我用两方修改的方法,
方法一:
sqlcon.connection.Open();
string cmd3 = string.Format("update student set photo='{0}' where stuno='{1}'", photoByte, FrmShowStuData.stuno);
command2 = new SqlCommand(cmd3, sqlcon.connection);
sqlcon.connection.Open();
用这种后存入的十六进制数很少,要显示出图片也显不了方法二:
sqlcon.connection.Open();
string comm = @"update student set photo = @stupho where stuno = @stuno";
command2 = new SqlCommand();
command2.CommandType = CommandType.Text;
command2.CommandText = comm;
command2.Connection = sqlcon.connection;
command2.Parameters.Add("@stupho", SqlDbType.Image);
command2.Parameters[0].Value = photoByte;
command2.Parameters.Add("@stuno", SqlDbType.VarChar);
command2.Parameters[1].Value = FrmShowStuData.stuno;
用这种就可以存入,而且显示的十六进制数也很长,也可以显示出图片,,请问要将方法二改成方法一的语句,该怎么改,急着用,谢谢啊,!!
方法一:
sqlcon.connection.Open();
string cmd3 = string.Format("update student set photo='{0}' where stuno='{1}'", photoByte, FrmShowStuData.stuno);
command2 = new SqlCommand(cmd3, sqlcon.connection);
sqlcon.connection.Open();
用这种后存入的十六进制数很少,要显示出图片也显不了方法二:
sqlcon.connection.Open();
string comm = @"update student set photo = @stupho where stuno = @stuno";
command2 = new SqlCommand();
command2.CommandType = CommandType.Text;
command2.CommandText = comm;
command2.Connection = sqlcon.connection;
command2.Parameters.Add("@stupho", SqlDbType.Image);
command2.Parameters[0].Value = photoByte;
command2.Parameters.Add("@stuno", SqlDbType.VarChar);
command2.Parameters[1].Value = FrmShowStuData.stuno;
用这种就可以存入,而且显示的十六进制数也很长,也可以显示出图片,,请问要将方法二改成方法一的语句,该怎么改,急着用,谢谢啊,!!
解决方案 »
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