Windows Forms

Luc Pattyn wrote:

You could adapt program.cs to have two Application.Run statements, one with the logon form, the next with your real main form.

crazylad wrote:

Will I be able to pass information

int main()
{
     // show the login form here. 
     if(loginValidated)
     {
         // starts the application message loop
         Application.Run(new ProcessForm());
         /* .... */
     }
}

N a v a n e e t h wrote:

int main(){ // show the login form here. if(loginValidated) { // starts the application message loop Application.Run(new ProcessForm()); /* .... */ }}

LogonForm lf=new LogonForm();
Application.Run(lf);
bool success=lf.Success;
lf.Dispose();
if (success) ...

N a v a n e e t h wrote:

message loop (Application.Run) only after validating login?

crazylad wrote:

If I use Application.Run(new LogonForm()) and try returning a boolean parameter to indicate successful/unsuccessful logon, the project doesn't build successfully.

int main()
{
     bool loginValidated = false;
     using (LoginWindow window = new LoginWindow())
     {
         loginValidated = window.ShowDialog() == DialogResult.OK;
     }

     if(loginValidated)
     {
         // starts the application message loop
         Application.Run(new ProcessForm());
         /* .... */
     }
}

N a v a n e e t h wrote:

IMO, you need to use Application.Run only when the login is successful. As I said, before calling Application.Run, validate the login and if successful call it. To get the validation result back, you can use DialogResult enum provided with .NET framework. You can do like this,int main(){ bool loginValidated = false; using (LoginWindow window = new LoginWindow()) { loginValidated = window.ShowDialog() == DialogResult.OK; } if(loginValidated) { // starts the application message loop Application.Run(new ProcessForm()); /* .... */ }}Inside the login form, once you validate all the details, write like

this.DialogResult = DialogResult.OK;

FileStream oFileStream = new FileStream("FilePath", FileMode.OpenOrCreate);
oFileStream.Write(YourByteArray,0,YourByteArraySize);