Re: Query help - add results then divide by
What about SELECT (SUM( ads.col * 1.91) * ads.depth ) ) / 131.77 FROM ads WHERE date = '2004-02-26' AND editionID = '13' AND ads.page = '16' Original Message On 2/25/04, 4:19:12 PM, Rogers, Dennis [EMAIL PROTECTED] wrote regarding Query help - add results then divide by : Good afternoon, How can I take the 3 results below add them together then divide by 131.77? Can it all be done in one SQL statement? Thanks in advance. mysql describe ads; +---+---+--+-+++ | Field | Type | Null | Key | Default| Extra | +---+---+--+-+++ | adID | int(11) | | PRI | NULL | auto_increment | | page | int(11) | | | 0 | | | adnum | varchar(20) | | || | | date | date | | | -00-00 | | | depth | decimal(3,2) | YES | | 0.00 | | | timestamp | timestamp(14) | YES | | NULL | | | col | int(11) | YES | | 0 | | | acc | varchar(50) | | || | | editionID | int(11) | | | 0 | | +---+---+--+-+++ 9 rows in set (0.00 sec) mysql SELECT ((ads.col * 1.91) * ads.depth) FROM ads Where date = '2004-02-26' AND editionID = '13' AND ads.page = '16'; +-+ | ((ads.col * 1.91) * ads.depth) | +-+ |7.64 | | 34.38 | |7.64 | +-+ 3 rows in set (0.01 sec) -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
RE: Query help - add results then divide by
Thanks so much!! SELECT (SUM(( ads.col * 1.91) * ads.depth ) / 131.77) * 100 FROM ads WHERE date = '2004-02-26' AND editionID = '13' AND ads.page = '16' -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Wednesday, February 25, 2004 5:55 PM To: Rogers, Dennis Cc: '[EMAIL PROTECTED]'; '[EMAIL PROTECTED]'; Hines, David Subject: Re: Query help - add results then divide by What about SELECT (SUM( ads.col * 1.91) * ads.depth ) ) / 131.77 FROM ads WHERE date = '2004-02-26' AND editionID = '13' AND ads.page = '16' Original Message On 2/25/04, 4:19:12 PM, Rogers, Dennis [EMAIL PROTECTED] wrote regarding Query help - add results then divide by : Good afternoon, How can I take the 3 results below add them together then divide by 131.77? Can it all be done in one SQL statement? Thanks in advance. mysql describe ads; +---+---+--+-+++ | Field | Type | Null | Key | Default| Extra | +---+---+--+-+++ | adID | int(11) | | PRI | NULL | auto_increment | | page | int(11) | | | 0 | | | adnum | varchar(20) | | || | | date | date | | | -00-00 | | | depth | decimal(3,2) | YES | | 0.00 | | | timestamp | timestamp(14) | YES | | NULL | | | col | int(11) | YES | | 0 | | | acc | varchar(50) | | || | | editionID | int(11) | | | 0 | | +---+---+--+-+++ 9 rows in set (0.00 sec) mysql SELECT ((ads.col * 1.91) * ads.depth) FROM ads Where date = '2004-02-26' AND editionID = '13' AND ads.page = '16'; +-+ | ((ads.col * 1.91) * ads.depth) | +-+ |7.64 | | 34.38 | |7.64 | +-+ 3 rows in set (0.01 sec)
Re: Query help - add results then divide by
I think that you can just do this: select sum(ads.col)*1.191*sum(ads.depth)/131.77 where date ='2004-02-26' AND editionID = '13' AND ads.page = '16'; because of the disttributive property of multiplication. (2 * 1.191) +(6*1.91) +(4*1.91)/131.77 = 12 *1.91/131.77 = (12*1.91)/131.77 = 12*(1.91/131.77) Test it to make sure I understand what you're asking, but it worked for my in my tests. bob Rogers, Dennis wrote: Good afternoon, How can I take the 3 results below add them together then divide by 131.77? Can it all be done in one SQL statement? Thanks in advance. mysql describe ads; +---+---+--+-+++ | Field | Type | Null | Key | Default| Extra | +---+---+--+-+++ | adID | int(11) | | PRI | NULL | auto_increment | | page | int(11) | | | 0 | | | adnum | varchar(20) | | || | | date | date | | | -00-00 | | | depth | decimal(3,2) | YES | | 0.00 | | | timestamp | timestamp(14) | YES | | NULL | | | col | int(11) | YES | | 0 | | | acc | varchar(50) | | || | | editionID | int(11) | | | 0 | | +---+---+--+-+++ 9 rows in set (0.00 sec) mysql SELECT ((ads.col * 1.91) * ads.depth) FROM ads Where date = '2004-02-26' AND editionID = '13' AND ads.page = '16'; +-+ | ((ads.col * 1.91) * ads.depth) | +-+ |7.64 | | 34.38 | |7.64 | +-+ 3 rows in set (0.01 sec) -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Query help - add results then divide by
Rogers, Dennis wrote: Good afternoon, How can I take the 3 results below add them together then divide by 131.77? Can it all be done in one SQL statement? Thanks in advance. mysql describe ads; +---+---+--+-+++ | Field | Type | Null | Key | Default| Extra | +---+---+--+-+++ | adID | int(11) | | PRI | NULL | auto_increment | | page | int(11) | | | 0 | | | adnum | varchar(20) | | || | | date | date | | | -00-00 | | | depth | decimal(3,2) | YES | | 0.00 | | | timestamp | timestamp(14) | YES | | NULL | | | col | int(11) | YES | | 0 | | | acc | varchar(50) | | || | | editionID | int(11) | | | 0 | | +---+---+--+-+++ 9 rows in set (0.00 sec) mysql SELECT ((ads.col * 1.91) * ads.depth) FROM ads Where date = '2004-02-26' AND editionID = '13' AND ads.page = '16'; +-+ | ((ads.col * 1.91) * ads.depth) | +-+ |7.64 | | 34.38 | |7.64 | +-+ 3 rows in set (0.01 sec) SELECT SUM((ads.col * 1.91) * ads.depth)/131.77 FROM ads Where date = '2004-02-26' AND editionID = '13' AND ads.page = '16'; -- Sasha Pachev Create online surveys at http://www.surveyz.com/ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
