一般性错误+

Question

一个有图片上传功能适用于本地主机正常，但当我尝试和Windows Server 2003下运行，我得到错误信息

这是代码..

现在任何人在我面前跳;）我已经看了前面的答案，我已经检查了所有的权限，他们似乎是正确的..文件夹/路径存在，等等。

ImageService imageService = new ImageService(); 

if (fileBase != null && fileBase.ContentLength > 0 && fileBase.ContentLength <= 2097152 && fileBase.ContentType.Contains("image/")) 
{ 
    var uploadedPath = "~/Assets/Images/"; 

    Path.GetExtension(fileBase.ContentType); 
    var extension = Path.GetExtension(fileBase.FileName); 

    if (extension.ToLower() != ".jpg" && extension.ToLower() != ".gif") // only allow these types 
    { 
     photoViewModel.ImageValid = "Not Valid"; 
     ModelState.AddModelError("Photo", "Wrong Image Type"); 
     return View(photoViewModel); 
    } 

    EncoderParameters encodingParameters = new EncoderParameters(1); 
    encodingParameters.Param[0] = new EncoderParameter(Encoder.Quality, 100L); 

    ImageCodecInfo jpgEncoder = imageService.GetEncoderInfo("image/jpeg"); 
    var uploadedimage = Image.FromStream(fileBase.InputStream, true, true); 

    Bitmap originalImage = new Bitmap(uploadedimage); 
    Bitmap newImage = new Bitmap(originalImage, 274, 354); 

    Graphics g = Graphics.FromImage(newImage); 
    g.InterpolationMode = InterpolationMode.HighQualityBilinear; 
    g.DrawImage(originalImage, 0, 0, newImage.Width, newImage.Height); 

    var streamLarge = new MemoryStream(); 
    newImage.Save(streamLarge, jpgEncoder, encodingParameters); 

    var fileExtension = Path.GetExtension(extension); 
    string newname; 
    if (photoViewModel.photoURL != null) 
    { newname = photoViewModel.photoURL; } 
    else 
    { newname = Guid.NewGuid() + fileExtension; } 

    var ImageName = newname; 
    newImage.Save(Server.MapPath(uploadedPath) + ImageName); 
    System.IO.File.WriteAllBytes(Server.MapPath(uploadedPath) + ImageName, streamLarge.ToArray()); 

    photoViewModel.uploadedPath = uploadedPath; 
    photoViewModel.photoURL = ImageName; 

    originalImage.Dispose(); 
    newImage.Dispose(); 
    streamLarge.Dispose(); 
    return View(photoViewModel); 
} 

Answer 1

随着Image.FromStream中，流必须在图像的整个生命周期内保持打开状态。这意味着如果流是一个文件，文件将保持打开状态。我认为您需要处理uploadedImage以允许关闭流（如果您等待GC清理uploadedImage，这将在未来发生的不确定时刻发生 - 或根本不会发生）。