тест 4 / SQL3

1

SELECT sNo, sName FROM s WHERE sNo IN ( SELECT sNo FROM sp WHERE pNo IN ( SELECT pNo FROM p WHERE color = 'Red')) ORDER BY sNo

2

SELECT DISTINCT s.sNo AS sNo, s.sName AS sName FROM s, sp, p WHERE s.sNo = sp.sNo AND sp.pNo = p.pNo AND p.color = 'Red' ORDER BY sNo

3

SELECT DISTINCT s.sNo AS sNo, s.sName AS sName FROM s LEFT JOIN sp ON (s.sNo = sp.sNo) LEFT JOIN p ON (sp.pNo = p.pNo) WHERE p.color = 'Red' ORDER BY sNo

4

SELECT sX.sNo AS sNo1, sX.sName AS sName1, sY.sNo AS sNo2, sY.sName AS sName2, sX.city AS city FROM s sX INNER JOIN s sY ON (sX.city = sY.city AND sX.sNo <> sY.sNo) ORDER BY sNo1, sNo2

5

SELECT DISTINCT s.sNo AS sNo, s.sName AS sName FROM s LEFT JOIN sp ON (s.sNo = sp.sNo) INNER JOIN p ON (sp.pNo = p.pNo) WHERE p.color = 'Red' ORDER BY sNo

6

SELECT sName, SUM (qty) AS sumQty FROM s LEFT JOIN sp USING (sNo) WHERE status > 20 GROUP BY sName ORDER BY sName

7

SELECT city, COUNT (*) AS cntS, SUM (qty) AS sumQty FROM s LEFT JOIN sp USING (sNo) WHERE status > 20 GROUP BY city ORDER BY city

8

SELECT sNo, sName FROM s WHERE NOT EXISTS ( SELECT * FROM p WHERE pNo NOT IN ( SELECT * FROM sp WHERE sp.pNo = p.pNo AND sp.sNo = s.sNo)) ORDER BY sNo

9

SELECT sNo, sName FROM s WHERE EXISTS ( SELECT * FROM sp WHERE s.sNo = sp.sNo AND EXISTS ( SELECT * FROM p WHERE sp.pNo = p.pNo AND color = 'Red')) ORDER BY sNo

10

SELECT DISTINCT s.sNo AS sNo, s.sName AS sName FROM s INNER JOIN sp ON (s.sNo = sp.sNo) INNER JOIN p ON (sp.pNo = p.pNo) WHERE p.color = 'Red' ORDER BY sNo