me1000 Posted February 3, 2008 Share Posted February 3, 2008 if ($store == '1' && !((isset($unsafe['in'])) || (isset($unsafe['buy'])) ) ){ so this is what it does, if $store == 1 keep going else return false if $in isset return false or if $buy isset return false so in layman's terms, is neither $in nor $buy are set and $store == 1 return true... What is wrong with my conditional? Thanks, Quote Link to comment https://forums.phpfreaks.com/topic/89273-solved-complex-conditional/ Share on other sites More sharing options...
CerealBH Posted February 3, 2008 Share Posted February 3, 2008 need a ! infront of isset($unsafe['buy']) don't ya? Quote Link to comment https://forums.phpfreaks.com/topic/89273-solved-complex-conditional/#findComment-457122 Share on other sites More sharing options...
trq Posted February 3, 2008 Share Posted February 3, 2008 if ($store == 1 && (!isset($unsafe['in']) || !isset($unsafe['buy']))) { Quote Link to comment https://forums.phpfreaks.com/topic/89273-solved-complex-conditional/#findComment-457124 Share on other sites More sharing options...
CerealBH Posted February 3, 2008 Share Posted February 3, 2008 if ($store == 1 && !isset($unsafe['in']) || !isset($unsafe['buy'])) { had 2 extra ( in there Quote Link to comment https://forums.phpfreaks.com/topic/89273-solved-complex-conditional/#findComment-457130 Share on other sites More sharing options...
me1000 Posted February 3, 2008 Author Share Posted February 3, 2008 thank you, after too many parentheses I get confused haha! Quote Link to comment https://forums.phpfreaks.com/topic/89273-solved-complex-conditional/#findComment-457134 Share on other sites More sharing options...
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