如何按特殊条件分组

Oh *_*oon 9 sql oracle group-by

目前,当我发出这个SQL时,它会获得不同的用户名.

我有一些不同的用户名,代表组,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
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我可以实现这样的目标:

GRP_OTHERS 20
GRP_BSN 226
GRP_ASN 243
GRP_DSC 93
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编辑:从答案修改查询

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;
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zer*_*kms 12

@bfavaretto很好(给他+1),但是如果你不知道username前缀或它们是不同的,你可以使用类似的东西:

GROUP BY CASE
             WHEN REGEXP_LIKE(username, '^\d+$') THEN 'GRP_OTHERS'
                                                 ELSE username
         END
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bfa*_*tto 6

效率不是很高,但应该可以:

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;
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