net
Why Do Some People Forget To Mark as Answer .If It Helps.
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Check "Do not interpret HTML tags (good for code snippets)" 
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 sir,
please tell me that how we can call a dll from another dll when both are in same folder.
thanks
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How this is related with asp.net.
Add Dll as reference, Add namespaces. use the classes 
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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thank u sir
but my issue is still remaining.
actually this is my first dll
using System;
namespace myfirst
{
public class demo
{
public string getmessage()
{
return "Hi!How Are You";
}
}
}
in folder e:\prac
and more dll is
using System;
using myfirst;
namespace mysecond
{
public class demo1
{
public string getmessage1()
{
demo d=new demo();
string str1=d.getmessage();
Console.WriteLine(str1);
}
}
}
and now i am using csc /t:library mysecond.cs
then i have following error and problem was same.please help me
mysecond.cs(2,7): error CS0246: The type or namespace name 'myfirst' could not
be found (are you missing a using directive or an assembly reference?)
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This has NOTHING to do with ASP.NET. Ask it in the right forum.
Christian Graus
Driven to the arms of OSX by Vista.
Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.
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Hi Experts,
I want to pass the data from first.aspx page to Third.aspx.....however my sequence is first.aspx then on button click second.aspx and then navigates to Third.aspx....
I KNOW ONLY THE PREVIOUS PAGE PROPERTY...
Sharing Of Knowledge Provides Solution....
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You can pass values on the URL, or place them in the session. You can also make all your 'pages' controls on the one page and let the values persist in viewstate.
Christian Graus
Driven to the arms of OSX by Vista.
Read my blog to find out how I've worked around bugs in Microsoft tools and frameworks.
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Put in the querystring or HiddenField to pass data during submit. The choice depends on how you call the page (GET / POST)
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rhtbhegade wrote: however my sequence is first.aspx then on button click second.aspx and then navigates to Third.aspx....
As suggesed by others, You can store the data inside Session Variable and can access any where from your application.
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 i want to add stylesheet in my content page.
i written code like this
<br />
<asp:Content ID="Content2" ContentPlaceHolderID="ContentPlaceHolder1" runat="Server" ><br />
<link href="../App_Themes/Blue/DefStyles.css" rel="stylesheet" type="text/css" /><br />
</asp:Content><br />
if i switch design window below error is coming
"
Error 40
Cannot switch views: Validation (XHTML 1.0 Transitional):
Element 'link' cannot be nested within element 'td'.
"
how to solve this problem?
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this should be like this
<asp:content id="Content2" contentplaceholderid="ContentPlaceHolder1" runat="Server" >
<="" asp:content="">
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if i write like that this error coming
Error 39 Only Content controls are allowed directly in a content page that contains Content controls.
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Include your css file using ScriptManager
ScriptManager.RegisterClientScriptBlock(this,
typeof(nameOfPageOrControl),
"typeAUniqueScriptNameHere3",
"<link href='" +
ResolveUrl("~") +
"mystyle.css' rel='stylesheet' type='text/css' />",
false);
Remove the <link /> tag from your content PlaceHolder, then, it should work.
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below is the sample content page which consists of style sheet..!
one button and a textbox..!
<asp:Content ID="Content1" ContentPlaceHolderID="head" Runat="Server">
<link href="App_Themes/StyleSheet.css" rel="stylesheet" type="text/css" />
Close <asp:Content>
<asp:Content ID="Content2" ContentPlaceHolderID="ContentPlaceHolder1" Runat="Server">
<asp:Button ID="Button1" runat="server" Text="Button" /> <asp:TextBox ID="TextBox1" runat="server"></asp:TextBox>Close<asp:Content>
Close<asp:Content>--</asp:Content>
LatestArticle :Log4Net
Why Do Some People Forget To Mark as Answer .If It Helps.
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Friend thnx for ur replay.
Now run time this error coming
Cannot find ContentPlaceHolder 'head' in the master page '../common/MasterPage.master', verify content control's ContentPlaceHolderID attribute in the content page.
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pss.srinivasan wrote: Cannot find ContentPlaceHolder 'head' in the master page '../common/MasterPage.master', verify content control's ContentPlaceHolderID attribute in the content page.
What is the ContentPlaceHolder name in master Page ? You should have to mention the Same name over there.
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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After Searchin google i am getting different types of solution..!
Below is one..!
[^]
[^]
LatestArticle :Log4Net
Why Do Some People Forget To Mark as Answer .If It Helps.
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Create a ContentPlaceHolder in the master page between the <head> and </head> tags with the id 'head'
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Hi All,
<br />
<br />
public MySqlDataReader ExecuteReader(String query)<br />
{<br />
MySqlConnection con = new MySqlConnection((connectionString));<br />
<br />
MySqlCommand cmd = new MySqlCommand(query, con);<br />
<br />
con.Open();<br />
<br />
MySqlDataReader reader =cmd.ExecuteReader();<br />
con.Close();<br />
<br />
return reader;<br />
}<br />
<br />
I am returning reader but connction gets close so it become null it is in DAl (Data Access layer) i know it is expected beheviour so how to solve this problem.
thanks and regards,
Amit Patel
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Amit Patel1985 wrote: I am returning reader but connction gets close so it become null
A fairly basic issue, you must keep the connection open while you are using the reader object.
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You need to have active open connection to use DataReader. Use DataSet instead for disconnected approach. 
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