当我的图像从网络服务器上的数据库加载时,我看到以下错误:
GDI+ 中发生一般错误。在
System.Drawing.Image.Save(Stream流,ImageCodecInfo编码器,
编码器参数(encoderParams) at
System.Drawing.Image.Save(Stream 流,ImageFormat 格式) at
MyWeb.Helpers.ImageHandler.ProcessRequest(HttpContext上下文)
我所有的代码试图做的就是加载图像,任何人都可以看一下并让我知道我做错了什么吗?
注意 - 如果我在本地计算机上测试它,则此方法有效,但当我将其部署到 Web 服务器时,此方法无效。
public void ProcessRequest(HttpContext context)
{
context.Response.Clear();
if (!String.IsNullOrEmpty(context.Request.QueryString["imageid"]))
{
int imageID = Convert.ToInt32(context.Request.QueryString["imageid"]);
int isThumbnail = Convert.ToInt32(context.Request.QueryString["thumbnail"]);
// Retrieve this image from the database
Image image = GetImage(imageID);
// Make it a thumbmail if requested
if (isThumbnail == 1)
{
Image.GetThumbnailImageAbort myCallback = new Image.GetThumbnailImageAbort(ThumbnailCallback);
image = image.GetThumbnailImage(200, 200, myCallback, IntPtr.Zero);
}
context.Response.ContentType = "image/png";
// Save the image to the OutputStream
image.Save(context.Response.OutputStream, ImageFormat.Png);
}
else
{
context.Response.ContentType = "text/html";
context.Response.Write("<p>Error: Image ID is not valid - image may have been deleted from the database.</p>");
}
}
该行发生错误:
image.Save(context.Response.OutputStream, ImageFormat.Png);
UPDATE
我已将代码更改为这样,但问题仍然发生:
var db = new MyWebEntities();
var screenshotData = (from screenshots in db.screenshots
where screenshots.id == imageID
select new ImageModel
{
ID = screenshots.id,
Language = screenshots.language,
ScreenshotByte = screenshots.screen_shot,
ProjectID = screenshots.projects_ID
});
foreach (ImageModel info in screenshotData)
{
using (MemoryStream ms = new MemoryStream(info.ScreenshotByte))
{
Image image = Image.FromStream(ms);
// Make it a thumbmail if requested
if (isThumbnail == 1)
{
Image.GetThumbnailImageAbort myCallback = new Image.GetThumbnailImageAbort(ThumbnailCallback);
image = image.GetThumbnailImage(200, 200, myCallback, IntPtr.Zero);
}
context.Response.ContentType = "image/png";
// Save the image to the OutputStream
image.Save(context.Response.OutputStream, ImageFormat.Png);
} }
Thanks.