postgresql中如何实现group by range
更新:HHH   时间:2023-1-7


小编给大家分享一下postgresql中如何实现group by range,相信大部分人都还不怎么了解,因此分享这篇文章给大家参考一下,希望大家阅读完这篇文章后大有收获,下面让我们一起去了解一下吧!

  1. 建立测试表

testdb=# CREATE TEMP TABLE team (
id serial,
name text,
birth_year integer,
salary integer
);

  1. 插入记录

testdb=# INSERT INTO team (name, birth_year, salary)
VALUES ('Gabriel', 1970, 44000),
('Tom', 1972, 36000),
('Bill', 1978, 39500),
('Bob', 1980, 29000),
('Roger', 1976, 26800),
('Lucas', 1965, 56900),
('Jerome', 1984, 33500),
('Andrew', 1992, 41600),
('John', 1991, 40000),
('Paul', 1964, 39400),
('Richard', 1986, 23000),
('Joseph', 1988, 87000),
('Jason', 1990, 55000);

  1. 查询结果

testdb=# WITH series AS (  
SELECT generate_series(1950, 2000, 10) AS time_start -- 1950 = min, 2010 = max, 10 = 10 year interval
), range AS (
SELECT time_start, (time_start + 9) AS time_end FROM series -- 9 = interval (10 years) minus 1
)
SELECT time_start, time_end,
(SELECT count(*) FROM team WHERE birth_year BETWEEN time_start AND time_end) as team_members,
round((SELECT AVG(salary) FROM team WHERE birth_year BETWEEN time_start AND time_end), 2) as salary_avg,
(SELECT MIN(salary) FROM team WHERE birth_year BETWEEN time_start AND time_end) as salary_min,
(SELECT MAX(salary) FROM team WHERE birth_year BETWEEN time_start AND time_end) as salary_max
FROM range;

输出结果:

time_start | time_end | team_members | salary_avg | salary_min | salary_max
------------+----------+--------------+------------+------------+------------
1950 |     1959 |            0 |            |            |          
1960 |     1969 |            2 |   48150.00 |      39400 |      56900
1970 |     1979 |            4 |   36575.00 |      26800 |      44000
1980 |     1989 |            4 |   43125.00 |      23000 |      87000
1990 |     1999 |            3 |   45533.33 |      40000 |      55000
2000 |     2009 |            0 |            |            |          
(6 rows)

testdb=#

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