show form on connection

Question

0.00/5 (No votes)

See more:

Okay this is my connection code

C#

class myConnection
{
    public static string datasource = "";
    public static string initialcatalog = "";
    public static string uid = "";
    public static string pwd = "";
    public static MySqlConnection GetConnection()
    {
        string str = "SERVER='" + datasource + "';DATABASE='" + initialcatalog + "';UID='" + uid + "';PASSWORD='" + pwd + "'";
        MySqlConnection con = new MySqlConnection(str);
        con.Open();
        return con;
    }
}

i cant seem to figure out how to make it Show form2 if it actually connects to the database, i have done this befor but im completely drawing a blank in my mind and i cant find anything on google (i suck at searching lol)

Any help would be apreciated

Posted 26-Oct-10 17:12pm

Tunacha

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Karthik. A · Accepted Answer · 2010-10-26T18:33:00

Assuming Form2 is the form you wish to show and you are in Form1, you could do this:

public partial class Form1 : Form
{
...
public void ShowForm2()
{
    MySqlConnection connection = myConnection.GetConnection();

    if (connection != null)
    {
         Form2 frm2 = new Form2();
         frm2.Show();
         // or 
         // frm2.ShowDialog();
    }
}
...
}

Note - I am assuming you haven't posted all of your "myConnection" class and so didn't verify that. But this is how you show another form to the user