2つのクエリを結合する - Postgresql

Question

私は2つのクエリの結果が必要なSQLクエリで作業しています。最初のクエリはこれです：

これは以下になる

select s.code, count(s.code), s.name from course_staff c join courses e on 
c.course = e.id 
join subjects s on e.subject = s.id 
join semesters x on e.semester = x.id 
where c.staff = 5033690 
group by s.code, s.name 
having count(s.code) > 8; 

：

code | count |    name 
----------+-------+-------------------------------- 
    MATS1464 | 10 | Professional Comm&Presentation 
    MATS6605 |  9 | Prof Comm & Presentation 

2のクエリ：

count 
------- 
    30 

：これは以下になる

select count(distinct s.code) from course_staff c join courses e 
on c.course = e.id 
join subjects s on e.subject = s.id 
join semesters x on e.semester = x.id 
where c.staff = 5033690 
having count(s.code)>20; 

私は両方のカウントを組み合わせる必要があります上記2つのクエリの条件を1つのクエリにまとめて、関数またはビューで使用できるようにします。両方の条件が満たされるようにこれらの2つのクエリをどのように結合することができるかについてアドバイスしてください。ヘルプは非常に感謝しています。おかげ

Answer 1

はここ（Postgresのウィンドウの機能を使用して）考えです：

WITH 
    code_name (code, name) AS ( 
    SELECT s.code, s.name 
    FROM course_staff c 
    JOIN courses e ON c.course = e.id 
    JOIN subjects s ON e.subject = s.id 
    JOIN semesters x ON e.semester = x.id 
    WHERE c.staff = 5033690 
), 
    code_name_counts (code, name, count_both, count_total) AS ( 
    SELECT DISTINCT code, 
        name, 
        COUNT(code) OVER (PARTITION BY code, name),          
        COUNT(code) OVER (PARTITION BY code)             
    FROM code_name 
) 
SELECT code, name, count_both, count_total 
FROM code_name_counts 
WHERE count_both > 8 AND count_total > 20 ; 

があなたの要件についての更なる知識がなく暗闇の中でショットを検討してください。

と、（WHEREなし）テスト目的のために

：

WITH 
    code_name (code, name) AS ( 
    VALUES (1, 'a'), (1, 'b'), (2, 'b'), (2, 'b'), (3, 'c') 
), 
    code_name_distinct (code, name) AS ( 
    SELECT DISTINCT code, name 
    FROM code_name 
), 
    code_name_counts (code, name, count_both, count_total) AS ( 
    SELECT code, 
      name, 
      (SELECT COUNT(*) FROM code_name a 
      WHERE a.code = x.code AND a.name = x.name), 
      (SELECT COUNT(*) FROM code_name b 
      WHERE b.code = x.code) 
    FROM code_name_distinct x 
) 
SELECT code, name, count_both, count_total 
FROM code_name_counts 

返し：

WITH 
    code_name (code, name) AS ( 
    VALUES (1, 'a'), (1, 'b'), (2, 'b'), (2, 'b'), (3, 'c') 
), 
    code_name_counts (code, name, count_both, count_total) AS ( 
    SELECT DISTINCT code, 
      name, 
      COUNT(code) OVER (PARTITION BY code, name), 
      COUNT(code) OVER (PARTITION BY code) 
    FROM code_name 
) 
SELECT code, name, count_both, count_total 
FROM code_name_counts ; 

ここ

code | name | count_both | count_total 
------+------+------------+------------- 
    1 | a |   1 |   2 
    2 | b |   2 |   2 
    1 | b |   1 |   2 
    3 | c |   1 |   1 

を与え、DISTINCTとPARTITION BY無し例であります上記と同じ結果です。