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See more: C# Windows Forms
C# Windows Form Application project. I have a menu form which has buttons to load other already created forms.

private void cmdReporting_Click(object sender, EventArgs e)
{
    reporting frm = null;
    if ((frm = (reporting)openCheck(typeof(reporting))) == null)
    {
        frm = new reporting();
        frm.Show();
    }
    else
        frm.Select();
}


The form it is trying to load is named reporting. This code is identical to WORKING code that loads other forms. This is the only one out of 5 that won't load. It does not throw an error either.

Is there a setting I'm missing in the GUI?

Added Code for openCheck method
public static Form openCheck(Type FormType)
{
    foreach (Form OpenForm in Application.OpenForms)
    {
        if (OpenForm.GetType() == FormType)
            return OpenForm;
    }
 
    return null;
}
Posted 25-Aug-12 9:23am
Edited 25-Aug-12 9:31am
v2
Comments
Zoltán Zörgő at 25-Aug-12 15:28pm
   
What is "loading a form"? Show method will show a visible or hidden form, Select activates a visible form. But Show is not modal. What is the source of openCheck? How are the forms closed?
Wes Aday at 25-Aug-12 15:30pm
   
Did you debug it to find if the form is being created? If not created then what error is causing the creation to fail?
Christian Graus at 25-Aug-12 20:54pm
   
No - he thinks the debugger only works when your code blows up.
Wes Aday at 25-Aug-12 21:31pm
   
OMG, I see... Well CG... it's hard to help people that do not listen. I like the "working code does not trip the debugger"....
Zoltán Zörgő at 25-Aug-12 16:01pm
   
I would rather use the singleton pattern, like here: http://hashfactor.wordpress.com/2009/03/31/c-winforms-create-a-single-instance-form/
joeswindell at 25-Aug-12 21:51pm
   
Thank you Zoltan, that would be a much better solution.

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