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在我的应用程序中,我将位图转换为 png 并将其存储在文件夹内的文件中。当我尝试再次打开文件进行第二次写入时,我得到一个UnauthorisedAccessException说法"Access is denied"。单击保存按钮时,将调用以下函数。

private async void SaveClicked(object sender, RoutedEventArgs e)
        {
            WriteableBitmap wb = new WriteableBitmap(InkCanvas2, null);
            Image image = new Image();
            image.Height = 150;
            image.Width = 450;
            image.Source = wb;
            await SaveToStorage(wb, image);
            TransparentLayer.Visibility = System.Windows.Visibility.Collapsed;   

        }

SaveToStorage 有以下代码

private async Task SaveToStorage(WriteableBitmap i, Image im)
        {
            try
            {    
                  var dataFolder = await local.CreateFolderAsync("Page", CreationCollisionOption.OpenIfExists);                    
                  using (var testpng = await dataFolder.OpenStreamForWriteAsync("testpng.png", CreationCollisionOption.ReplaceExisting)) 
// HITS EXCEPTION AND GOES TO CATCH BLOCK
                     {    
                       i.WritePNG(testpng);    
                       testpng.Flush();
                       testpng.Close();
                     }                    
            }  
            catch(Exception e)
            {
            string txt = e.Message;
            }
        }

它保存第一次没有错误,第二次抛出异常。知道为什么会这样吗?

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1 回答 1

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我想通了!这显然是因为我在尝试打开它之前没有包含文件存在检查。

我已经重新编写了这样的代码

       IStorageFolder dataFolder = await local.CreateFolderAsync("Page", CreationCollisionOption.OpenIfExists);
            StorageFile Ink_File = null;
            //Using try catch to check if a file exists or not as there is no inbuilt function yet
            try
            {
                Ink_File = await dataFolder.GetFileAsync("testpng.png");
            }
            catch (FileNotFoundException)
            {
                return false;
            }

            try
            {
                if (Ink_File != null)
                {

                    using (var testpng = await Ink_File.OpenStreamForWriteAsync())
                    {

                        i.WritePNG(testpng);

                        testpng.Flush();
                        testpng.Close();
                        return true;
                    }
                }

            }
            catch(Exception e)
            {
                string txt = e.Message;
                return false;
            }
            return false;
于 2013-02-06T12:07:13.783 回答