Question

我已将图片作为byte[]上传到我的数据库，现在我试图将其显示出来。

出现错误 - 用户代码Parameter is not valid

未处理参数异常

在这一行

newImage = System.Drawing.Image.FromStream(stream);

这是我的代码

for (int i = 0; i < topRatingList.Count; i++)
            {
                int commentRating = topRatingList[i].CommentRating;
                int drinkID = topRatingList[i].DrinkID;

                if (i == 0)
                {
                    DrinkMenuDAO drink = DrinkMenuBLL.getDrinkMenu(drinkID);
                    LinkButton1.Text = drink.DrinkName; 
                    ImageButton1.ImageUrl = byteArrayToImage(drink.DrinkImage);
                }
            }


private string byteArrayToImage(byte[] byteArrayIn)
    {
        System.Drawing.Image newImage;
        string strFileName = Server.MapPath("~/Temp/images/") + ".jpg";
        if (byteArrayIn != null)
        {
            using (MemoryStream stream = new MemoryStream(byteArrayIn))
            {
                newImage = System.Drawing.Image.FromStream(stream);
                newImage.Save(strFileName);
            }
            return strFileName;
        }
        else
        {
            return "";
        }
    }

我用来存储图片的代码

protected void bn_upload_Click(object sender, EventArgs e)
    {
        lbl_msg.Text = "";
        Stream imgStream = FileUpload1.PostedFile.InputStream;
        BinaryReader imgBinary = new BinaryReader(imgStream);
        bytes = imgBinary.ReadBytes((Int32)imgStream.Length);
        imgBinary.Close();
        imgStream.Close();
        string src = byteArrayToImage(bytes);
        if (src.Equals(""))
        {
        }
        else
        {
            Image1.ImageUrl = "~/Temp/UploadedImage.jpg";
        }
    }

    private string byteArrayToImage(byte[] byteArrayIn)
    {
        System.Drawing.Image newImage;
        string strFileName = Server.MapPath("~/Temp/") + "UploadedImage.jpg";
        if (byteArrayIn != null)
        {
            using (MemoryStream stream = new MemoryStream(byteArrayIn))
            {
                newImage = System.Drawing.Image.FromStream(stream);
                newImage.Save(strFileName);
            }
            return strFileName;
        }
        else
        {
            return "";
        }
    }

    protected void btn_submit_Click(object sender, EventArgs e)
    {
        DrinkMenuBLL.uploadImg(bytes);
        lbl_msg.Text = "uploaded";
    }

我在数据库和我使用的sql中使用了varbinary(MAX)

public static void UploadImg(byte[] drinkImage)
    {

        SqlConnection con = new SqlConnection(Constring.getConString());

        string sql = "Update DrinkMenu set drinkImage = (@imgData) where drinkID = 4";

        SqlCommand cmd = new SqlCommand(sql, con);
        cmd.Parameters.Add("@imgData", SqlDbType.Binary).Value = drinkImage;
        try
        {
            con.Open();
            cmd.ExecuteNonQuery();
        }
        finally
        {
            con.Close();
            cmd.Dispose();
            con.Dispose();
        }
    }

Answer 1

一个问题可能是以下问题：

bytes = imgBinary.ReadBytes((Int32)imgStream.Length);

如果流长度大于Int32.MaxValue，因为它是Int64会怎么样？也许你正在截断你的图像...使用这种方法来将流读取到缓冲区：

public static Byte[] ReadStream(Stream input)
{
    Byte[] buffer = new Byte[(16 * 1024)];

    using (MemoryStream stream = new MemoryStream())
    {
        Int32 read;

        while ((read = input.Read(buffer, 0, buffer.Length)) > 0)
            stream.Write(buffer, 0, read);

        return stream.ToArray();
    }
}

protected void bn_upload_Click(object sender, EventArgs e)
{
    lbl_msg.Text = "";

    Byte[] bytes = ReadStream(FileUpload1.PostedFile.InputStream);
    String src = byteArrayToImage(bytes);

    if (!src.Equals(""))
        Image1.ImageUrl = "~/Temp/UploadedImage.jpg";
}

将byte []转换为图像

1 个答案: