将图像/图片格式保存在数据库中 [英] save image/picture format in Database

问题描述

我需要保存图像，但出现错误.如何解决此错误?

Hi,
I need to save image but i got error.How to solved this error?

byte[] pic = new byte[1];
                pic[0] = 0;
                if (pcbPhoto.Image != null)
                {
                    stream = new MemoryStream();
                    pcbPhoto.Image.Save(stream, System.Drawing.Imaging.ImageFormat.Jpeg);// error occur here...
                    pic = stream.ToArray();
                }

错误:GDI +中发生一般错误.

问候，
Karthikeyan，
Bangalore.

error : A generic error occurred in GDI+.

Regards,
Karthikeyan,
Bangalore.

推荐答案

通常，发生此错误是因为尝试保存时使用了输出流，但是由于输出的是MemoryStream而不是这种情况.

抱歉，问题不在您显示的代码中-往后很多...

问题可能在于，您最初用于创建图像的流不再可用-在您进入Save方法时，该流已被关闭或处置.那是行不通的！
试试这个:
首先加载图像时，创建一个新图像作为实际图片的副本，并使用该副本-您可以处置读入的原始版本.例如:

Normally, this error occurs because the output stream is in use when you try to save, but since you are outputting to a MemoryStream that is not the case.

Sorry, but the problem is not in teh code you show - it''s a lot further back...

The problem is probably that the stream you used to create the image in the first place is no longer available - it has been closed or disposed by the time you get to the Save method. And that doesn''t work!
Try this:
When you load the image in the first place, create a new image as a copy of the actual picture, and use that - you can dispose of the original version you read in. For example:

Image myImage;
using (Image inputImage = Image.FromFile(@"D:\Temp\MyPic.jpg"))
    {
    myImage = new Bitmap(inputImage);
    }

由于新图像未与文件或流关联，因此可以毫无问题地将其保存.

Because the new image is not associated with a file or stream, it can be saved with no problem.

您好，
使用此代码将图片读取为Bytes数组

Hi,
Use this code to read picture as Bytes array

MemoryStream stream=new MemoryStream();

            pictureBox1.Image.Save(stream,System.Drawing.Imaging.ImageFormat.Jpeg);
            
            byte[] pic=stream.ToArray();

然后将pic插入数据库

then insert pic into the database

     SqlConnection con = new SqlConnection
                            ("Data Source=localhost;Initial Catalog=Northwind;Persist Security Info=True;User ID=sa;Password=123456");

         SqlCommand com = new SqlCommand("insert into Employees(LastName,FirstName,Photo) values('aaaa','bbb',@Photo)", con);

com.Parameters.AddWithValue("@Photo", pic);

希望这对您有帮助，
A.Mandour

Hope this helps you,
A.Mandour

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