Question

我在数据库中有3个表，如下所示：

airlines
+----+-----------------+-------+---------+ 
| Id | Name            | Abbr  | Country |
+----+-----------------+-------+---------+
| 1  | United Airlines | UAL   | USA     |
| 2  | US Airways      | USAir | USA     |
|... | ...             | ...   | ...     |
+----+-----------------+-------+---------+

airports
+--------------+------+-------------+---------------+-------+
| City         | Code | Name        | Country       | CAbbr |
+--------------+------+-------------+---------------+-------+
| Apalachicola | AAF  | Municipal   | United States | US    |
| Abilene      | ABI  | US Airways  | United States | US    |
| ...          | ...  | ...         | ...           | ...   |
+--------------+------+-------------+---------------+-------+

flights
+---------+----------+--------+-------------+
| Airline | FlightNo | Source | Destination |
+---------+----------+--------+-------------+
| 1       | 28       | APG    | ASY         |
| 1       | 44       | CVO    | ACV         |
| ...     | ...      | ...    | ...         |
+---------+----------+--------+-------------+

我正在尝试报告Frontier和JetBlue服务的所有机场对。每个机场对必须准确地报告一次（如果报告了X，Y对，那么Y，X对就是多余的，不应报告）。

我相信我知道如何分别获得Frontier航班和JetBlue航班：

SELECT  ap1.Code, ap2.Code
FROM    flights f, airports ap1, airports ap2, airlines al
WHERE   ap1.Code = f.Source
AND     ap2.Code = f.Destination
AND     f.Airline = al.Id
AND     al.Abbr = 'Frontier'
ORDER BY ap1.Code ASC;

SELECT  ap1.Code, ap2.Code
FROM    flights f, airports ap1, airports ap2, airlines al
WHERE   ap1.Code = f.Source
AND     ap2.Code = f.Destination
AND     f.Airline = al.Id
AND     al.Abbr = 'JetBlue'
ORDER BY ap1.Code ASC;

但是我不明白如何获得它们之间的交点。我已经尝试了LEFT JOIN和WHERE EXISTS，但是我做错了，因为我得到的看起来更像是一个Union，而不是一个Intersection。这是我最近的尝试（肯定是错误的）。

SELECT  ap1.Code, ap2.Code AS code
FROM    airports ap1, airports ap2
WHERE 
EXISTS (SELECT  ap1.Code, ap2.Code
        FROM    flights f, airports ap1, airports ap2, airlines al
        WHERE   ap1.Code = f.Source
        AND     ap2.Code = f.Destination
        AND     f.Airline = al.Id
        AND     al.Abbr = 'Frontier') AND
EXISTS (SELECT  ap1.Code, ap2.Code
        FROM    flights f, airports ap1, airports ap2, airlines al
        WHERE   ap1.Code = f.Source
        AND     ap2.Code = f.Destination
        AND     f.Airline = al.Id
        AND     al.Abbr = 'JetBlue')
ORDER BY ap1.Code ASC;

一旦找到交集，我就不太确定如何区分X，Y对和Y，X对，因为DISTINCT不会将元组识别为相同的元组。

任何帮助将不胜感激。

Answer 1

如果同时需要两者，则可以尝试使用IN航空公司

SELECT  ap1.Code, ap2.Code
FROM    flights f
INNER JOIN airports ap1 ON  ap1.Code = f.Source
INNER JOIN airports ap2 ON ap2.Code = f.Destination
INNER JOIN airlines al ON  f.Airline = al.Id
WHERE  al.Abbr IN ( 'Frontier', 'JetBlue')
ORDER BY ap1.Code ASC;

您应该避免基于位置而使用old（1992）隐式联接sintax并使用显式联接sintax

Answer 2

SELECT apPairs.*, ap1.stuff, ap2.stuff
FROM (
   SELECT LEAST(f.Source, f.Destination) AS ap1Code
      , GREATEST(f.Source, f.Destination) AS ap2Code
   FROM airlines AS al
   INNER JOIN flights AS f ON al.Id = f.Airline
   WHERE al.Abbr IN ('Frontier', 'JetBlue')
   GROUP BY ap1Code, ap2Code
   HAVING COUNT(DISTINCT al.Abbr) = 2  -- Served by both airlines
) AS apPairs
INNER JOIN airports AS ap1 ON apPairs.ap1Code = ap1.Code
INNER JOIN airports AS ap2 ON apPairs.ap1Code = ap2.Code
;

我使用LEAST和GREATEST合并了所有（X，Y）和（Y，X）冗余；这些功能是特定于MySQL的，因此在其他RDBMS中，您可能需要创建自己的版本。注意：根据数据的使用方式，您可能需要仔细检查以下假设：x-> y和y-> x对确实是多余的；航空公司可能（至少在现实中）拥有X-> Y-> Z-> X航线，但没有直接的Y-> X。

HAVING子句中的

COUNT（DISTINCT）确保仅对每个航空公司的缩写计数一次，因此100个航班不会使数字膨胀。它在查询中的使用可确保您获得的是两家航空公司提供的机场，而不是两家航空公司。

由于apPairs中的查询已经有了代码，因此您甚至不需要执行外部查询。但是我补充说，以防您需要机场表中的信息，而您又不需要在问题中指定该信息。

如何在MySQL中相交两个查询

2 个答案: