RE: Image Upload fails after Migration 4.0 --> 4.3...

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My first inclination would be to ask you what your php.ini setting is for
register_globals. See this page http://us3.php.net/register_globals for
background info. Problem could be any number of things, but this would be
the first thing that comes to mind.

> -----Original Message-----
> From: lynch [mailto:lynch@telda.net]
> Sent: Tuesday, July 29, 2003 5:00 PM
> To: php-db@lists.php.net
> Subject:  Image Upload fails after Migration 4.0 --> 4.3...
> 
> 
> hello everybody,
> 
> my script fror uploading a number of images to a 777 mode 
> directory worked fine all the time - until i decided to 
> rebuild my machine from scratch with
> 
> apache 2.0 (running virtual hosts)
> redhat 9.0
> mysql 4.0.12
> php 4.3
> 
> now when attempting to upload an image i keep getting the 
> "could not copy image... to..." message, on some occasions 
> even a confirmation of successful script execution but with 
> no image actually uploaded..
> 
> i have checked my php.ini settings including upload 
> permissions, max upload file size, location of tmp folder 
> safe_mode etc. to no effect.
> 
> does anybody have an idea that may stop me from buying a rope...?
> 
> kind regards
> philip
> 
> attached please find the delinquent for your scrutiny
> ---------
> $link_identifier = mysql_connect($hostname,$username,$passwd);
> 
> if ($link_identifier<=0)
> 
> {
> 
> echo mysql_errno().": ".mysql_error()."<br>";
> 
> exit;
> 
> }
> 
> $now = date("YmdHis");
> 
> $SQLStatement =
> 
> "insert into $databasename.hotel 
> 
> (
> 
> name,
> 
> ort,
> 
> land,
> 
> kategorie,
> 
> link,
> 
> beschreibung,
> 
> thumb1,  // -- image thumbnail to upload
> 
> bild1, // -- image to upload
> 
> thumb2,  // -- image thumbnail to upload
> 
> bild2, // -- image to upload
> 
> timestamp
> 
> )
> 
> values
> 
> (
> 
> '$name',
> 
> '$ort',
> 
> '$land',
> 
> '$kategorie',
> 
> '$link', 
> 
> '$beschreibung',
> 
> '$thumb1path',
> 
> '$bild1path',
> 
> '$thumb2path',
> 
> '$bild2path',
> 
> '$now'
> 
> )";
> 
> $result = mysql_query($SQLStatement,$link_identifier);
> 
> if (!$result)
> 
> {
> 
> echo mysql_errno().": ".mysql_error()."<br>";
> 
> exit;
> 
> }
> 
> $id = mysql_insert_id($link_identifier);
> 
> mysql_close($link_identifier);
> 

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