C# 将图像作为 blob 保存到 MySql 数据库

Question

由于某种原因，当我尝试更新用户的图像时，我的代码失败。图像未正确保存。例如，38 kib 的图像在数据库中保存为 13 字节。

这是我的代码:

    public void UploadImage(Image img)
    {
        OpenConnection();
        MySqlCommand command = new MySqlCommand("", conn);
        command.CommandText = "UPDATE User SET UserImage = '@UserImage' WHERE UserID = '" + UserID.globalUserID + "';";
        byte[] data = imageToByte(img);
        MySqlParameter blob = new MySqlParameter("@UserImage", MySqlDbType.Blob, data.Length);
        blob.Value = data;

        command.Parameters.Add(blob);

        command.ExecuteNonQuery();
        CloseConnection();
    }

    public byte[] imageToByte(Image img)
    {
        using (var ms = new MemoryStream())
        {
            img.Save(ms, System.Drawing.Imaging.ImageFormat.Jpeg);
            return ms.ToArray();
        }
    }

OpenConnection 和 closeconnection 只是 conn.Open() 和 conn.Close()。

但是转换不会失败:

但是在数据库中我看到了这个:

有谁知道这里发生了什么吗？

Answer 1

替换此代码:

OpenConnection();
MySqlCommand command = new MySqlCommand("", conn);
command.CommandText = "UPDATE User SET UserImage = '@UserImage' WHERE UserID = '" + UserID.globalUserID + "';";
byte[] data = imageToByte(img);
MySqlParameter blob = new MySqlParameter("@UserImage", MySqlDbType.Blob, data.Length);
blob.Value = data;

command.Parameters.Add(blob);

command.ExecuteNonQuery();
CloseConnection();

与

var userImage = imageToByte(img);

OpenConnection();

var command = new MySqlCommand("", conn);

command.CommandText = "UPDATE User SET UserImage = @userImage WHERE UserID = @userId;";

var paramUserImage = new MySqlParameter("@userImage", MySqlDbType.Blob, userImage.Length);
var paramUserId = new MySqlParameter("@userId", MySqlDbType.VarChar, 256);  

paramUserImage.Value = userImage;
paramUserId.Value = UserID.globalUserID;    

command.Parameters.Add(paramUserImage);
command.Parameters.Add(paramUserId);

command.ExecuteNonQuery();  

CloseConnection();

您发送的是 '@UserImage'，它是一个 10 字节长的字符串，删除引号即可。

上面的代码还使用两个变量的参数 which you should always do .

无论如何希望这对您有帮助。