Re: populate form input option dropdown box from existing data

Martin Scotta wrote: > It is a sintax error > > if (in_array($ex, $selected) <--- missing ) > echo "<br />yes"; > else echo "<br />no"; > > On Thu, Jun 18, 2009 at 10:13 AM, PJ <af.gourmet@videotron.ca > <mailto:af.gourmet@videotron.ca>> wrote: > > Ford, Mike wrote: > > On 17 June 2009 14:30, PJ advised: > > > > > > > >> For the moment, I am trying to resolve the problem of > >> extracting a value > >> from a string returned by a query. I thought that in_array() > would do > >> it, but the tests I have run on it are 100% negative. The only > thing I > >> have not used in the tests is third parameter bool $strict > which only > >> affects case-sensitivity if the $needle is a string. > >> > > > > $strict has nothing whatsoever at all in any way to do with case > > sensitivity -- $strict controls whether the values are compared > using > > equality (==) or identity (===) tests. As is stated quite > clearly on the > > manual page, in_array() is always case-sensitive. > > > > > >> This leads me to > >> believe that in_array() is either inappropriately defined in > >> the manual > >> and not effective on associative arrays > >> > > > > Complete rubbish -- the in_array() manual page is excellent and > totally > > accurate, complete with 3 working examples. > > > > > I would really like to understand why my attempts to reproduce the > first > example just did not work. > Note also that the examples do not show in_array($string, $array) > My array was Array ([0]=>6[1]=>14....), so when I tried if > (in_array($string, $array) , echo $string did not return 14 as I had > expected; neither did if(in_array("14", $array) ... nor > if(in_array(14, > $array). It still does not... actually, the screen goes blank. > So, what am I doing wrong? > Here's what is not working... I'm trying to reproduce the example from > php.net <http://php.net>: > > $selected = array(); > if ( ( $results = mysql_query($sql, $db) ) ) { > while ( $row = mysql_fetch_assoc($results) ) { > $selected[] = $row['id']; > } > } > print_r($selected); > $ex = 14; > if (in_array($ex, $selected) > echo "<br />yes"; > else echo "<br />no"; > > Regardless if I put 14 into $ex or "14" or '14' or even if I put > the 14 > instead of the $ex into the if line, I get a blank screen. It seems tp > me, from what I see in the manual that this should work... or am I > supposed to know something that is not clear in the examples... ? > > > -- > Hervé Kempf: "Pour sauver la planète, sortez du capitalisme." > ------------------------------------------------------------- > Phil Jourdan --- pj@ptahhotep.com > <mailto:pj@ptahhotep.com> > http://www.ptahhotep.com > http://www.chiccantine.com/andypantry.php > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > -- > Martin Scotta Actually, we have to call it a typo because i fixed that and the results are exactly the same = blank screen. :-( -- Hervé Kempf: "Pour sauver la planète, sortez du capitalisme." ------------------------------------------------------------- Phil Jourdan --- pj@ptahhotep.com http://www.ptahhotep.com http://www.chiccantine.com/andypantry.php