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 | | | 0000-00-00 | | | 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)
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