如何在PostgreSQL中找到重复的记录

基本思想将使用一个嵌套查询计数聚合:

select * from yourTable ou
where (select count(*) from yourTable inr
where inr.sid = ou.sid) > 1

您可以调整内部查询中的where子句来缩小搜索范围。

评论中提到了另一个很好的解决方案，(但不是每个人都读它们):

select Column1, Column2, count(*)
from yourTable
group by Column1, Column2
HAVING count(*) > 1

或更短:

SELECT (yourTable.*)::text, count(*)
FROM yourTable
GROUP BY yourTable.*
HAVING count(*) > 1

您可以在将被复制的字段上连接到同一个表，然后在id字段上反连接。从第一个表别名(tn1)中选择id字段，然后对第二个表别名的id字段使用array_agg函数。最后，为了使array_agg函数正常工作，您将根据tn1对结果进行分组。id字段。这将产生一个结果集，其中包含记录的id和符合连接条件的所有id的数组。

select tn1.id,
array_agg(tn2.id) as duplicate_entries,
from table_name tn1 join table_name tn2 on
tn1.year = tn2.year
and tn1.sid = tn2.sid
and tn1.user_id = tn2.user_id
and tn1.cid = tn2.cid
and tn1.id <> tn2.id
group by tn1.id;

显然，在duplicate_entries数组中的id在结果集中也有自己的条目。你必须使用这个结果集来决定你想要哪个id成为“真相”的来源。唯一不应该被删除的记录。也许你可以这样做:

with dupe_set as (
select tn1.id,
array_agg(tn2.id) as duplicate_entries,
from table_name tn1 join table_name tn2 on
tn1.year = tn2.year
and tn1.sid = tn2.sid
and tn1.user_id = tn2.user_id
and tn1.cid = tn2.cid
and tn1.id <> tn2.id
group by tn1.id
order by tn1.id asc)
select ds.id from dupe_set ds where not exists
(select de from unnest(ds.duplicate_entries) as de where de < ds.id)

选择具有重复的最小数字ID(假设ID在PK中递增)。这些就是你要保存的ID。

从"使用PostgreSQL查找重复的行"这里有一个聪明的解决方案:

select * from (
SELECT id,
ROW_NUMBER() OVER(PARTITION BY column1, column2 ORDER BY id asc) AS Row
FROM tbl
) dups
where
dups.Row > 1

为了方便起见，我假设您希望仅对列year应用唯一约束，并且主键是名为id的列。

为了找到重复的值，您应该运行，

SELECT year, COUNT(id)
FROM YOUR_TABLE
GROUP BY year
HAVING COUNT(id) > 1
ORDER BY COUNT(id);

使用上面的sql语句，您将得到一个包含表中所有重复年份的表。为了删除除最新重复项外的所有重复项，你应该使用上面的sql语句。

DELETE
FROM YOUR_TABLE A USING YOUR_TABLE_AGAIN B
WHERE A.year=B.year AND A.id<B.id;

在您的情况下，由于限制，您需要删除重复的记录。

1. 查找重复的行
2. 根据created_at日期组织它们——在这种情况下，我保留了最老的日期
3. 删除带有USING的记录以过滤正确的行

WITH duplicated AS (
SELECT id,
count(*)
FROM products
GROUP BY id
HAVING count(*) > 1),
ordered AS (
SELECT p.id,
created_at,
rank() OVER (partition BY p.id ORDER BY p.created_at) AS rnk
FROM products o
JOIN     duplicated d ON d.id = p.id ),
products_to_delete AS (
SELECT id,
created_at
FROM   ordered
WHERE  rnk = 2
)
DELETE
FROM products
USING products_to_delete
WHERE products.id = products_to_delete.id
AND products.created_at = products_to_delete.created_at;

受到Sandro Wiggers的启发，我做了一些类似的事情

WITH ordered AS (
SELECT id,year, user_id, sid, cid,
rank() OVER (PARTITION BY year, user_id, sid, cid ORDER BY id) AS rnk
FROM user_links
),
to_delete AS (
SELECT id
FROM   ordered
WHERE  rnk > 1
)
DELETE
FROM user_links
USING to_delete
WHERE user_link.id = to_delete.id;

如果你想测试它，稍微改变一下:

WITH ordered AS (
SELECT id,year, user_id, sid, cid,
rank() OVER (PARTITION BY year, user_id, sid, cid ORDER BY id) AS rnk
FROM user_links
),
to_delete AS (
SELECT id,year,user_id,sid, cid
FROM   ordered
WHERE  rnk > 1
)
SELECT * FROM to_delete;

这将给出将要删除的内容的概述(在运行删除时，在to_delete查询中保留year,user_id,sid,cid是没有问题的，但随后它们就不需要了)

begin;
create table user_links(id serial,year bigint, user_id bigint, sid bigint, cid bigint);
insert into  user_links(year, user_id, sid, cid) values (null,null,null,null),
(null,null,null,null), (null,null,null,null),
(1,2,3,4), (1,2,3,4),
(1,2,3,4),(1,1,3,8),
(1,1,3,9),
(1,null,null,null),(1,null,null,null);
commit;

使用distinct和except设置操作。

(select id, year, user_id, sid, cid from user_links order by 1)
except
select distinct on (year, user_id, sid, cid) id, year, user_id, sid, cid
from user_links order by 1;

除了所有工作。因为id serial使所有行都是唯一的。

(select id, year, user_id, sid, cid from user_links order by 1)
except all
select distinct on (year, user_id, sid, cid)
id, year, user_id, sid, cid  from user_links order by 1;

到目前为止工作为空值和非空值。
删除:< / >强

with a as(
(select id, year, user_id, sid, cid from user_links order by 1)
except all
select distinct on (year, user_id, sid, cid)
id, year, user_id, sid, cid  from user_links order by 1)
delete from user_links using a  where user_links.id = a.id returning *;