如何在窗口电话8.1中改进性能方法GetThumbnailAsync

Question

我在窗口电话8.1中写了一个函数来显示文件夹上的图像(假设我在这个文件夹中有大约60个图像).问题是函数GetThumbnailAsync()在创建流来获取bitmapImage时需要很长时间.这是我的代码

       //getFileInPicture is function get all file in picture folder 
       List<StorageFile> lstPicture = await getFileInPicture();
       foreach (var file in lstPicture)
       {
            var thumbnail = await        file.GetThumbnailAsync(Windows.Storage.FileProperties.ThumbnailMode.PicturesView,50);
            var bitmapImage = new BitmapImage();
            bitmapImage.DecodePixelWidth = (int)(ScreenWidth / 4);
            await bitmapImage.SetSourceAsync(thumbnail);
            g_listBitmapImage.Add(bitmapImage);
           //g_listBitmapImage is a list of bitmapImage
        }

我是测试,发现问题是函数GetThumbnailAsync花了这么长时间.如果我有大约60张图片,则需要大约15秒来完成此功能(我在lumia 730中测试). 有没有人遇到这个问题以及如何让这段代码运行得更快？.

非常感谢您的支持

Answer 1

您当前正在等待file.GetThumbnailAsync每个文件,这意味着虽然该函数是为每个文件异步执行的,但它是按顺序执行的,而不是并行执行的.

尝试将每个返回的异步操作转换file.GetThumbnailAsync为a Task,然后将其存储在列表中,然后将await所有任务存储起来Task.WhenAll.

List<StorageFile> lstPicture = await getFileInPicture();
List<Task<StorageItemThumbnail>> thumbnailOperations =  List<Task<StorageItemThumbnail>>();

   foreach (var file in lstPicture)
   {
        thumbnailOperations.Add(file.GetThumbnailAsync(Windows.Storage.FileProperties.ThumbnailMode.PicturesView,50).AsTask());
   }

   // wait for all operations in parallel
   await Task.WhenAll(thumbnailOperations);

   foreach (var task in thumbnailOperations)
   {
       var thumbnail = task.Result;
        var bitmapImage = new BitmapImage();
        bitmapImage.DecodePixelWidth = (int)(ScreenWidth / 4);
        await bitmapImage.SetSourceAsync(thumbnail);
        g_listBitmapImage.Add(bitmapImage);
       //g_listBitmapImage is a list of bitmapImage
   }