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);
}
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;
Related
I Am programming in C# and I Already have it that I can select my files from my phone with this text:
private void Button_Click_2(object sender, RoutedEventArgs e)
{
var FileOpenPicker = new Windows.Storage.Pickers.FileOpenPicker();
FileOpenPicker.ViewMode = Windows.Storage.Pickers.PickerViewMode.List;
FileOpenPicker.FileTypeFilter.Add(".jpg");
FileOpenPicker.FileTypeFilter.Add(".jpeg");
FileOpenPicker.FileTypeFilter.Add(".png");
FileOpenPicker.PickSingleFileAndContinue();
How can I let the picture popup in the picturebox?
Use a FileDialog and load a bitmap from the selected path:
using (OpenFileDialog dlg = new OpenFileDialog())
{
dlg.Title = "Open Image";
dlg.Filter = "bmp files (*.bmp)|*.bmp";
if (dlg.ShowDialog() == DialogResult.OK)
{
// 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);
}
}
I am following this tutorial and downloaded source code to practice, and it works. The problem occurs when I rewrite the code: just one image is added instead of all the selected images. What am I doing wrong here?
private void button1_Click(object sender, EventArgs e)
{
ofd.Filter = "Images (*.BMP;*.JPG;*.GIF,*.PNG,*.TIFF)|*.BMP;*.JPG;*.GIF;*.PNG;*.TIFF|" +
"All files (*.*)|*.*";
ofd.Multiselect = true;
if (ofd.ShowDialog() == DialogResult.OK)
{
foreach (string name in ofd.FileNames)
{
PictureBox imageControl = new PictureBox();
imageControl.Width = 100;
imageControl.Height = 100;
Image.GetThumbnailImageAbort CallBck = new Image.GetThumbnailImageAbort(ThumbnailCallback);
Bitmap myBitmap = new Bitmap(name);
Image img = myBitmap.GetThumbnailImage(97, 97, CallBck, IntPtr.Zero);
imageControl.Image = img;
panel1.Controls.Add(imageControl);
}
}
}
I'll bet they are all being added but they're just all going on top of each other at location (0,0) in the panel (you should step through your code to check this though).
The solution: Either manually specify a location for each new PictureBox, or use a layout control such as a FlowLayoutPanel.
I have a form with OpenFileDialog for selecting image and showing it in pictureBox. Until the form is open the user can open and then save the opened image as many times as he wants. What I want to do is, after each new selection-save, to delete the previously saved image if there is such. The problem is that as I implemented the code now I am able to delete the image the first time, if I keep on saving images with the currently open form I get an error that the resource is being used. What I do is Dispose() the image but I guess I don't do it in the right place.
This is how I open and load the image:
private void btnExplorer_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Title = "Select file";
openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = fileNameFilter;
openFileDialog1.FilterIndex = 2;
openFileDialog1.RestoreDirectory = true;
openFileDialog1.FileName = prefixFilter;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
try
{
pictureBox1.InitialImage = new Bitmap(openFileDialog1.FileName);
pictureBox1.ImageLocation = openFileDialog1.FileName;
selectedFile = pictureBox1.ImageLocation;
selectedFileName = openFileDialog1.SafeFileName;
pictureBox1.Load();
}
catch (Exception ex)
{
logger.Error(ex.ToString());
MessageBox.Show("Error loading image!", "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);
}
}
}
And in the same class I have this method which I call if I need to delete an old image:
public void DeleteImage(AppConfig imagePath, string ImageName)
{
pictureBox1.InitialImage.Dispose();//Release the image before trying to delete it
string imgPath = imagePath.ConfigValue.ToString();
File.Delete(imgPath + "\\" + ImageName);
}
As you can see. The Dispose() method is here which I though will ensure that the resource will be disposed before trying to delete it but as I said this only work once and then I get the error as many times as attempts to save image I make.
P.S
The exact error I get is:
The process cannot access the file 'C:\Images\ME_083a210e1a7644198fe1ecaceb80af52.jpg' because it is being used by another process.
