2

我有一个庞大的 MySQL 查询，它依赖于 JOIN。

SELECT m.id, l.name as location, CONCAT(u.firstName, " ", u.lastName) AS matchee, u.email AS mEmail, u.description AS description, m.time AS meetingTime
FROM matches AS m
LEFT JOIN locations AS l ON locationID=l.id 
LEFT JOIN users AS u ON (u.id=m.user1ID)
WHERE m.user2ID=2

UNION

SELECT m.id, l.name as location, CONCAT(u.firstName, " ", u.lastName) AS matchee, u.email AS mEmail, u.description AS description, m.time AS meetingTime
FROM matches AS m
LEFT JOIN locations AS l ON locationID=l.id 
LEFT JOIN users AS u ON (u.id=m.user2ID)
WHERE m.user1ID=2

每个子语句的前 3 行除以 UNION 是相同的。如何遵守 DRY 原则，不重复这三行，让这个查询更简洁？

4

3 回答 3

1

以这种方式尝试，应该可以工作：

SELECT m.id, l.name as location, CONCAT(u.firstName, " ", u.lastName) AS matchee, u.email AS mEmail, u.description AS description, m.time AS meetingTime
FROM matches AS m
LEFT JOIN locations AS l ON locationID=l.id 
LEFT JOIN users AS u 
ON ((u.id=m.user1ID AND m.user2ID=2) OR (u.id=m.user2ID AND  m.user1ID=2))
WHERE (m.user1ID=2 OR m.user2ID=2)
于 2012-04-08T18:17:50.713 回答
0 
SELECT
  s.id,
  s.location,
  CONCAT(u.firstName, " ", u.lastName) AS matchee,
  u.email AS mEmail,
  u.description AS description,
  s.meetingTime
FROM (
  SELECT
    m.id,
    l.name AS location,
    m.time AS meetingTime,
    CASE m.user1ID when @userID THEN m.user2ID ELSE m.user1ID END AS userID
  FROM matches AS m
  LEFT JOIN locations AS l ON m.locationID = l.id
  WHERE m.user1ID = @userID OR m.user2ID = @userID
) AS s
LEFT JOIN users AS u ON s.userID = u.id

子查询过滤matches指定用户 ID 上的表，将其连接到locations，从两个表中提取必要的列并准备单个userID列（使用user1ID或user2ID，取决于两者中的哪一个不是指定的 ID）与 连接users。外部查询将内部 SELECT 的结果集连接到users并提取作为连接结果获得的其余列。

于 2012-04-08T23:09:29.290 回答
-1

1）我看到您正在匹配用户，并且您保留了与该匹配相关的一些信息。

设计起来有点复杂。简单的方法就是使匹配中的数据冗余。通过这种方式，您将始终获得 user1ID=2 的匹配。意味着您需要将每个匹配项存储 2 次，一个是 user1ID 和 user2ID，另一个是 user1ID 和 user2ID 的相反值。但是，正如我所说，这是多余的，如果您期望有很多记录，我不推荐它。

2）其他方法是规范您的设计。创建另一个表 matchUser(matchId, userId) （主键都是 attrs）

SELECT ... FROM matches AS m
LEFT JOIN locations AS l ON locationID=l.id 
LEFT JOIN users AS u ON (u.id=mu2.userId)
INNER JOIN matchUser mu1 ON mu1.userId=2
INNER JOIN matchUser mu2 ON mu2.matchId=mu1.matchId AND mu2.userId != 2

像这样，查询可能是错误的，我没有测试它，但你会明白的。

于 2012-04-08T20:08:24.677 回答