0

Please, help me with following.

I have a WPF application. There are ListView with Images as items and one Image - DisplayedImage. When item clicked - its image displayed in the DisplayedImage Image Control.

When Image Control have some transformations I need to save them to source file.

How I can to free needed file, if it uses as source for ListView items and DisplayedImage Image Control?

Exception

The process cannot access the file "<Path>" because it is being used by another process

XAML

<Image x:Name="DisplayedImage" Source="{Binding Path=TheImageViewItem.DisplayedPath, IsAsync=True, Mode=OneWay}"/> <ListView x:Name="listView" ItemsSource="{Binding Path=Images, IsAsync=True}" />

C#

JpegBitmapEncoder jpegBitmapEncoder = new JpegBitmapEncoder(); jpegBitmapEncoder.QualityLevel = 100; jpegBitmapEncoder.Frames.Add(BitmapFrame.Create(DisplayedImage.Source as BitmapSource)); if (File.Exists(file)) File.Delete(file); FileStream fileStream = new FileStream(file, FileMode.CreateNew); jpegBitmapEncoder.Save(fileStream); fileStream.Close();

THANKS TO ALL

Improve this question
3
  • As long as the image exist in the ListView with Images, it is kept open, and you cannot overwrite it. Commented Jan 24, 2016 at 12:07
  • Set the control to null which should free the image. Commented Jan 24, 2016 at 12:12
  • You may change your Image.Source binding from string in TheImageViewItem.DisplayedPath to a BitmapImage instance. For example like here stackoverflow.com/a/20648/5574010 Commented Jan 24, 2016 at 12:29
Load 7 more related questions Show fewer related questions

0

Reset to default

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.