Re: populate form input option dropdown box from existing data

2009/6/18 PJ <af.gourmet@videotron.ca>: > 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. :-( If you're getting a blank screen instead of an error message I highly recommend turning display_errors on and setting error_reporting to E_ALL in you dev php.ini - that way you'll actually see error messages rather than having to hunt through the code by hand. -Stuart -- http://stut.net/