如何检查动态创建的winform实例是否已经在c#中打开

Question

I know how to check for the form if it is already open. But, I am having a hard time finding the same if the form was created dynamically. How can I find the type of dynamic object? In the code below, I am trying to find what goes in Application.OpenForms.OfType< ??? >().Any(). I am sure it is something silly but I can't crack it.

public static bool OpenForm(string formName)
{
    try
    {
        string namespace = MethodInfo.GetCurrentMethod().ReflectedType.Namespace;
        Assembly assembly = Assembly.GetExecutingAssembly();
        var _form = assembly.CreateInstance($"{namespace}.{formName}") as Form;

        if (Application.OpenForms.OfType<???>().Any())
        {
            //Form is already open, do nothing
            return true;
        }            

        if (_form != null)
            _form.Show();

        return true;
    }
    catch (Exception ex)
    {
        return false;
    }            
}

Answer 1

If you mean to use a generic method to verify whether an instance of a Form type exists and, if it's not, then run it, you can use a generic method and to perform both the check and create the instance:

Check whether an instance of the type exists:

Application.OpenForms.OfType<T>() 

Create the instance and show the Form if none that type is already created:

public static bool OpenForm<T>() where T : Form, new()
{
    if (Application.OpenForms.OfType<T>().Any()) return true;

    try {
        var f = new T();
        f.Show();
        return true;
    }
    catch (Exception) {
        return false;
    }
}

---

Since you're passing the name of a Form in your code snippet, you might instead check whether an instance of a Form with that name already exists:

  Application.OpenForms.OfType<Form>().Any(f => f.Name.Equals(formName))

• If it doesn't, first check whether the assembly actually contains a type with that name (passing true , causes GetType() to throw if it fails to find a match, so it goes to the same exception handler):

   var formType = assembly.GetType(formAsmName, true);

Otherwise, it goes on and creates the instance.
Next time you try to create an instance of a Form with that name, the first check will return true and won't create a new instance.

---

public static bool OpenForm(string formName)
{
    if (Application.OpenForms.OfType<Form>().Any(f => f.Name.Equals(formName))) return true;

    try {
        var assembly = Assembly.GetExecutingAssembly();
        var formAsmName = $"{assembly.GetName().Name}.{formName}";
        var formType = assembly.GetType(formAsmName, true); // <- Throws

        var form = assembly.CreateInstance(formType.FullName) as Form;
        form.Show();
        return true;
    }
    catch (Exception) {
        // Throw a specific exception instead?
        return false;
    }
}