HTTP Question ...?

[eezmo]

I user an ENCTYPE form to upload an image from client computers
When I tested my script "localy" whith the addresse 127.0.0.1... I could get all informations I need : $image, $image_name, $image_size and $image_type
so i could put the image in a database.

But when I use the script by using the IP address of my server, somme thing like 192.16.5.1, nothing is working. (some times it works and sometimes no ???).

I got this warning message:
"File Upload Error - No Mime boundary found after start of file header in Unknown on line 0"

What's wrong ??? Thanks for your help.

Sofiane

[tinuviel]

Can you show us your code you are using?
Tinuviel

[Anon]

hello,

Here is the code of my program "imgputdb.php"

thanks for help ;-)

---

---

<?
if ($action=="valider")
{
// sauvegarder l'image en base
// "$to" fichier cible

$date=0;

if ($to)
{
copy($to,"c:\temp\".$to_name);
echo " <br> copy $to c:\temp\$to_name <br>";
}

$purl_file=addslashes("$to");
$insert_file="LOAD_FILE(\"$purl_file\")";

$querytab = array();
$querytab[]="delete from tmp_etat_img where id_e=$id_session_cook and titre_img='$titre_img'";

$querytab[]="INSERT INTO tmp_etat_img
(id_img, id_e, titre_img, txt_img, img, name_img, type_img, usr_maj, date_maj)
VALUES
('', $id_session_cook, '$titre_img', '$txt_img', $insert_file, '$to_name', '$to_type', '$auteur', '$date')";

$taille=count($querytab);
$mesg="<br><hr>";
for ($index=0 ; $index<$taille ; $index++)
{
$res=mysql_query($querytab[$index],$link);
if ($res)
{
$code="<font color=\"green\"><b> OK </b></font>";
}
else
{
$code="<font color=\"red\"><b> ERROR ! </b></font>";
}

$mesg.="<br>query = $querytab[$index] $code ";
}

$mesg.="<p><br> file_svr = $to";
$mesg.="<br> size = $to_size";
$mesg.="<br> type = $to_type";
$mesg.="<br> Name = $to_name";
$mesg.="<br> titre_img = $titre_img";

$id= mysql_insert_id($link);
$mesg.="<p>Cette objet porte l'ID: <b>$id</b>";
}//////// Fin if ($action = add)

//*********************************************************************************
// afficher l'image si elle existe
//******************************************************************************
$query="select from tmp_etat_img where id_e=$id_session_cook and titre_img='$titre_img'";
$res=mysql_query($query,$link);
while($val=mysql_fetch_array($res,$link))
{
$txt_img = $val['txt_img'];
$image = $val['img'];
$name_img =$val['name_img'];
}
$num=mysql_num_rows($res);
if ($num!=0)
{
echo " <img src=\"imgdbget.php?titre_img=$titre_img\" alt=\"$name_img\" > ";

}
else
{
echo "<b><font color='red'> Pas d'image </font></b><br>";
}
//**********************************************************************************
//<input name="to" type="file" VALUE="default" size="50,2" >

?>

<FORM ENCTYPE="multipart/form-data" action="<? echo "creer_img.php?titre_img=$titre_img"; ?>" METHOD="post" >
Description de l'image :<br>
<textarea cols=50 rows=3 name="txt_img" wrap=auto><? echo $txt_img ?></textarea>

<INPUT TYPE="hidden" name="MAX_FILE_SIZE" value="100000">
<br>Url de l'image :<br>
<input name="to" type="file" VALUE="toto" size="50,2" >

<br>
<input type="hidden" name="action" value="valider">
<input type="submit" name="bt" value="Valider">
</form>

<?
echo "$mesg";

?>
</BODY>

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[tinuviel]

From what I can see of this form..it looks like it is self submitting...you say in the form to set the field action to valider and then you test for it at the top of the same script. If this is so...shouldn't your action on your form be <? echo $PHP_SELF.
"?titre_img=".$titre_img;?>....or at least:
<? echo "imgputdb.php?titre_img=".$titre_img;?>

Not sure If I'm understanding it right 🙂