Question

有人可以解释如何做以下事情吗？我有一张桌子人和房子。

人：

    | Person id | Person Name |
    | 1         | One         |
    | 2         | Two         |

房屋：

    | House id | Person Id | House type |
    | 1        | 1         | Small      |
    | 2        | 1         | Big        |
    | 3        | 1         | Undefined  |
    | 4        | 2         | Big        |
    | 5        | 2         | Undefined  |

我有以下情况：我需要选择拥有“Big”类型房子的人，如果他有“小”类型的房子，则不选择此人，如果他有2个类型为“big”和“Undefined”的房子，请选择此人，但不选择房屋“未定义”类型的房子的ID。

所以结果应该是（只有这一行）

| Person id | House id |
| 2         | 4        |

Answer 1

一步一步地采取行动

我需要选择拥有类型＆＃34; Big＆＃34;
的房子的人

SELECT DISTINCT PersonID, HouseID
FROM HOUSES
WHERE HouseType = 'Big'

如果他的房子类型为＆＃34;小＆＃34;，
，请不要选择此人

SELECT DISTINCT PersonID, HouseID
FROM HOUSES
WHERE HouseType = 'Big'
  AND PersonID NOT IN (SELECT PersonID FROM HOUSES WHERE HouseType = 'small')

如果他有2个类型为＆＃34; big＆＃34;和＆＃34;未定义＆＃34;

(SELECT PersonID FROM HOUSES WHERE HouseType = 'big')
UNION 
(SELECT PersonID FROM HOUSES WHERE HouseType = 'undefined')

但不选择类型＆＃34;未定义＆＃34;的房子的房屋ID。

SELECT DISTINCT PersonID, HouseID
FROM HOUSES
WHERE PersonID IN (
  (SELECT PersonID FROM HOUSES WHERE HouseType = 'big')
  UNION 
  (SELECT PersonID FROM HOUSES WHERE HouseType = 'undefined')
)
  AND HouseType != 'undefined'

Answer 2

您可以使用以下查询：

SELECT PersonId, HouseId
FROM Houses
WHERE HouseType = 'Big' AND 
      PersonId IN (SELECT PersonId
                   FROM Houses
                   GROUP BY PersonId
                   HAVING COUNT(CASE WHEN HouseType = 'Big' THEN 1 END) >= 1 AND
                          COUNT(CASE WHEN HouseType = 'Small' THEN 1 END) = 0 AND
                          COUNT(CASE WHEN HouseType = 'Undefined' THEN 1 END) >= 1)

IN运算符使用的子查询返回与“＆＃39; Big＆＃39;”相关的PersonId值。和＆＃39;未定义＆＃39;房屋类型和不与＆＃39; Small＆＃39;的。

Demo here

Answer 3

这应该有效

select case2.Person_ID, case2.House_ID
 from (
        ( -- Removes persons with Big and Small houses
            select Person_ID, count(*) as Cnt from Houses
                where House_Type in ('Big', 'Small')
            group by Person_ID
            having count(*) = 1
        ) case1
        inner join 
        (
            select Person_ID, House_ID, House_Type from Houses 
                where House_Type in ('Big', 'Undefined')
        ) case2 on case1.Person_ID = case2.Person_ID
    ) 
where House_Type <> 'Undefined'

当其中一行满足条件时，请勿选择行

3 个答案: