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I have created a web form with the login control. Now I want to validate the UserName & the Password with the data on a table in a SQL database (username & password as the fields of the table). I am new to ASP.Net & I need your help thanks.

I am using VS 2010 .net Framework 4.0 in win 7 professional 64 bit! SQl server 2008 R2 Developer
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Comments
[no name] 5-May-13 14:24pm    
Okay so connect to your database, write a query and execute it.
Chiranthaka Sampath 5-May-13 20:45pm    
Ok then what method of that login control to use?

You can have a look at How to implement forms authentication and Managing users with roles. There are functionalities available in Asp.NET for user authentication.

Possible method using database:

C#
SqlConnection conn = new SqlConnection();
conn.ConnectionString = System.Configuration.ConfigurationManager.ConnectionStrings["leave"].ConnectionString;
conn.Open();
string uname = TextBox1.Text;
string ps = Textbox2.Text;
SqlCommand cmd = new SqlCommand("Select uname, pswd from emp_details where uname =@uname and pswd =@ps", conn);
cmd.Parameters.Add(new SqlParameter("@uname", "id here"));
cmd.Parameters.Add(new SqlParameter("@ps", "password here"));
SqlDataReader dr = cmd.ExecuteReader();
if (dr.Read())
{
       /*Perform operations for authenticated users*/
}

-Additional note:

Its advisable to store password in encrypted format in database(Not as plain text). You can make use of algorithm. For example, have a look at Simple String Encryption and Decryption
 
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v5
Comments
Chiranthaka Sampath 6-May-13 3:21am    
Ok I have visited that and then I want to validate that login control's username and the password from a table that in a SQL database. Then how am I going to do that? Please help me!
Thanks7872 6-May-13 3:29am    
Refer to updated answer. You can accept that answer if it helps you.
Chiranthaka Sampath 6-May-13 4:01am    
In here you have develop the code using some sort of new textboxes and buttons but in my case I want to use the Login control of the tools of the VS 2010 and I need the help to add the database to this Login control. Remember that this as ASP.Net form ASP.Net uses the Login control!
Thanks7872 6-May-13 4:43am    
Refer to this link Asp.net Login Controls
try
{
SqlConnection con = new SqlConnection(ConfigurationManager.ConnectionStrings["SampleConnectionString"].ToString());
con.Open();
SqlCommand com = new SqlCommand("select password from Registration where username='" + Txtusername.Text + "'", con);
SqlDataReader dr = com.ExecuteReader();
if (!dr.Read())
{
labLogin.Text = ("Invalid User");
}
else
{
if (dr[0].ToString() == Txtpassword.Text)
{
Response.Redirect("Select.aspx");
}
else
{
labLogin.Text = ("Wrong Password");
Txtpassword.Focus();
}
}
}
catch (Exception ex)
{
labLogin.Text = "ex" + ex.Message;
}
 
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For Your Question comment
>Ok I have visited that and then I want to validate that login control's username and >the password from a table that in a SQL database. Then how am I going to do that? >Please help me!
I guess is this ur requirement.


C#
SqlConnection conn = new SqlConnection();
conn.ConnectionString = System.Configuration.ConfigurationManager.ConnectionStrings["connection"].ConnectionString;
conn.Open();
string username = TextBox1.Text;
string password = Textbox2.Text;
SqlCommand cmd = new SqlCommand("Select uname, pswd from emp_details where uname =@uname and pswd =@ps", conn);
cmd.Parameters.Add(new SqlParameter("@uname", "username"));
cmd.Parameters.Add(new SqlParameter("@ps", "password"));
SqlDataReader dr = cmd.ExecuteReader();
dr.Read();
if (dr.HasRows)
{
       dr.close();
       //Your Homepage or Ur need;
//Eg:// Response.Redirect("Home.aspx");
// As a beginner the above one is enough in future u have to show the user name in the further pages like  Hi Chiranthaka Sampath in the next pages u have to move with Session.
Session["username"] = txtusername.Text;
 Session["password"] = txtpassword.Text;
//My means of Session wherever u show the name u can get from the Session..

}
else
{
 dr.close();
msg.Showalert("Invalid username or password");
}
con.close();
}
 
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v5

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