getting row position from query..

Sending this over to php-general because I think I may be more of a logic issue than a query issue.. Original post: CD> I'm trying to determine the percentile value of where a row lies in a CD> query.. hard to explain, but I'll try. CD> I do a select all of my table, sorting it by SCORE. I want to find the CD> position of a certain row in that select all, get that row #, and then I can CD> calculate the percentile fine by dividing that position into the total # of CD> rows, then multiplying by 100. I'm guessing I need to use mysql_data_seek somehow but I can't wrap my head around how I'm supposed to use it in this situation, and how I can get the position of the row to seek to.. the thing I need to get out of this is the position of where that row lies in that sorted query. (Alex.. I tried your 2nd query below as it looked like what I needed, but returned a syntax error which I couldn't figure out.. :( ) Thanks, Chad -----Original Message----- From: Alexey Borzov [mailto:borz_off@rdw.ru] Sent: Wednesday, December 06, 2000 3:10 AM To: Chad Day Cc: 'php-db@lists.php.net' Subject: Re: [PHP-DB] auto incrementing rows from query.. Greetings, Chad! At 06.12.2000, 11:00, you wrote: CD> I'm trying to determine the percentile value of where a row lies in a CD> query.. hard to explain, but I'll try. CD> I do a select all of my table, sorting it by SCORE. I want to find the CD> position of a certain row in that select all, get that row #, and then I can CD> calculate the percentile fine by dividing that position into the total # of CD> rows, then multiplying by 100. CD> my code is like: CD> $querysetup = "SELECT SCORE, PID from pictures where VOTES > 100 order by CD> SCORE DESC"; Ugh. I don't think that I quite get you, but: If you want to find "that row" by SCORE, then you should issue something like SELECT COUNT(*) FROM pictures WHERE VOTES>100 AND SCORE>$your_needed_SCORE If you want to find "that row" by PID, then SELECT COUNT(*) FROM pictures WHERE VOTES>100 AND SCORE>(SELECT SCORE FROM pictures WHERE PID=$your_needed_PID) Thus you get the number of rows before "that row" Then issue SELECT COUNT(*) FROM pictures WHERE VOTES>100 to get _total_ number of rows then divide one by the other, multiply by 100 and you're all done. :] I'm almost sure that can be done in a more elegant way, though... -- Yours, Alexey V. Borzov, Webmaster of RDW -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net For additional commands, e-mail: php-db-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net