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asp.net 将图片上传到mysql数据库的方法

2009-06-17 00:00 711 查看
这是页面上的按钮单击事件
  protected void Button1_Click(object sender, EventArgs e) 
  { 
  string tid = Utils.getRandom(32); 
  Stream mystream = this.FileUpload1.PostedFile.InputStream; 
  int length = this.FileUpload1.PostedFile.ContentLength; 
  byte[] pic = new byte[length]; 
  mystream.Read(pic, 0, length); 
  bool flg = insert(tid, pic); 
  }

  这是执行插入的方法
  
 public bool insert(string tid,byte[] pic) 
  { 
  DBConn db = new DBConn(); 
  StringBuilder sql = new StringBuilder(); 
  sql.Append("insert into teacher(TID,TPHOTO,TDELETE) values (?tid,?pic,?flg)"); 
  int flg = 0; 
  try 
  { 
  myConnection = db.getConnection(); 
  MySqlCommand myCommand = new MySqlCommand(sql.ToString(), myConnection); 
  myCommand.Parameters.Add(new MySqlParameter("?tid", MySqlDbType.String, 32)); 
  myCommand.Parameters["?tid"].Value = tid; 
  myCommand.Parameters.Add(new MySqlParameter("?pic", MySqlDbType.Blob)); 
  myCommand.Parameters["?pic"].Value = pic; 
  myCommand.Parameters.Add(new MySqlParameter("?flg", MySqlDbType.Int16)); 
  myCommand.Parameters["?flg"].Value = 0; 
  myConnection.Open(); 
  flg = myCommand.ExecuteNonQuery(); 
  } 
  catch (Exception ex) 
  { 
  return false; 
  } 
  finally 
  { 
  if (myConnection != null) 
  { 
  myConnection.Close(); 
  } 
  } 
  if (flg > 0) 
  { 
  return true; 
  } 
  return false; 
  }
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