2009/6/18 PJ <af.gour...@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.gour...@videotron.ca >> <mailto:af.gour...@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 --- p...@ptahhotep.com <mailto:p...@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/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
