Dear George thankīs again but  I donīt know how to save the image in another
file. I tryed this way. But my question is: This is the best way to do that?
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
 <title>Upload.php</title>
</head>
<BODY BGCOLOR=#3366FF>
<CENTER>
   <?php
      echo "<FORM ENCTYPE=multipart/form-data ACTION=Upload1.php
METHOD=POST>";
         echo "<h1> Kompatscher & Cia Ltda </h1>";
         echo "<hr>";
         $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'] ;
            $num=rand(0,9999);
            $nome='temp'.$num.'.jpg';
            $myfile = fopen($nome,"w");
            $fp = fwrite($myfile,$i);
            fclose($myfile);
            $len=strlen($i);
            echo "<input type=image src=$nome border=0>";
          echo "</td>";
         echo "</tr>";
   echo "</TABLE>";
   mysql_close($con);
   ?>
   </FORM>
</CENTER>
</BODY>
</html>
----- Original Message -----
From: "George Nicolae" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Saturday, April 27, 2002 2:10 PM
Subject: [PHP-WIN] 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" <[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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