Question

How do I position a Windows Form to the bottom-right of the screen when it gets open, instead of top-left?

Situation: I have a Form1 which doesn’t actually do anything as a form, I just used it for its context menu (my app works from the tray only). So, most of the main running code goes into the Form1 class. When a context menu is clicked, it will do some processing and in the end it will show Form2. So Form2 gets opened/called by a context menu item of Form1. How do I change Form2’s position in this case?

Form1.cs (part where Form2 gets triggered)

private void menu_upload_file_Click(object sender, EventArgs e)
{
    DialogResult dialogOpened = openFileDialog1.ShowDialog();
    if (dialogOpened == DialogResult.OK)
    {
        string filename = openFileDialog1.FileName;

        using (var client = new WebClient())
        {
            var response = client.UploadFile("http://localhost/imgitv3/upload.php?submit=true&action=upload&request=app", "POST", filename);
            // string path = Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments) + Path.DirectorySeparatorChar + "response.txt";

            if (response != null)
            {
                string responseContent = System.Text.Encoding.ASCII.GetString(response);
                Form2 linkWindow = new Form2();

                if (isURL(responseContent))
                {
                    linkWindow.toTextBox(responseContent);
                    linkWindow.Show();
                }
            }
        }
    }
}

Form2.Designer.cs

// 
            // Form2
            // 
            this.AutoScaleDimensions = new System.Drawing.SizeF(6F, 13F);
            this.AutoScaleMode = System.Windows.Forms.AutoScaleMode.Font;
            this.CausesValidation = false;
            this.ClientSize = new System.Drawing.Size(419, 163);
            this.Controls.Add(this.textBox2);
            this.Controls.Add(this.textBox1);
            this.Controls.Add(this.label3);
            this.Controls.Add(this.label2);
            this.Controls.Add(this.label1);
            this.Icon = ((System.Drawing.Icon)(resources.GetObject("$this.Icon")));
            this.MaximizeBox = false;
            this.MaximumSize = new System.Drawing.Size(435, 202);
            this.MinimizeBox = false;
            this.MinimumSize = new System.Drawing.Size(435, 202);
            this.Name = "Form2";
            this.ShowInTaskbar = false;
            this.StartPosition = System.Windows.Forms.FormStartPosition.Manual;
            this.Text = "IMGit Image Uploader";
            this.TopMost = true;
            this.WindowState = System.Windows.Forms.FormWindowState.Maximized;
            this.Load += new System.EventHandler(this.Form2_Load);
            this.ResumeLayout(false);
            this.PerformLayout();

Answer 1

There are two things you need to know. First is the working area of the screen on which you are going to display the form. The working area is the size of the screen minus the task bars displayed on that screen. You can use the Screen.WorkingArea property for that.

Second is the actual size of the window. Which is not in general the design size of the form, your user might have altered the size of the text in the window titlebar or may be running the video adapter in a different DPI setting from yours. You have to wait until the form’s Load event fires before you know that size.

So make your code look like this, assuming that you want to display the form on the primary monitor:

        var frm = new Form2();
        frm.Load += (s, ea) => {
            var wa = Screen.PrimaryScreen.WorkingArea;
            frm.Location = new Point(wa.Right - frm.Width, wa.Bottom - frm.Height);
        };
        frm.Show();

Which relocates the window just before it becomes visible. The form’s StartPosition property doesn’t matter.