marmite Posted May 16, 2007 Share Posted May 16, 2007 When I print_r an array (which I've got from an SQL query), I get "Array ( [0] => 2 )" All I need to do is set a variable to "2" or whatever this Array produces. There will only ever be one result as it is returning a count(distinct) How do I do this??? I am going INSANE trying to figure this out. Please help! Link to comment https://forums.phpfreaks.com/topic/51628-solved-ridiculously-simple-php-array-question/ Share on other sites More sharing options...
Orio Posted May 16, 2007 Share Posted May 16, 2007 I think this example will explain the best (I used "AS" in my query): <?php $result = mysql_query("SELECT COUNT(*) AS 'c' FROM some_table"); $array = mysql_fetch_array($result); $count = $array['c']; echo $count; ?> You could also do (without "AS"): <?php $result = mysql_query("SELECT COUNT(*) FROM some_table"); $array = mysql_fetch_assoc($result); $count = list($array); echo $count; ?> Orio. Link to comment https://forums.phpfreaks.com/topic/51628-solved-ridiculously-simple-php-array-question/#findComment-254332 Share on other sites More sharing options...
marmite Posted May 16, 2007 Author Share Posted May 16, 2007 THANKYOU The second one didn't seem to work (?) but when I selected with "as" it worked fine. That's a useful technique. Cheers Link to comment https://forums.phpfreaks.com/topic/51628-solved-ridiculously-simple-php-array-question/#findComment-254348 Share on other sites More sharing options...
Orio Posted May 16, 2007 Share Posted May 16, 2007 Woops I used list() incorrectly. It's supposed to be: <?php $result = mysql_query("SELECT COUNT(*) FROM some_table"); $array = mysql_fetch_assoc($result); list($count) = $array; echo $count; ?> Orio. Link to comment https://forums.phpfreaks.com/topic/51628-solved-ridiculously-simple-php-array-question/#findComment-254350 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.