怎样把图片的路径存入数据库,然后再把图片显示在picturebox里

Question

怎样把图片的路径存入数据库,然后再把图片显示在picturebox里

Answer 1

Stream ms;
  byte[] picbyte;
  OpenFileDialog ofdSelectPic = new OpenFileDialog();
  if (ofdSelectPic.ShowDialog() == DialogResult.OK)
  {
  if ((ms = ofdSelectPic.OpenFile()) != null)
  {
  picbyte = new byte[ms.Length];
  ms.Position = 0;
  ms.Read(picbyte, 0, Convert.ToInt32(ms.Length));
  SqlConnection conn = new SqlConnection();
  conn.ConnectionString = "";  sql = "Insert into Person(Photo) values(@Image)";
  SqlCommand cmd = new SqlCommand(sql, conn);  cmd.Parameters.Add("@Image", SqlDbType.VarBinary);
  cmd.Parameters["@Image"].Value = picbyte;  conn.Open();
  cmd.ExecuteNonQuery();
  conn.Close();  ms.Close();
  }
  } SqlConnection conn=new SqlConnection();   
 conn.ConnectionString="";   
 string strSql="";   
 SqlCommand cmd=new SqlCommand(strSql,conn);   
 conn.Open();   
 SqlDataReader reader=cmd.ExecuteReader();   
  reader.Read();   
 MemoryStream ms=new MemoryStream((byte[])reader["Photo"]);   
 Image image=Image.FromStream(ms,true);   
 reader.Close();   
 conn.Close();   
 picturebox1.Image=image;   

Answer 2

OpenFileDialog opendialog = new OpenFileDialog();             //打开 文件对话框
            opendialog.Filter = "JpegAndGif|*.jpg;*.Gif;*.png";       //筛选格式
            opendialog.RestoreDirectory = true;                               
            opendialog.FilterIndex = 1;
           if (opendialog.ShowDialog() == DialogResult.OK)           
            {
               pictureaddress = opendialog.FileName;                                          //  获取地址
                pictureBox1.Image = Image.FromFile(pictureaddress, false);       //   用picturebox1显示图片            }
       db是连接数据库的类
      
     class db
    {
        public static void ExecSql(string strsql, string image)
        {
            FileStream fs = File.OpenRead(image);
            byte[] imageb = new byte[fs.Length];
            fs.Read(imageb, 0, imageb.Length);
            fs.Close();
            SqlConnection con = Getconn();
            SqlCommand cmd = new SqlCommand(strsql, con);
            cmd.Parameters.Add("@Content", SqlDbType.Image);
            cmd.Parameters["@Content"].Value = imageb;
            con.Open();
            cmd.ExecuteNonQuery();
            con.Close();
    }