在winform中将图片存入sqlserver,并从sqlserver中读取并显示在picturebox中

1,存入sqlserver
  数据库结构
  create table test
  {
     id identity(1,1),
     FImage image
  }
 1.1,在窗体中加一个OpenFileDialog控件,命名为ofdSelectPic
 1.2,在窗体上添加一个打开文件按钮,添加如下单击事件代码:
    Stream ms;
  byte[] picbyte;
  //ofdSelectPic.ShowDialog();
  if (ofdSelectPic.ShowDialog()==DialogResult.OK)
  {
   if ((ms=ofdSelectPic.OpenFile())!=null)
   {
    //MessageBox.Show("ok");
    picbyte=new byte[ms.Length];
    ms.Position=0;
    ms.Read(picbyte,0,Convert.ToInt32(ms.Length));
    //MessageBox.Show("读取完毕!");

    //连接数据库
    SqlConnection conn=new SqlConnection();
    conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

    SqlCommand cmd=new SqlCommand("UpdateImage",conn);
    cmd.CommandType=CommandType.StoredProcedure;

    cmd.Parameters.Add("@UpdateImage",SqlDbType.Image);
    cmd.Parameters["@UpdateImage"].Value=picbyte;

    conn.Open();
    cmd.ExecuteNonQuery();
    conn.Close();

    ms.Close();
    }
   }

2,读取并显示在picturebox中
 2.1 添加一个picturebox,名为ptbShow
 2.2 添加一个按钮,添加如下响应事件:
      SqlConnection conn=new SqlConnection();
 conn.ConnectionString="Data Source=localhost;Database=test;User Id=sa;Pwd=sa";

 string strSql="select FImage from test where id=1";

 SqlCommand cmd=new SqlCommand(strSql,conn);

 conn.Open();
 SqlDataReader reader=cmd.ExecuteReader();
        reader.Read();

 MemoryStream ms=new MemoryStream((byte[])reader["FImage"]);


 Image image=Image.FromStream(ms,true);

        reader.Close();
        conn.Close();

 ptbShow.Image=image;

posted @ 2004-05-10 14:20  kwklover  阅读(5277)  评论(10编辑  收藏  举报