Sql Postgres:选择字段计数大于1的所有行
我有一个存储产品价格信息的表,该表类似于(否是主键) 现在我想选择“name”字段在表中多次出现的所有行。基本上,我希望查询返回前三行 我试过:Sql Postgres:选择字段计数大于1的所有行,sql,postgresql,Sql,Postgresql,我有一个存储产品价格信息的表,该表类似于(否是主键) 现在我想选择“name”字段在表中多次出现的所有行。基本上,我希望查询返回前三行 我试过: SELECT * FROM product_price_info GROUP BY name HAVING COUNT(*) > 1 但我得到一个错误,说: 列“product\u price\u info.no”必须出现在GROUP BY子句中,或在聚合函数中使用 试试这个: SELECT no, name, price, "date"
SELECT * FROM product_price_info GROUP BY name HAVING COUNT(*) > 1
但我得到一个错误,说:
列“product\u price\u info.no”必须出现在GROUP BY子句中,或在聚合函数中使用
试试这个:
SELECT no, name, price, "date"
FROM (
SELECT no, name, price, "date",
COUNT(*) OVER (PARTITION BY name) AS cnt
FROM product_price_info ) AS t
WHERE t.cnt > 1
您可以使用
COUNT
的窗口版本来获取每个name
分区的填充。然后,在外部查询中,筛选出填充数小于2的分区。自连接版本,使用返回多次出现的名称的子查询
select t1.*
from tablename t1
join (select name from tablename group by name having count(*) > 1) t2
on t1.name = t2.name
基本上与/中的相同
存在
版本,但可能快一点。非常适合这一点
SELECT *
FROM product_price_info
WHERE name IN (SELECT name
FROM product_price_info
GROUP BY name HAVING COUNT(*) > 1)
SELECT p.*, count(*) OVER (PARTITION BY name) FROM product p;
举一个完整的例子:
CREATE TABLE product (no SERIAL, name text, price NUMERIC(8,2), date DATE);
INSERT INTO product(name, price, date) values
('paper', 1.99, '2017-03-23'),
('paper', 2.99, '2017-05-25'),
('paper', 1.99, '2017-05-29'),
('orange', 4.56, '2017-04-23'),
('apple', 3.43, '2017-03-11')
;
WITH report AS (
SELECT p.*, count(*) OVER (PARTITION BY name) as count FROM product p
)
SELECT * FROM report WHERE count > 1;
给出:
no | name | price | date | count
----+--------+-------+------------+-------
1 | paper | 1.99 | 2017-03-23 | 3
2 | paper | 2.99 | 2017-05-25 | 3
3 | paper | 1.99 | 2017-05-29 | 3
(3 rows)
最好(如在更快的情况下)根据官方文档使用例如
COUNT(id)
。不错,但这并不能完全回答op的问题question@Madbreaks谢谢你接电话。我更新了答案。
no | name | price | date | count
----+--------+-------+------------+-------
1 | paper | 1.99 | 2017-03-23 | 3
2 | paper | 2.99 | 2017-05-25 | 3
3 | paper | 1.99 | 2017-05-29 | 3
(3 rows)