Mohamed Rafi

Mohamed Rafi

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Login Details Code Error

Mar 22 2022 2:36 AM

Sir, I have studied and tried to put login page but it shows code error please tell me the correction sir
It shows error: System.InvalidOperationException: 'The connection is already and also code error

private void button14_Click(object sender, EventArgs e)
{
if (textBox9.Text != "" && textBox10.Text != "")
{
string connectionString;
MySqlConnection cnn;
connectionString = @"Data Source=localhost;Initial Catalog=testDB;User ID=root;Password=mysql";
cnn = new MySqlConnection(connectionString);
cnn.Open();
string id = textBox9.Text;
string password = textBox10.Text;
textBox9.Text = "";
textBox10.Text = "";
string query = "select * from login where userid=@userid,password=@password,confirmpassword=@confirmpassword where loginid=@loginid is same";
//string query = "update employee set employee_name=@employee_name,employee_salary=@employee_salary where employee_id=@employee_id";
using (MySqlCommand cmd = new MySqlCommand(query))
{
cmd.Parameters.AddWithValue("@userid", id);
//cmd.Parameters.AddWithValue("@employee_id", Convert.ToInt32(id));
cmd.Parameters.AddWithValue("@password", password);
//cmd.Parameters.AddWithValue("@confirmpassword", confirmpassword);
cmd.Connection = cnn;
cnn.Open();
cmd.ExecuteNonQuery();
DialogResult dr = MessageBox.Show("Are you sure to Login now?", "Confirmation", MessageBoxButtons.YesNo);
if (dr == DialogResult.Yes)
{
MessageBox.Show("Login Successfully");
cnn.Close();
this.Hide();
Form2 f2 = new Form2();
f2.ShowDialog();
}
else if (dr == DialogResult.No)
{
MessageBox.Show("Please Enter Correct Login details");
}
}
}
else
{
MessageBox.Show("Please Enter details to Login");
}
}
}

 


Answers (8)