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LiveInternet.ru:
: 16.03.2006
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DISTINCT

, 21 2019 . 17:49 +
!


SELECT , COUNT(*) AS 
FROM (SELECT DISTINCT ,  FROM )
GROUP BY


, .
, .

: "" 

PARAMETERS []![___2]![fil_status] Short, []![___2]![fil_filial] Short;
SELECT Count(.) AS , MonthName(Month([])) AS , ., Year([]) AS , ., ., .
FROM  RIGHT JOIN  ON . = .
GROUP BY MonthName(Month([])), ., Year([]), ., ., .
HAVING (((Year([]))=Year(Now())) AND ((.)=[]![___2]![fil_status]) AND ((.)=[]![___2]![fil_filial]));

https://www.sql.ru/forum/1312740/kak-poschitat-unikalnye-znacheniya-pro-pomoshhi-distinct


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