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在仅使用单一公司的客户中，找到飞行最多的不同乘客。结果集：乘客姓名，旅行次数

[详情] http://i.stack.imgur.com/cBr7E.jpg [查看此内容]

我有这个查询

SELECT P.name, COUNT(T.trip_no) AS journeys, C.name as namecompany
FROM
Passenger P, Trip T, Company C, Pass_in_trip PIT
WHERE
P.ID_psg = PIT.ID_psg AND PIT.trip_no = T.trip_no
AND T.ID_comp = C.ID_comp
AND
NOT EXISTS (SELECT * FROM
Passenger, Trip, Company, Pass_in_trip
WHERE Pass_in_trip.ID_psg = P.ID_psg AND Pass_in_trip.trip_no = Trip.trip_no
AND Trip.ID_comp <> C.ID_comp)
GROUP BY P.name, C.name;

我有这个结果集：

# name, journeys, namecompany
'Alan Rickman', '1', 'Don_avia'
'George Clooney', '1', 'Dale_avia'
'Harrison Ford', '3', 'British_AW'
'Michael Caine', '4', 'British_AW'
'Mullah Omar', '4', 'British_AW'
'Nikole Kidman', '3', 'Don_avia'
'Russell Crowe', '1', 'British_AW'
'Steve Martin', '2', 'British_AW'

我想将结果分为一个同伴一个像

# name, journeys, namecompany
'Michael Caine', '4', 'British_AW'
'Nikole Kidman', '3', 'Don_avia'
'George Clooney', '1', 'Dale_avia'

Answer 1

请注意，接受的答案在技术上是错误的，因为它包含SELECT子句中未在GROUP BY中指定为聚合的列。以下解决方案使用两个子查询方法。第二个子查询获取每个公司的最大行程数，第一个子查询将每个公司名称与乘客连接。

另请注意，我重写了原始查询以使用显式连接，使其更易于阅读和维护。

SELECT t1.name, t2.jouneys, t2.namecompany
FROM
(
    SELECT P.name, C.name AS namecompany, COUNT(T.trip_no) AS journeys
    FROM Passenger P INNER JOIN Pass_in_trip PIT
        ON P.ID_psg = PIT.ID_psg
    INNER JOIN Trip T
        ON T.trip_no = PIT.trip_no
    INNER JOIN Company C
        ON C.ID_comp = T.ID_comp
    WHERE NOT EXISTS
    (
        SELECT *
        FROM
        Passenger, Trip, Company, Pass_in_trip
        WHERE Pass_in_trip.ID_psg = P.ID_psg AND Pass_in_trip.trip_no = Trip.trip_no AND
        Trip.ID_comp <> C.ID_comp
    )
    GROUP BY P.name, C.name
) t1
INNER JOIN
(
    SELECT t.namecompany, MAX(t.journeys) AS maxJourneys
    FROM
    (
        SELECT P.name, C.name AS namecompany, COUNT(T.trip_no) AS journeys
        FROM Passenger P INNER JOIN Pass_in_trip PIT
            ON P.ID_psg = PIT.ID_psg
        INNER JOIN Trip T
            ON T.trip_no = PIT.trip_no
        INNER JOIN Company C
            ON C.ID_comp = T.ID_comp
        WHERE NOT EXISTS
        (
            SELECT *
            FROM
            Passenger, Trip, Company, Pass_in_trip
            WHERE Pass_in_trip.ID_psg = P.ID_psg AND Pass_in_trip.trip_no = Trip.trip_no AND
            Trip.ID_comp <> C.ID_comp
        )
        GROUP BY P.name, C.name
    ) t
    GROUP BY t.namecompany
) t2
    ON t1.namecompany = t2.namecompany AND t1.journeys = t2.maxJourneys

Answer 2

在此查询中，您可以按公司名称获取最大值。试试这个

SELECT subquery.name, MAX(subquery.journeys),subquery.namecompany FROM
(
    SELECT P.name, COUNT(T.trip_no) AS journeys, C.name as namecompany
    FROM Passenger P, Trip T, Company C, Pass_in_trip PIT
    WHERE P.ID_psg = PIT.ID_psg AND PIT.trip_no = T.trip_no
    AND T.ID_comp = C.ID_comp
    AND NOT EXISTS (SELECT * FROM
       Passenger, Trip, Company, Pass_in_trip
       WHERE Pass_in_trip.ID_psg = P.ID_psg 
       AND Pass_in_trip.trip_no =    Trip.trip_no
       AND Trip.ID_comp <> C.ID_comp)
       GROUP BY P.name, C.name
)AS subquery GROUP BY subquery.namecompany;

选择列确定分组类型的最大计数

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2 个答案: