Gotharious Posted November 7, 2011 Share Posted November 7, 2011 Hey all, I get many warnings saying "Warning: extract() [function.extract]: First argument should be an array in /home/www/waw-eg.com/Users.php on line 93 Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/www/waw-eg.com/Users.php on line 93" while($row = mysql_fetch_array($result)) { $query = "SELECT ". " (SELECT COUNT(*) FROM users WHERE recruiteris = ".$row['id'].") AS count_1,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_1) AS count_2,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_2) AS count_3,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_3) AS count_4,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_4) AS count_5,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_5) AS count_6,". " (SELECT COUNT(*) FROM users WHERE recruiteris = count_6) AS count_7,"; $res = mysql_query($query); extract( mysql_fetch_assoc($res) ); $total = $count_1 + $count_2 + $count_3 + $count_4 + $count_5 + $count_6; $var3 = '10'; $commission = $total * $var3; Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted November 7, 2011 Share Posted November 7, 2011 That query is not going to work as it's written. What are you trying to do there? Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 7, 2011 Author Share Posted November 7, 2011 I want the result of the first query to be used in the second and so on for example, search for user id from users where recruiterid = 1 so I get 2 results 2 and 3 so the second query search for user id from users where reruiterid= 2 and so on and the result of each query would be used in the query after Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted November 7, 2011 Share Posted November 7, 2011 I have to admit I'm not completely clear on what your end goal is, but I don't think COUNT() is what you're after here. It appears you're executing a query in a loop also, which should be avoided as it's (almost) never necessary or wise. Maybe if you post some sample data and expected results, someone can help you get this figured out. This seems like it's going to be a MySQL issue rather than php, so I'll move the thread there . . . Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 7, 2011 Author Share Posted November 7, 2011 Ok, thanks anyway Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 7, 2011 Author Share Posted November 7, 2011 What I want to do is, calculate a commission of a user by summing the number of people he recruited and the number of people they recruited and so on till the last group recruited who will be the 13th group if we consider the first group he recruited is the 1st and the group they recruited is the 2nd and then multiply the number by 10 and echo the result Quote Link to comment Share on other sites More sharing options...
mikosiko Posted November 7, 2011 Share Posted November 7, 2011 post your table(s) description(s) Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 7, 2011 Author Share Posted November 7, 2011 Field Typr Null Default Key Extra id int(11) NO NULL PRI auto_increment fname text NO NULL mname text NO NULL lname text NO NULL mobile text NO NULL tel text NO NULL job text NO NULL company text NO NULL address text NO NULL email text NO NULL recruiteris int(11) NO NULL nid text NO NULL password text NO NULL type text NO NULL position varchar(35) NO NULL Quote Link to comment Share on other sites More sharing options...
mikosiko Posted November 8, 2011 Share Posted November 8, 2011 here is a couple links that could help you: http://www.sitepoint.com/hierarchical-data-database/ and this presentation is a must... http://www.slideshare.net/billkarwin/sql-antipatterns-strike-back slides 48 to 77 show you alternatives... pay attention to the "Closure Table" implementation... the implementation should allow you to solve your query easily (and many more) Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 8, 2011 Author Share Posted November 8, 2011 Thanks A lot, mate Will post here when I figure this out or have any more questions Quote Link to comment Share on other sites More sharing options...
fenway Posted November 8, 2011 Share Posted November 8, 2011 I'm sorry, you want to do what/ Quote Link to comment Share on other sites More sharing options...
Gotharious Posted November 8, 2011 Author Share Posted November 8, 2011 Ok, this is a multi level marketing thingy, wher each person recruits 2 people, and each of those two recruit 2 people and so on each person gets commission of the 2 he recruited and every new recruit down his downline so if I recruited 2 people, each recruited 2, and then each of the ones they recruited recruited 2 thta's 2 + 4 + 8 = 14 So to calculate my comission, lets say it's $10 per recruit, then it would be $140 for me, and for the 2 people below me it would be $60 each now I have only one table where all this information is stored in user id and recruiter id the recruiter id of the 2 people i recruit would be my user id what I want to do now is to query the table to count the number of people in my downline I figured to do that, I have to query my userid in the recruiter id column and then use the result of the two people and query their user ids in the recruiter id column to find out the number they recruited and so on is it clear now? Quote Link to comment Share on other sites More sharing options...
fenway Posted November 8, 2011 Share Posted November 8, 2011 Yup, the previous comments were correct -- you're dealing with a hierarchy here. If you would store the "depth" when you store the original data, life would be easy. Quote Link to comment Share on other sites More sharing options...
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