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" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> 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" <[EMAIL PROTECTED]> wrote in message
> 005f01c1edf9$7ca7dce0$[EMAIL PROTECTED]">news:005f01c1edf9$7ca7dce0$[EMAIL PROTECTED]...
> 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>
>
>
>
>



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