目前,当我发出此SQL时,它会获得不同的用户名。
我有一些不同的用户名,代表群组,例如GRP_BSN
。
我想将所有其他用户名(恰好是数字)分组到一个组中,例如GRP_OTHERS
select username, count(*)
from host
where seq between 0 and 2000
group by username;
63149 1
63732 1
64110 2
70987 12
76841 4
GRP_BSN 226
GRP_ASN 243
GRP_DSC 93
我可以实现这样的目标:
GRP_OTHERS 20
GRP_BSN 226
GRP_ASN 243
GRP_DSC 93
编辑:来自回答的修改后的查询
select username, count(*)
from host
where created_dt
-- date selection
between to_date('2012-may-23 00:00:00', 'yyyy-mon-dd hh24:mi:ss')
and to_date('2012-may-23 23:59:59', 'yyyy-mon-dd hh24:mi:ss')
GROUP BY CASE
WHEN REGEXP_LIKE(username, '^\d+$') THEN 'GRP_OTHERS'
ELSE username
END;
答案 0 :(得分:10)
@bfavaretto很好(给他+1),但是如果你不知道username
前缀或者它们不同,你可以选择以下内容:
GROUP BY CASE
WHEN REGEXP_LIKE(username, '^\d+$') THEN 'GRP_OTHERS'
ELSE username
END
答案 1 :(得分:3)
效率不高,但应该有效:
SELECT
CASE WHEN username LIKE 'GRP%' THEN username ELSE 'GRP_OTHERS' END AS username,
COUNT(*)
FROM host
WHERE seq BETWEEN 0 AND 2000
GROUP BY CASE WHEN username LIKE 'GRP%' THEN username ELSE 'GRP_OTHERS' END;
答案 2 :(得分:2)
如果您想通过将小组放入一个存储桶而不是特定的名称模式来实现,则可以使用:
select (case when cnt > 100 then username else 'OTHER' end), sum(cnt) as cnt
from (select username, count(*) as cnt
from host
where seq between 0 and 2000
group by username
) t
group by (case when cnt > 100 then username else 'OTHER' end)