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 | | | 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) -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
