Как объединить три несколько SQL запросов в один выходной

Question

Я пытаюсь объединить под SQL запросов, чтобы получить один единственный выходной

SELECT reg.REGION_ID, count(*) as totalDist FROM REGION reg LEFT OUTER JOIN ORGANIZATION org 
ON org.FK_REGION=reg.REGION_ID group by reg.REGION_ID; 

SELECT reg.REGION_ID, count(*) as activeTotal FROM REGION reg LEFT OUTER JOIN ORGANIZATION org 
ON org.FK_REGION=reg.REGION_ID and org.ACTIVE=1 group by reg.REGION_ID; 

select u.Region_ID, count(*) as userCount from USER u where u.ORG_ID IN(
Select org.ORG_ID from ORGANIZATION org where org.FK_REGION IN(select REGION_ID from REGION reg)) group by ORG_ID; 

Answer 1

Добавить UNION между ними:

SELECT reg.REGION_ID, COUNT(*) AS totalDist 
FROM REGION reg 
LEFT OUTER JOIN ORGANIZATION org 
    ON org.FK_REGION=reg.REGION_ID 
GROUP BY reg.REGION_ID 
UNION 
SELECT reg.REGION_ID, COUNT(*) 
FROM REGION reg 
LEFT OUTER JOIN ORGANIZATION org 
    ON org.FK_REGION=reg.REGION_ID 
AND org.ACTIVE=1 
GROUP BY reg.REGION_ID 
UNION 
SELECT u.Region_ID, COUNT(*) 
FROM `USER` u 
WHERE u.ORG_ID IN(SELECT org.ORG_ID 
        FROM ORGANIZATION org 
        WHERE org.FK_REGION IN(SELECT REGION_ID FROM REGION reg)) 
GROUP BY ORG_ID; 

EDIT:

SELECT r.REGION_ID, s1.totalDist, s2.activeTotal, s3.useCount 
FROM REGION r 
LEFT JOIN 
(
    SELECT reg.REGION_ID, COUNT(*) AS totalDist 
    FROM REGION reg 
    LEFT OUTER JOIN ORGANIZATION org 
    ON org.FK_REGION=reg.REGION_ID 
    GROUP BY reg.REGION_ID 
) AS s1 
ON r.REGION_ID = s1.REGION_ID 
LEFT JOIN 
(
    SELECT reg.REGION_ID, COUNT(*) AS activeTotal 
    FROM REGION reg 
    LEFT OUTER JOIN ORGANIZATION org 
    ON org.FK_REGION=reg.REGION_ID 
    AND org.ACTIVE=1 
    GROUP BY reg.REGION_ID 
) AS s2 
    ON r.REGION_ID = s2.REGION_ID 
LEFT JOIN 
(
    SELECT u.Region_ID, COUNT(*) AS useCount 
    FROM `USER` u 
    WHERE u.ORG_ID IN(SELECT org.ORG_ID 
        FROM ORGANIZATION org 
        WHERE org.FK_REGION IN(SELECT REGION_ID FROM REGION reg)) 
) AS s3 
    ON r.REGION_ID = s3.REGION_ID;