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Musica Compilation Disco Anni 70 80 90 tulbirc
musica compilation disco anni 70 80 90
musica compilation disco anni 70 80 90
Anni 70 80 90 disco anni 70 80 90 Category:Italian dance music groupsQ: MySQL combining columns and performing a comparison I have a MySQL table with the following structure: table id (INT) month (DATE) percent_conversion (DECIMAL(10,2)) For the first three months of the year, I want to count how many unique days have occurred. For example: January is January 1 - 28. February is February 1 - 31. March is March 1 - 31. So in January we have 28 unique days, then in February we have 30 unique days and so on. The months are not selected randomly, they are specific months. So for example, months containing 31 unique days are March, September and November. I'm having trouble formulating the following query in SQL. SELECT COUNT(DISTINCT DATE) AS `unique_days_in_this_month` FROM `table` WHERE `month` = {month}; I'm pretty much stuck on this problem. A: SELECT COUNT(*) AS `unique_days_in_this_month`, SUM(CASE WHEN MONTH(DATE) = 1 THEN 1 ELSE 0 END) AS `first_month`, SUM(CASE WHEN MONTH(DATE) = 2 THEN 1 ELSE 0 END) AS `second_month`, SUM(CASE WHEN MONTH(DATE) = 3 THEN 1 ELSE 0 END) AS `third_month`, SUM(CASE WHEN MONTH(DATE) = 4 THEN 1 ELSE 0 END) AS `fourth_month`, SUM(CASE WHEN MONTH(DATE) = 5 THEN 1 ELSE 0 END) AS `fifth_month`, SUM(CASE WHEN MONTH(DATE) = 6 THEN 1 ELSE 0 END) AS `sixth_month`, SUM(CASE WHEN MONTH(DATE) = 7 THEN 1 ELSE 0 END) AS `seventh_month`, SUM(CASE WHEN MONTH(DATE) = 8 THEN 1 ELSE 0 END) AS `eighth_month`, SUM(CASE WHEN MONTH(DATE) = 9 THEN 1 ELSE 0 END
Musica Compilation Disco Anni 70 80 90 Serial Windows Software Full Key
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