Question

loginid       Status
========================
jo            deleted
jo            active
jo            discard
kir           active
user1         active
user1         deleted

结果应该只给我

kir      active

休息登录ID不应该在那里。

您能否请求查询

Answer 1

这应该可以解决问题：

select *
from yourtable
where loginid in
(select loginid
 from yourtable
 group by loginid
 having count([Status]) = 1)

HAVING子句确保只选择映射了loginid个Status的{{1}}个值。请注意，此查询仅选择所需的行。如果要删除有问题的行，可以这样做：

delete 
from yourtable
where loginid in
(select loginid
 from yourtable
 group by loginid
 having count([Status]) > 1)

Answer 2

试试这个：

DELETE
FROM login_table
WHERE login_id IN
  (SELECT login_id FROM login_table GROUP BY login_id HAVING COUNT(login_id)>1
  );

Answer 3

缺少规则，但试试这个：

它返回没有其他状态而不是活动状态的所有loginid。

SELECT
    t.loginid
    ,t.Status
FROM
    table t
    LEFT JOIN 
        (SELECT
            loginid       
        FROM
            table
        WHERE
            Status <> 'active') loginDuplicatedAndNotActive
    ON t.loginid = loginDuplicatedAndNotActive.loginid
WHERE
    loginDuplicatedAndNotActive.loginid IS NULL
    AND t.Status = 'active'

Answer 4

试试这个

DELETE  si
FROM    Tbl si
JOIN    (
        SELECT  login_id,Count(status)
        FROM    tbl
        GROUP BY
                login_id 
        HAVING  COUNT(status) > 1
        ) q
ON      q.login_id = si.login_id

<强> FIDDLE DEMO

如果同一登录ID有两个值，如何删除某些行

4 个答案: