Postgres:选择具有大于1的字段数的所有行

Question

我有表存储产品价格信息,表看起来类似,(不是主键)

no   name    price    date
1    paper   1.99     3-23
2    paper   2.99     5-25
3    paper   1.99     5-29
4    orange  4.56     4-23
5    apple   3.43     3-11

现在我想选择"name"字段在表格中多次出现的所有行.基本上,我希望我的查询返回前三行.

我试过了:

SELECT * FROM product_price_info GROUP BY name HAVING COUNT(*) > 1  

但我得到一个错误说:

列"product_price_info.no​​"必须出现在GROUP BY子句中或用于聚合函数

Answer 1

SELECT * 
FROM product_price_info 
WHERE name IN (SELECT name 
               FROM product_price_info 
               GROUP BY name HAVING COUNT(*) > 1)

Answer 2

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

您可以使用窗口版本COUNT来获取每个name分区的填充.然后,在外部查询中,筛选出name具有小于2的填充的分区.

Answer 3

窗口函数对此非常好。

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)