C#重复打开表单代码冗余

Question

我有一个代码来打开我的表单没有任何重复,但我想为此创建一个方法,以避免代码冗余.

我的代码:

    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();
    }

我想要的是:

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);
}

我对参数有一些问题.谢谢你的回答!

Answer 1

那样的东西？:

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();
}

OfType<T> 扩展方法要求 using System.Linq;

用法

formOpener<Form1>();

如果有任何打开,这将显示表单.否则,将创建一个新的.

如果您可以打开多种形式的T类型,则使用Name属性来区分它们.

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