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Sir, I have removed First given open connection now the error is coming like
MySql.Data.MySqlClient.MySqlException: 'Fatal error encountered during command execution.'
MySqlException: Parameter '@loginid' must be defined.
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Look at your query:
Quote:
select * from login where userid=@userid, password=@password, confirmpassword=@confirmpassword where loginid=@loginid is same
You have two where clauses, which is not valid. And as far as I can see, that is same on the end is also not valid. And putting a comma between conditions is also not valid - you need to use AND instead.
Aside from that, you have four parameters: @userid, @password, @confirmpassword, and @loginid.
Now look at the parameters you are passing to the command:
Quote:
cmd.Parameters.AddWithValue("@userid", id);
cmd.Parameters.AddWithValue("@password", password);
You are passing TWO parameters: @userid and @password.
Either fix you query to use the parameters you are passing, or fix your code to pass the parameters required by the query.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Sir, Now i have passed three parameters but it shows login successfull with wrong userid password also sir
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);
string id = textBox9.Text;
string password = textBox10.Text;
string loginid = "";
textBox9.Text = "";
textBox10.Text = "";
string query = "select * from login where userid=@userid and password=@password and loginid=@loginid";
using (MySqlCommand cmd = new MySqlCommand(query))
{
cmd.Parameters.AddWithValue("@userid", id);
cmd.Parameters.AddWithValue("@password", password);
cmd.Parameters.AddWithValue("@loginid", loginid);
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");
}
}
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Because as the other Richard said below, you are not checking the result of your query! 
And as I said above, you are storing passwords insecurely. If you're intending to use this code in a real application, then I hope you've got deep pockets, because you're going to get hit with a multi-million dollar fine PDQ.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Sir, how to save password securely? pls suggest in my code and mysql query also wrong bcz it login with wrong userid and password also
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string connectionString;
MySqlConnection cnn;
connectionString = @"Data Source=localhost;Initial Catalog=testDB;User ID=root;Password=mysql";
cnn = new MySqlConnection(connectionString);
string id = textBox9.Text;
string password = textBox10.Text;
string loginid = "";
textBox9.Text = "";
textBox10.Text = "";
string query = "select * from login where userid=@userid and password=@password and loginid=@loginid";
using (MySqlCommand cmd = new MySqlCommand(query))
{
cmd.Parameters.AddWithValue("@userid", id);
cmd.Parameters.AddWithValue("@password", password);
cmd.Parameters.AddWithValue("@loginid", loginid);
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");
}
}
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You really can't be bothered to pay attention, can you?
Programming is not about throwing some random code together from a couple of internet searches, and then pestering other people to fix it for you.
If you can't think for yourself, then you have chosen the wrong career.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Sir, I pay attention sir, you said i am passing 2parameters now i am passing 3parameters and login means it shows login successful with wrong userid password also i think my sql query should be change here
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You are clearly not paying attention, neither here nor in your class.
You have repeatedly been told that you need to check the results of your query. You have been told how to do that. And yet you continue to ask how to do what you have already been told how to do, and insist that you need to change your query rather than your code.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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How many times do I need to say it: Do not post a success message when you have not checked the result of your Database query.
This is the way you should do it:
cnn.Open();
int result = cmd.ExecuteNonQuery(); cnn.Close();
if (result == 1)
{
MessageBox.Show("Login was successful");
}
else
{
MessageBox.Show("The entered details were not correct");
}
Also, why do you need two ids (userid and loginid)? You should only need a single id and a password.
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How to change code in my coding like this sir?
because my code is logging in if i enter wrong username and password also
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Put the same code in my code that why i can understand easily sir
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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);
string id = textBox9.Text;
string password = textBox10.Text;
textBox9.Text = "";
textBox10.Text = "";
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();
}
else if (dr == DialogResult.No)
{
MessageBox.Show("Please Enter Correct Login details");
}
}
}
else
{
MessageBox.Show("Please Enter With Correct Login Details");
}
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There really is no point in continuing this thread. It does not matter how many times we tell you how to correct your code you insist on ignoring our advice. My previous reply showed you exactly how to do it correctly, and yet you just repeat the same bad code. I suggest you consider a different career path.
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Why you are not correcting my code and sending me? i said the error you know the code and sending some examples also then you can send correct code it is useful to me please
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Sir, How to write mysql query if stored userid password is correct means then it enter successfull suppose not stored values entered means the messagebox is wrong.....
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Why do you ignore everything we tell you? Go back to the code sample I gave you, it shows exactly how to do it correctly. This is my last message on this subject.
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Pls correct and send me in my code sir that why i can understand sir send me please
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How many times do you have to be told "nobody is going to do your work for you" before it sinks in?
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And you are still not checking the result of ExecuteNonQuery, but putting the success message based on the user pressing the "Yes" button.
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How to check it. Please give step procedure sir or video link
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I have already explained more than once. When you call cmd.ExecuteNonQuery(); you must check the return value, to see whether the SQL command succeeded. Only when you have a success return can you post a message that tells the user that his action has worked. You must do this on all database access commands, never assume that your code has worked - most of the time it has not.
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No sir mysql command was not succeeded, when i am giving wrong userid passwor dit shows login successfull i think sql query to be change
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