value validation problem

sachi vasishta

Hi,

I have a winform which contains three text boxes and a button. The textboxes are username,password and confirm password. I have entered the values for the three text boxes and clicked the ok button but it is not properly validating

under button1_click method I have the following code

string uname=textBox1.Text.ToString();

string pass=textBox2.Text.Tostring();

string confirm=textBox3.Text.ToString();

if(pass!=confirm)

MessageBox.Show("Please retype password");

here if I input different values for pass and confirm it is just displaying the message in messagebox but not prompting me back to the textBox3 and also that diffrent values are being inserted into the database. How to avoid this??

9 Replies

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Vimal Kandasamy17y ago

Accepted answer

Hi Sachi,

Try the following
add the following code in MouseDoubleClick event of datagridview..

private void dataGridView1_MouseDoubleClick(object sender, MouseEventArgs e)

            DataGridViewRow dr=dataGridView1.CurrentRow;
              
             //copy 1st cell of row into string var "user"

            String user = dr.Cells[0].Value.ToString();
            Form2 f=new Form2(user);

            f.show();

        }
when a user double click a row... its just retriew its 1st cell [username] and send it to form2 via constructor

add the following code in form2

// its a constructor
public form2(String usr)
{
InitializeComponent();
textBox1.Text=usr;
}

this will store passed value into textbox..

try it..

Vimal Kandasamy17y ago

Hi,

try this,

            this.Dispose();

            Form1 f = new Form1();

            f.Show();

sachi vasishta17y ago
Hi,

Vimal I have a winform namely form1 which contains two radio buttons gotoform2 and gotoform3. Now if I click on the gotoform2 radiobutton then it should close the form1 and open the new winform form2.In form2 I have a button called `back` whick on clicking should reopen the winform form1. In button1_click method of form2 I have written the following code

Form1.ActiveForm.Show();

but if I click on `back` button no action is performed.
sachi vasishta17y ago
Thanks brother,you helped me a lot.

vimal now I am having trouble with datagridview control. I think you have understood the code that i have sent. Now what I need is I will retrive the info from a table(containing fields username,password,etc)and display it on the datagridview.Now if i click on a particular row then it should take me into the new winform and should display the username of the selected row of first form into the textbox of new winform. I did search in the google, and other forums but didn't get any solution.Plz help

Thanks

Vimal Kandasamy17y ago

Hi,

This is right structure

  private void button1_Click(object sender, EventArgs e)


            string uname = textBox1.Text.ToString();


            string paswd = textBox2.Text.ToString();


            string confpasswd = textBox5.Text.ToString();


            string fname = textBox6.Text.ToString();


            string lname = textBox7.Text.ToString();


            string utype = comboBox1.SelectedItem.ToString();


            if (paswd != confpasswd)


                MessageBox.Show(" Invalid password Please retype");


                textBox5.Focus();

            else

                string qry = string.Format("insert into employee(username,password,emp_type) values('{0}','{1}','{2}')", uname, paswd, utype);


                OdbcConnection oConnection = new OdbcConnection("Dsn=Medoffice"); /* connection for 'employee' table */


                oConnection.Open();


                OdbcCommand oCommand = new OdbcCommand(qry);


                oCommand.Connection = oConnection;


                oCommand.ExecuteNonQuery();


                oConnection.Close();


                string qry1 = string.Format("insert into usersprofile(userid,firstname,lastname)values('{0}','{1}','{2}')", utype, fname, lname);


                OdbcConnection oConnection1 = new OdbcConnection("Dsn=Medoffice"); /* connection for 'usersprofile' table */


                oConnection1.Open();


                OdbcCommand oCommand1 = new OdbcCommand(qry1);


                oCommand1.Connection = oConnection1;


                oCommand1.ExecuteNonQuery();


                oConnection1.Close();


                if (utype == "Reception")


                    MessageBox.Show(" New Reception created");


                else if (utype == "Doctor")


                    MessageBox.Show(" New Doctor created");


                else


                    MessageBox.Show(" New Administrator created");


                Form8.ActiveForm.Close();

sachi vasishta17y ago
here is the code

private void button1_Click(object sender, EventArgs e)

{

string uname = textBox1.Text.ToString();

string paswd = textBox2.Text.ToString();

string confpasswd = textBox5.Text.ToString();

if (paswd != confpasswd)

{

MessageBox.Show(" Invalid password Please retype");

textBox5.Focus();

}

string fname = textBox6.Text.ToString();

string lname = textBox7.Text.ToString();

string utype = comboBox1.SelectedItem.ToString();

string qry = string.Format("insert into employee(username,password,emp_type) values('{0}','{1}','{2}')", uname, paswd, utype);

OdbcConnection oConnection = new OdbcConnection("Dsn=Medoffice"); /* connection for 'employee' table */

oConnection.Open();

OdbcCommand oCommand = new OdbcCommand(qry);

oCommand.Connection = oConnection;

oCommand.ExecuteNonQuery();

oConnection.Close();

string qry1 = string.Format("insert into usersprofile(userid,firstname,lastname)values('{0}','{1}','{2}')",utype,fname,lname);

OdbcConnection oConnection1 = new OdbcConnection("Dsn=Medoffice"); /* connection for 'usersprofile' table */

oConnection1.Open();

OdbcCommand oCommand1 = new OdbcCommand(qry1);

oCommand1.Connection = oConnection1;

oCommand1.ExecuteNonQuery();

oConnection1.Close();

if (utype == "Reception")

MessageBox.Show(" New Reception created");

else if (utype == "Doctor")

MessageBox.Show(" New Doctor created");

else

MessageBox.Show(" New Administrator created");

Form8.ActiveForm.Close();

}
Vimal Kandasamy17y ago
Can you post your full code which is under button1_click method....

in your code, where you insert your values into database?... whether its in button1_click method or some in other method?..

you have to ensure that your insert code [insert value into database] will be executed after all inputs are correct...if they are not correct you have to show error msg.
sachi vasishta17y ago
the textBox3.Focus(); is not working ,I mean the indifferent data(pass and con pass) are still being inserted into the database
Vimal Kandasamy17y ago
Hi,

you have to re arrange conditions..

like below

string uname=textBox1.Text.ToString();

string pass=textBox2.Text.Tostring();

string confirm=textBox3.Text.ToString();

if(pass==confirm)
{

//your code to add inputs to database
}
else
{
MessageBox.Show("Please retype password");
textbox3.Focus();
}