Question

I’m using the following code to open and display image in one of my forms using fileDialog :

private void btnExplorer_Click(object sender, EventArgs e)
        {
            OpenFileDialog openFileDialog1 = new OpenFileDialog();

            openFileDialog1.InitialDirectory = "c:\\";
            openFileDialog1.Filter = "All files (*.*)|*.*";
            openFileDialog1.FilterIndex = 2;
            openFileDialog1.RestoreDirectory = true;

            if (openFileDialog1.ShowDialog() == DialogResult.OK)
            {
                try
                {
                    PictureBox PictureBox1 = new PictureBox();
                    PictureBox1.Image = new Bitmap(openFileDialog1.FileName);
                    // Add the new control to its parent's controls collection
                    this.Controls.Add(PictureBox1);
                }
                catch (Exception ex)
                {
                    MessageBox.Show("Error loading image" + ex.Message);
                }
            }
        }

The problem is that my image is shown at the top left corner of my form, when I have left almost quarter of my down-right side for this purpose. How can I show it there?

Answer 1

Like I said in my comment, here’s how: How to: Position Controls on Windows Forms.

PictureBox PictureBox1 = new PictureBox();
PictureBox1.Image = new Bitmap(openFileDialog1.FileName);
PictureBox1.Location = new Point(20, 100); //20 from left and 100 from top
this.Controls.Add(PictureBox1);

Or change it afterwards:

PictureBox1.Top += 50; //increase distance from top with 50