GDI 中发生一般错误；

Question

A 有一个图像上传功能，可以在 localhost 上正常工作，但是当我尝试在 Windows Server 2003 下运行时，我收到错误消息“

This is the code..”，

现在在有人攻击我之前；）我已经查看了之前的答案，并且我已检查所有权限，它们似乎是正确的..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

使用Image.FromStream，流必须在图像的生命周期内保持打开状态。这意味着如果流是一个文件，该文件将保持打开状态。我认为您需要处置 uploadedImage 以允许关闭流（如果您等待 GC 清理 uploadedImage，这将在未来的不确定点发生 - 或者可能根本不会发生）。