Question

我有这个查询会给我那些字段“username”重复的行：

SELECT id, jos_users.username, email, password, lastvisitDate 
FROM jos_users
INNER JOIN (SELECT username 
            FROM jos_users
            GROUP BY username HAVING count(id) > 1) dup 
           ON jos_users.username = dup.username;

我需要让那些lastvisitDate更低。

例如：

id  |  username  |  email  |  password  |      lastvisitDate      |
1   |  mylogin   |         |            |   2014-10-15 16:42:42   |
2   |  mylogin   |         |            |   2014-10-16 16:42:42   |

如您所见，id = 1的行具有最低的lastvisitDate。我怎么能把这句话放在查询上？我想要这个，因为我将使用此选项执行删除查询以删除重复的行。

Answer 1

SELECT id, jos_users.username, email, password, lastvisitDate
  FROM jos_users
 INNER JOIN (SELECT username
               FROM jos_users jb
              where jb.lastvisitDate =
                    (select min(jb1.lastvisitDate)
                       from jos_users jb1
                      where jb1.username = jb.username)
              GROUP BY username
             HAVING count(id) > 1) dup
    ON jos_users.username = dup.username;

Answer 2

我可能会把它写成：

SELECT u.id, u.username, u.email, u.password, u.lastvisitDate 
FROM jos_users u
WHERE EXISTS (SELECT 1 FROM jos_users WHERE username = u.username GROUP BY username HAVING count(id) > 1)
    AND EXISTS (SELECT 1 FROM jos_users WHERE username = u.username GROUP BY username HAVING min(lastvisitDate) = u.lastvisitDate)

Answer 3

您可以获得最短的上次访问日期并进行删除

DELETE J FROM jobusers J
INNER JOIN (
            SELECT username, MIN(lastvisitDate)  as lastVisitDate
            FROM job_users
            GROUP BY username HAVING count(id) > 1 
)T
ON J.username = T.username
AND J.lastvisitDate = T.lastVisitDate

查找重复行只获取较旧的行

3 个答案: