dodgeitorelse Posted July 27, 2011 Share Posted July 27, 2011 my database has 2 columns which are clan and site. I have put both into arrays as such: while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ $clanname[]= $row['clan']; $siteaddy[]= $row['site']; } I then look through $clanname array to see if I have a match and if I do I want to output the $siteaddy result. I tried this $found2 = array_keys($clanname, $server['o']['comment']); if(is_array($found2) && !empty($found2)){ foreach($found2 as $key3){ if($server['o']['comment'] == $clanname[$key3]) { $server2 = $siteaddy; } } } but all it returns is ARRAY and not the actual data for the $siteaddy. I would appreciate any suggestions. Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/ Share on other sites More sharing options...
AyKay47 Posted July 27, 2011 Share Posted July 27, 2011 how are you outputting your array? Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/#findComment-1248051 Share on other sites More sharing options...
xyph Posted July 27, 2011 Share Posted July 27, 2011 So I'm guessing $server['o']['comment'] will store a name of the clan that might be in the $clanname array? You could do something like this <?php while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ $clan_data[]= array( $row['clan'], $row['site'] ); } // ... foreach( $clan_data as $clan ) { if( $server['o']['comment'] == $clan[0] ) { $server2 = $clan[1]; } } ?> Or if $server['o']['comment'] is defined before you loop your SQL results, you could also just do while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ if( $server['o']['comment'] == $row['clan'] ) { $server2 = $row['site']; } } Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/#findComment-1248055 Share on other sites More sharing options...
dodgeitorelse Posted July 27, 2011 Author Share Posted July 27, 2011 I should have more clear.... my database has 2 columns which are clan and site. I have put both into arrays as such in a file called web_sites.php: while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ $clanname[]= $row['clan']; $siteaddy[]= $row['site']; } then in another file called zone.php I have included web_sites.php I then look through $clanname array to see if I have a match and if I do I want to output the $siteaddy result. I tried this $found2 = array_keys($clanname, $server['o']['comment']); if(is_array($found2) && !empty($found2)){ foreach($found2 as $key3){ if($server['o']['comment'] == $clanname[$key3]) { $server2 = $siteaddy; } } } $output .= " <td style='padding-top:5px; padding-bottom:5px; vertical-align:top; text-align:center'> <table style='width:{$zone_width}; margin:auto; text-align:center' cellpadding='0' cellspacing='2'> <tr> <td title='{$server['s']['name']}' style='padding:0px; text-align:center'> <div style='left:0px; right:0px; top:0px; bottom:0px; width:{$zone_width}; white-space:nowrap; overflow:hidden; text-align:center'> $server2 </div> </td> </tr> The $server['o']['comment'] does store a name of the clan that might be in the $clanname array which is in a 3rd file called class.php which is included in the zone.php file. So basically if $server['o']['comment'] is someclanname and is found in the $clanname array I want the web site address that is stored in the $siteaddy array to display in the table. I hope this is clearer picture of what I have. Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/#findComment-1248156 Share on other sites More sharing options...
xyph Posted July 27, 2011 Share Posted July 27, 2011 My first example will help you. Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/#findComment-1248168 Share on other sites More sharing options...
dodgeitorelse Posted July 28, 2011 Author Share Posted July 28, 2011 Thank you Quote Link to comment https://forums.phpfreaks.com/topic/242991-array-problem/#findComment-1248264 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.