c# database login

raluca mindrut

Hi, I made this login form and even if the username and pass are correct, my program still show "NO". Can you tell me where I do wrong?
here is the code

I am getting this error: "data type mismatch in criteria expression" Can you tell me why?

here's the code

try

{

string constring = @"Provider = Microsoft.ACE.OLEDB.12.0; Data source= bla.accdb;Persist Security Info=True;";

OleDbConnection conexiune = new OleDbConnection(constring);

OleDbCommand cmdDataBase = new OleDbCommand("Select * from Table1 where username =' " + this.textBox1.Text+"' and pass ='"+this.textBox2.Text+"';", conexiune);

OleDbDataReader myReader;

conexiune.Open();

myReader = cmdDataBase.ExecuteReader();

int count = 0;

while (myReader.Read())

{ count = count + 1; }

if (count == 1)

{

MessageBox.Show("FS");

}

else

{

MessageBox.Show("no");

}

catch (Exception ex)

{ MessageBox.Show(ex.Message); }

}

2 Replies

Know the answer? Post it — somebody with the same question will find it here.

Munesh Sharma12y ago
private void login_txt_Click(object sender, EventArgs e)
{
try
{
string str = "server = MUNESH\\SQL2008R2;Database=datastore;UID=sa;Password=123;";
SqlConnection con = new SqlConnection(str);
SqlCommand cmd = new SqlCommand("Select * from login where username = '" + this.textBox1.Text + "' and password = '" + this.textBox2.Text + "'", con);
SqlDataReader myreader;
con.Open();
myreader = cmd.ExecuteReader();
int count = 0;
while (myreader.Read())
{
count = count + 1;
}

if (count == 1)
{
MessageBox.Show("Username and password correct");
}

else if (count > 1)
{
MessageBox.Show("Duplicate username and password");
}
else
MessageBox.Show(" username and password incorrect");

con.Close();
}
catch (Exception es)
{
MessageBox.Show(es.Message);

}
}
Vulpes12y ago
I don't understand how you can be getting a 'data type mismatch' error (presumably when you set myReader) but the program is still managing to continue and display "no" in the MessageBox.

Also, it's difficult to see how the username and password columns can be anything other than strings in the database which is the usual reason for this error.