Re: Showing a blob filed part II

My mistake. The correct code is: "<input type=image src=image.php border=0>"; -- Best regards, George Nicolae IT Manager ___________________ PaginiWeb.com - Professional Web Design www.PaginiWeb.com "George Nicolae" <geomasterweb@hotmail.com> wrote in message news:20020427155607.428.qmail@pb1.pair.com... > try to echo $i in a separate file (lets say image.php) and in the main file > do > echo "<input type=image src=".image.php." border=0>"; > > -- > > > Best regards, > George Nicolae > IT Manager > ___________________ > PaginiWeb.com - Professional Web Design > www.PaginiWeb.com > > > "Waldemar Brand Neto" <wbn@softall.com.br> wrote in message > news:005f01c1edf9$7ca7dce0$0b01a8c0@komptascher.org... > Thank you for the help. I put the header in the top ot the document and now > it works. > I need one more tip. The result is not the image itself, but the content of > the Blob Field, how can I fix that? > > Thanks. > Waldemar > The source code: > <?php > header("Content-type: image/jpeg"); > header("Content-Disposition: inline; filename=temp.jpg"); > ?> > <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> > <html> > <head> > <title>Teste.php</title> > </head> > <BODY BGCOLOR=#3366FF> > <CENTER> > <FORM> > <h1> Kompatscher & Cia Ltda </h1> > <hr> > <?php > $con = mysql_connect("**************************) or die ("Erro ao > conectar!"); > mysql_select_db("***********", $con) or die ("Erro ao selecionar a > database!"); > $comando="SELECT > codigo,nome,material,temperatura,pressao,fluido,descricao,foto FROM cadgrupo > WHERE codigo=010"; > $res = mysql_query($comando,$con); > $linha = mysql_fetch_array($res); > echo "<TABLE ALIGN=CENTER>"; > echo "<tr>"; > echo "<td>"; > echo "Foto"; > echo "</td>"; > echo "<td>"; > $i= $linha['foto'] ; > echo "<input type=image src=$i border=0>"; > echo "</td>"; > echo "</tr>"; > echo "</TABLE>"; > mysql_close($con); > ?> > </FORM> > </CENTER> > </BODY> > </html> > > > >