My SQl Query Error

Question

question

MohamedRafiN-6400 asked Mar 22, '22 YijingSun-MSFT commented Mar 23, '22

My SQl Query Error

I have stored userid and password in mysql database, this query is showing login successfully for not stored data, so please correct it

string query = "select * from login where userid=@userid and password=@password";
using (MySqlCommand cmd = new MySqlCommand(query))
{
cmd.Parameters.AddWithValue("@userid", id);
cmd.Parameters.AddWithValue("@password", password);
cmd.Connection = cnn;
cnn.Open();
cmd.ExecuteNonQuery();
DialogResult dr = MessageBox.Show("Are you sure to Login now?", "Confirmation Message", MessageBoxButtons.YesNo);
if (dr == DialogResult.Yes)
{
MessageBox.Show("Login Successfully");
cnn.Close();
this.Hide();
Form2 f2 = new Form2();
f2.ShowDialog();

dotnet-csharp dotnet-aspnet-webforms

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Answer 1 · 2022-03-23T02:56:10.107Z

YijingSun-MSFT answered Mar 23, '22 YijingSun-MSFT commented Mar 23, '22

Hi @MohamedRafiN-6400 ,
I think your userid and password are unique. So I suggest you could count the same userid and the password. If the count is more than 1, you could login in.

        string strSql = "SELECT COUNT(*) from login where userid=@userid and password=@password";
       cmd.Connection = cnn;
       cnn.Open();
         int count = Convert.ToInt32(command.ExecuteScalar());
         if (count >= 1)
             MessageBox.Show("Login Successfully");
         else
              MessageBox.Show("Please register");

Best regards,
Yijing Sun

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· 2

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MohamedRafiN-6400 · Mar 23 at 05:01 AM

if(textBox9.Text!="" && textBox10.Text!= "")
{
string connectionString;
MySqlConnection cnn;
MySqlCommand cmd;
connectionString = @"Data Source=localhost;Initial Catalog=testDB;User ID=root;Password=mysql";
cnn = new MySqlConnection(connectionString);
string query = "select * from login where userid='" + textBox9.Text + "' and password='" + textBox10.Text + "'";
int count = Convert.ToInt32(cmd.ExecuteScalar());
if (count >= 1)
{
MessageBox.Show("Login Successfully");
cnn.Close();
this.Hide();
Form2 f2 = new Form2();
f2.ShowDialog();
}
else
{
MessageBox.Show("Please Enter Correct Login details or Register User");
}

         }
         else
         {
             MessageBox.Show("Please Enter Details to Login");
         }

}
Error shows like: Use of unassigned local variable 'cmd'

0 ·

YijingSun-MSFT MohamedRafiN-6400 · Mar 23 at 07:14 AM

Hi @MohamedRafiN-6400 ,
You need to add like this:

 cmd = new SqlCommand(query , cnn);

Best regards,
Yijing Sun

1 ·

Answer 2 · 2022-03-22T18:32:49.717Z

karenpayneoregon answered Mar 22, '22 MohamedRafiN-6400 commented Mar 23, '22

You need to use ExecuteReader rather than ExecuteNonQuery.

var results = cmd.ExecuteReader();

Then check results e.g. results.HasRows which returns true if a record was found, false otherwise.

· 1

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Attachments: Up to 10 attachments (including images) can be used with a maximum of 3.0 MiB each and 30.0 MiB total.

MohamedRafiN-6400 · Mar 23 at 05:41 AM

Error: the name cmd does not exist in current context

string query = "select count (*) from login where userid='" + textBox9.Text + "' and password='" + textBox10.Text + "'";
int count = Convert.ToInt32(cmd.ExecuteScalar());
using (MySqlCommand cmd = new MySqlCommand(query))
if (count >= 1)
{
cmd.Parameters.AddWithValue("@userid", id);
cmd.Parameters.AddWithValue("@password", password);
cmd.Connection = cnn;
cnn.Open();
cmd.ExecuteNonQuery();
MessageBox.Show("Login Successfully");
cnn.Close();
this.Hide();
Form2 f2 = new Form2();
f2.ShowDialog();
}
else
{
MessageBox.Show("Please Enter Correct Login details");
}

         }
         else
         {
             MessageBox.Show("Please Enter Details to Login");
         }

0 ·

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