There is a better way to do it. Load the image using FileStream and than assign it to the pictureBox
FileStream bmp = new FileStream(openFileDialog1.FileName, FileMode.Open, FileAccess.Read);
Image img = new Bitmap(bmp);
pictureBox1.Image = img;
bmp.Close();
and if you want to clear the picture box, simply
pictureBox1.Image = null;
If I understand this correctly you want to remove the once "used" image from it's picture box: Set
picturebox.InitialImage=null;
(By the way: You better use picturebox.Image ...)
"Dispose" is to force the garbage collector to remove an unused object from memory.
Your error has nothing to do with the disposal of the pictureBox-image but with the lock of the source file.
Maybe it already helps if you use a "using" block for the handling of the openFileDialog.
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Title = "Select file";
openFileDialog1.InitialDirectory = "c:\\";
openFileDialog1.Filter = "Jpeg Files(*.jpg)|*.jpg|All files (*.*)|*.*";
openFileDialog1.FilterIndex = 2;
openFileDialog1.RestoreDirectory = true;
if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
pictureBox1.InitialImage = new Bitmap(openFileDialog1.FileName);
pictureBox1.ImageLocation = openFileDialog1.FileName;
selectedFile = pictureBox1.ImageLocation;
selectedFileName = openFileDialog1.SafeFileName;
pictureBox1.Load();
}
}
public string selectedFileName { get; set; }
public string selectedFile { get; set; }
private void button2_Click(object sender, EventArgs e)
{
pictureBox1.ImageLocation = null;
}
I have been trying to save the image in pictureBox1. What am trying to do is to make it so people can "Save" the image from pictureBox1 by clicking on my "Save As" Button.
I am using Visual Studio 2010 with C#
This is what I have so far:
private void Button2_Click(System.Object sender, System.EventArgs e)
{
SaveFileDialog saveFileDialog1 = new SaveFileDialog();
saveFileDialog1.Filter = "Png Image|*.jpg";
saveFileDialog1.Title = "Save an Image File";
saveFileDialog1.ShowDialog();
if (saveFileDialog1.FileName != "")
pictureBox1.Image = new Bitmap ("c:/avatar.png");
{
SaveFileDialog dialog = new SaveFileDialog();
if (dialog.ShowDialog() == DialogResult.OK)
{
pictureBox1.Image.Save(#"C:\Documents and Settings\.Png", System.Drawing.Imaging.ImageFormat.Png);
}
}
}
I'm not sure what you are trying with your code, there seems to be quite a few things wrong in there. This is how I would do it:
private void Button2_Click(System.Object sender, System.EventArgs e)
{
if (pictureBox.Image != null)
{
using {var dialog = new SaveFileDialog())
{
dialog.Title = ...
saveFileDialog1.Filter = "Png Image|*.png";
...other properties...
if (dialog.ShowDialog == DialogResult.OK)
{
pictureBox.Image.Save(dialog.FileName, System.Drawing.Imaging.ImageFormat.Png)
}
}
}
}
I am working on form . I want small window to pop up when I click button and to will select XML file of my choice from various folder.
I guess, this OpenFileDialog will help me.
private void button3_Click(object sender, EventArgs e)
{
/
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
if (OpenFileDialog1.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(filed.FileName.ToString());
}
}
I tried using following code but when I click on the button there window doesn't pop up.
I am not geting what mistake I have made.
What is the problem with that?
Thanks!
You cant just open the file dialog from a console app. You will have to workaround it with some setting to single thread apartment (STA).
[STAThread]
static void Main(string[] args)
{
MessageBox.Show("Test");
}
--EDIT--
Following works on click event:
OpenFileDialog f = new OpenFileDialog();
f.Filter = "XML Files|*.xml";
f.InitialDirectory = "D:\\";
if(f.ShowDialog() == DialogResult.OK)
{
MessageBox.Show(f.FileName);
}
You cant open file fialog in console app.
You say I have button, so this must be Win app, use
openFileDialog1.ShowDialog(); is missing
private void button3_Click(object sender, EventArgs e)
{
OpenFileDialog OpenFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = " XML Files|*.xml";
openFileDialog1.InitialDirectory = #"D:\";
openFileDialog1.ShowDialog();
// Get file name and use OpenFileDialog1.FileName or something like that
}