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Load a bitmap image into Windows Forms using open file dialog

开发者 https://www.devze.com 2023-03-08 11:23 出处:网络
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.

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 the code I tried:

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

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

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

  开发者_如何学运维  dialog.Dispose();
}


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);
        }
    }
}


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;

    }
}


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();
}


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


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


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

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


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();
        }
    }


It's simple. Just add:

PictureBox1.BackgroundImageLayout = ImageLayout.Zoom;
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