保存并加载图像SQLite C#

Question

我正在尝试使用带有CF的WinForm上的应用程序使用SQLite保存和加载图像.我找到了一种方法将图像保存到数据库中,但我不知道它是否正确,因为我找不到加载存储在数据库中的图像的方法.

我有一个代码将我的图像转换为Base64:

public void ImageToBase64(Image image, System.Drawing.Imaging.ImageFormat format){
        using (MemoryStream ms = new MemoryStream()){
            // Convert Image to byte[]
            image.Save(ms, format);
            byte[] imageBytes = ms.ToArray();

            // Convert byte[] to Base64 String
            string base64String = Convert.ToBase64String(imageBytes);
            SaveImage(base64String);
        }
    }

这是我将图像保存到db中的代码:

void SaveImage(string pic){
        string query = "insert into Table (Photo) values (@pic);"; 
        string conString = @" Data Source = \Program Files\Users.s3db ";            
        SQLiteConnection con = new SQLiteConnection(conString); 
        SQLiteCommand cmd = new SQLiteCommand(query, con);
        cmd.Parameters.Add("@pic",DbType.String);
        con.Open(); 
        try{
            cmd.ExecuteNonQuery();
        }
        catch (Exception exc1){
            MessageBox.Show(exc1.Message);
        }
        con.Close();
    }

我有一个代码与ImageToBase64相反,但首先我需要从db加载图像.有没有想过这样做？

编辑我正在尝试使用blob来保存图像,正如查理建议的那样.我试过这段代码:

Image photo = new Bitmap(@"\Photos\Image20120601_1.jpeg");
SaveImage(photo);

void SaveImage(Image pic){
string conString = @" Data Source = \Program Files\Users.s3db ";            
SQLiteConnection con = new SQLiteConnection(conString);
SQLiteCommand cmd = con.CreateCommand();
cmd.CommandText = String.Format("INSERT INTO Table (Photo) VALUES (@0);");
SQLiteParameter param = new SQLiteParameter("@0", System.Data.DbType.Binary);
param.Value = pic;
cmd.Parameters.Add(param);
con.Open(); 
try{
    cmd.ExecuteNonQuery();
}
catch (Exception exc1){
    MessageBox.Show(exc1.Message);
}
con.Close();}

但是当我执行ExcecuteNonQuery()时,它会捕获InvalidCastException的错误.

有什么建议？

解决方案此代码将图像保存到数据库中,然后加载图像以在pictureBox中显示它:

namespace ImagenSQLite
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
            Image photo = new Bitmap(@"\Photos\Image20120601_1.jpeg");
            byte[] pic = ImageToByte(photo, System.Drawing.Imaging.ImageFormat.Jpeg);
            SaveImage(pic);
            LoadImage();
        }

        public byte[] ImageToByte(Image image, System.Drawing.Imaging.ImageFormat format){
            using (MemoryStream ms = new MemoryStream())
            {
                // Convert Image to byte[]
                image.Save(ms, format);
                byte[] imageBytes = ms.ToArray();
                return imageBytes;
            }
        }
        //public Image Base64ToImage(string base64String)
         public Image ByteToImage(byte[] imageBytes)
        {
            // Convert byte[] to Image
            MemoryStream ms = new MemoryStream(imageBytes, 0, imageBytes.Length);
            ms.Write(imageBytes, 0, imageBytes.Length);
            Image image = new Bitmap(ms);
            return image;
        }
        /***************** SQLite **************************/
        void SaveImage(byte[] imagen){
            string conStringDatosUsuarios = @" Data Source = \Program Files\GPS___CAM\Data\DatosUsuarios.s3db ";            
            SQLiteConnection con = new SQLiteConnection(conStringDatosUsuarios); 
            SQLiteCommand cmd = con.CreateCommand();
            cmd.CommandText = String.Format("INSERT INTO Empleados (Foto) VALUES (@0);");
            SQLiteParameter param = new SQLiteParameter("@0", System.Data.DbType.Binary);
            param.Value = imagen;
            cmd.Parameters.Add(param);
            con.Open();

            try
            {
                cmd.ExecuteNonQuery();
            }
            catch (Exception exc1)
            {
                MessageBox.Show(exc1.Message);
            }
            con.Close();
        }
        void LoadImage(){
            string query = "SELECT Photo FROM Table WHERE ID='5';";
            string conString = @" Data Source = \Program Files\Users.s3db ";
            SQLiteConnection con = new SQLiteConnection(conString); 
            SQLiteCommand cmd = new SQLiteCommand(query, con);            
            con.Open();
            try
            {
                IDataReader rdr = cmd.ExecuteReader();
                try
                {
                    while (rdr.Read())
                    {
                        byte[] a = (System.Byte[])rdr[0];
                        pictureBox1.Image = ByteToImage(a);
                    }
                }
                catch (Exception exc) { MessageBox.Show(exc.Message); }
            }
            catch (Exception ex) { MessageBox.Show(ex.Message); }
            con.Close();
        }       
    }
}

谢谢你的帮助!

Answer 1

要从数据库加载映像,可以使用SQL select语句来获取数据,就像使用任何其他类型的数据一样.base64编码的图像作为字符串存储在SQLite数据库中.因此,您将检索它作为字符串,就像您在数据库中存储任何其他字符串(例如,您的名字)一样.

string LoadImage() {
    string query = "select Photo from Table;"; 
    string conString = @" Data Source = \Program Files\Users.s3db ";            
    SQLiteConnection con = new SQLiteConnection(conString); 
    SQLiteCommand cmd = new SQLiteCommand(query, con);
    string base64EncodedImage = null;
    con.Open(); 
    try {
        IDataReader reader = cmd.ExecuteReader();
        reader.Read(); // advance the data reader to the first row
        base64EncodedImage = (string) reader["Photo"];
        reader.Close();
    }
    catch (Exception ex) {
        MessageBox.Show(ex.Message);
    }
    con.Close();
    return base64EncodedImage;
}

与保存代码一样,我加载图像的示例代码仅使用保存到表中的一个图像.要加载和保存多个映像,您需要在表中插入一个ID字段,然后where在SQL select语句中使用一个子句.

我不确定我个人会将图像存储为base64编码的字符串.图像的字符串表示将比二进制表示大得多.SQLite支持BLOB数据类型,所以我会考虑使用它.