Load a bitmap image into Windows Forms using open file dialog

问题

I need to open the bitmap image in the window form using open file dialog (i will load it from drive). The image should fit in the picture box. Here is some code i have tried but got error!

 private void button1_Click(object sender, EventArgs e)
        {
            OpenFileDialog dlg = new OpenFileDialog();

            dlg.Title = "Open Image";
            dlg.Filter = "bmp files (*.bmp)|*.bmp";

            if (dlg.ShowDialog() == DialogResult.OK)
            {                     
            PictureBox PictureBox1 = new PictureBox();                    
                PictureBox1.Image(dlg.FileName);
            }

            dlg.Dispose();
        }

回答1:

You have to create an instance of the Bitmap class, using the constructor overload that loads an image from a file on disk. As your code is written now, you're trying to use the PictureBox.Image property as if it were a method.

Change your code to look like this (also taking advantage of the using statement to ensure proper disposal, rather than manually calling the Dispose method):

private void button1_Click(object sender, EventArgs e)
{
    // Wrap the creation of the OpenFileDialog instance in a using statement,
    // rather than manually calling the Dispose method to ensure proper disposal
    using (OpenFileDialog dlg = new OpenFileDialog())
    {
        dlg.Title = "Open Image";
        dlg.Filter = "bmp files (*.bmp)|*.bmp";

        if (dlg.ShowDialog() == DialogResult.OK)
        {
            PictureBox PictureBox1 = new PictureBox();

            // Create a new Bitmap object from the picture file on disk,
            // and assign that to the PictureBox.Image property
            PictureBox1.Image = new Bitmap(dlg.FileName);
        }
    }
}

Of course, that's not going to display the image anywhere on your form because the picture box control that you've created hasn't been added to the form. You need to add the new picture box control that you've just created to the form's Controls collection using the Add method. Note the line added to the above code here:

private void button1_Click(object sender, EventArgs e)
{
    using (OpenFileDialog dlg = new OpenFileDialog())
    {
        dlg.Title = "Open Image";
        dlg.Filter = "bmp files (*.bmp)|*.bmp";

        if (dlg.ShowDialog() == DialogResult.OK)
        {
            PictureBox PictureBox1 = new PictureBox();
            PictureBox1.Image = new Bitmap(dlg.FileName);

            // Add the new control to its parent's controls collection
            this.Controls.Add(PictureBox1);
        }
    }
}

回答2:

Works Fine. Try this,

private void addImageButton_Click(object sender, EventArgs e)
{
    OpenFileDialog of = new OpenFileDialog();
    //For any other formats
    of.Filter = "Image Files (*.bmp;*.jpg;*.jpeg,*.png)|*.BMP;*.JPG;*.JPEG;*.PNG"; 
    if (of.ShowDialog() == DialogResult.OK)
    {
        pictureBox1.ImageLocation = of.FileName;

    }
}

回答3:

You should try to:

Create the picturebox visually in form (it's easier)
Set Dock property of picturebox to Fill (if you want image to fill form)
Set SizeMode of picturebox to StretchImage

Finally:

private void button1_Click(object sender, EventArgs e)
{
    OpenFileDialog dlg = new OpenFileDialog();
    dlg.Title = "Open Image";
    dlg.Filter = "bmp files (*.bmp)|*.bmp";
    if (dlg.ShowDialog() == DialogResult.OK)
    {                     
        PictureBox1.Image = Image.FromFile(dlg.Filename);
    }
    dlg.Dispose();
}

回答4:

private void button1_Click(object sender, EventArgs e)
{
    OpenFileDialog open = new OpenFileDialog();
    if (open.ShowDialog() == DialogResult.OK)
        pictureBox1.Image = Bitmap.FromFile(open.FileName);
}

回答5:

You, can also try like this, PictureBox1.Image = Image.FromFile("<your ImagePath>" or <Dialog box result>);

回答6:

PictureBox.Image is a property, not a method. You can set it like this:

PictureBox1.Image = System.Drawing.Image.FromFile(dlg.FileName);

回答7:

You can try the following:

private void button1_Click(object sender, EventArgs e)
    {
        OpenFileDialog fDialog = new OpenFileDialog();
        fDialog.Title = "Select file to be upload";
        fDialog.Filter = "All Files|*.*";
        //  fDialog.Filter = "PDF Files|*.pdf";
        if (fDialog.ShowDialog() == DialogResult.OK)
        {
            textBox1.Text = fDialog.FileName.ToString();
        }
    }

回答8:

It's simple. Just add:

PictureBox1.BackgroundImageLayout = ImageLayout.Zoom;

来源：https://stackoverflow.com/questions/6122984/load-a-bitmap-image-into-windows-forms-using-open-file-dialog

标签

winforms

bitmap

picturebox

openfiledialog