我有一张表author_data
:
author_id | author_name
----------+----------------
9 | ernest jordan
14 | k moribe
15 | ernest jordan
25 | william h nailon
79 | howard jason
36 | k moribe
现在我需要结果:
author_id | author_name
----------+----------------
9 | ernest jordan
15 | ernest jordan
14 | k moribe
36 | k moribe
也就是说,对于有重复出现的名称,我需要author_id
。我试过这句话:
select author_id,count(author_name)
from author_data
group by author_name
having count(author_name)>1
但它没有用。我怎么能得到这个?
答案 0 :(得分:9)
我建议在子查询中使用window function:
SELECT author_id, author_name -- omit the name here, if you just need ids
FROM (
SELECT author_id, author_name
, count(*) OVER (PARTITION BY author_name) AS ct
FROM author_data
) sub
WHERE ct > 1;
您将识别基本聚合函数count()
。可以通过附加OVER
子句将其转换为窗口函数 - 就像任何其他聚合函数一样。
这样它会计算每个分区的行 。瞧。
在没有窗口功能(v.8.3或更早版本)的旧版本中 - 或者通常 - 这种替代方案执行速度非常快:
SELECT author_id, author_name -- omit name, if you just need ids
FROM author_data a
WHERE EXISTS (
SELECT 1
FROM author_data a2
WHERE a2.author_name = a.author_name
AND a2.author_id <> a.author_id
);
如果您关注性能,请在author_name
上添加索引。
答案 1 :(得分:2)
您可以将表连接到自身,这可以通过以下任一查询实现:
SELECT a1.author_id, a1.author_name
FROM authors a1
CROSS JOIN authors a2
ON a1.author_id <> a2.author_id
AND a1.author_name = a2.author_name;
-- 9 |ernest jordan
-- 15|ernest jordan
-- 14|k moribe
-- 36|k moribe
--OR
SELECT a1.author_id, a1.author_name
FROM authors a1
INNER JOIN authors a2
WHERE a1.author_id <> a2.author_id
AND a1.author_name = a2.author_name;
-- 9 |ernest jordan
-- 15|ernest jordan
-- 14|k moribe
-- 36|k moribe
答案 2 :(得分:1)
你已经到了一半了。您只需使用标识的Author_IDs
并获取其余数据。
试试这个..
SELECT author_id, author_name
FROM author_data
WHERE author_id in (select author_id
from author_data
group by author_name
having count(author_name)>1)