C # duplicate open code form

Question

I have code to open my forms without duplication, but I want to create a method for this to avoid code redundancy.

My code is:

    private void form1ToolStripMenuItem_Click(object sender, EventArgs e)
    {
        foreach (Form openForm in Application.OpenForms)
        {
            if (openForm.GetType() == typeof(form1))
            {
                openForm.Activate();
                return;
            }
        }

        form1 f1 = new form1();
        f1.MdiParent = this;
        if (!f1.IsDisposed)
            f1.Show();
    }

What I want:

public void formOpener(Form form, string formName)
{
    foreach (Form openForm in Application.OpenForms)
    {
        if (openForm.GetType() == typeof(form))
        {
            openForm.Activate();
            return;
        }
    }

    form formName = new form();
    formName.MdiParent = this;
    if (!formName.IsDisposed)
        formName.Show();
}

private void form1ToolStripMenuItem_Click(object sender, EventArgs e)
{
    formOpener(form1);
}

I have some problems with the parameters. Thanks for any answer!

Answer 1

Something like that?:

public void formOpener<T>() where T : Form, new()
{
    var openedForm = Application.OpenForms.OfType<T>().FirstOrDefault();

    if (openedForm != null)
    {
        openedForm.Activate();
        return;
    }

    T newForm = new T();
    newForm.MdiParent = this;
    newForm.Show();
}

For extension method

OfType<T> required using System.Linq;

Using

formOpener<Form1>();

This will show the form, if any. Otherwise will create a new one.

If you can open several forms of type T, use the property Nameto distinguish them.

public void formOpener<T>(string formName) where T : Form, new()
{
    var openedForm = Application.OpenForms.OfType<T>()
        .Where(x => x.Name == formName).FirstOrDefault();
}