Question

我正在尝试在PostgreSQL 8.0中进行汇总。在最新版本的postgresql中，我们具有ROLLUP函数，但是如何在postgresql 8.0中实现汇总？任何人都有相同的经验吗？

我尝试了以下

SELECT
    EXTRACT (YEAR FROM rental_date) y,
    EXTRACT (MONTH FROM rental_date) M,
    EXTRACT (DAY FROM rental_date) d,
    COUNT (rental_id)
FROM
    rental
GROUP BY
    ROLLUP (
        EXTRACT (YEAR FROM rental_date),
        EXTRACT (MONTH FROM rental_date),
        EXTRACT (DAY FROM rental_date)
    );

但是出现以下错误：

42883: function rollup( integer, integer, integer) does not exist

来自http://www.postgresqltutorial.com/postgresql-rollup/

Answer 1

version 9.5中引入了GROUP BY ROLLUP，因此该查询没有任何工作机会。但是，如果您考虑一下它的作用，那么您可以很容易地想到一个产生相同结果的版本。

基本上，您希望拥有：

总和
每年的金额
以及每月的金额
每天的计数

我以特殊的方式编写了以上内容，以便清楚地知道您的实际需求：

产生每日计数
通过每日计数每月产生总金额
每年从月度总和或每日计数中产生总和
从年度总和，每月总和或每日计数中生成总计
UNION ALL以上按您想要的顺序

由于GROUP BY ROLLUP的默认设置是先写出总数，然后用NULLS LAST写出单独的分组集，因此以下查询将执行相同的操作：

WITH
    daily AS (
        SELECT EXTRACT (YEAR FROM rental_date) y, EXTRACT (MONTH FROM rental_date) M, EXTRACT (DAY FROM rental_date) d, COUNT (rental_id) AS count
        FROM rental
        GROUP BY 1, 2, 3
    ),
    monthly AS (
        SELECT y, M, NULL::double precision d, SUM (count) AS count
        FROM daily
        GROUP BY 1, 2
    ),
    yearly AS (
        SELECT y, NULL::double precision M, NULL::double precision d, SUM (count) AS count
        FROM monthly
        GROUP BY 1
    ),
    totals AS (
        SELECT NULL::double precision y, NULL::double precision M, NULL::double precision d, SUM (count) AS count
        FROM yearly
    )
SELECT * FROM totals
UNION ALL
SELECT * FROM daily
UNION ALL
SELECT * FROM monthly
UNION ALL
SELECT * FROM yearly
;

以上适用于PostgreSQL 8.4+。如果您甚至没有该版本，那么我们必须退回到传统的UNION，而不必重新使用汇总数据：

SELECT NULL::double precision y, NULL::double precision M, NULL::double precision d, COUNT (rental_id) AS count
FROM rental
UNION ALL
SELECT EXTRACT (YEAR FROM rental_date) y, EXTRACT (MONTH FROM rental_date) M, EXTRACT (DAY FROM rental_date) d, COUNT (rental_id) AS count
FROM rental
GROUP BY 1, 2, 3
UNION ALL
SELECT EXTRACT (YEAR FROM rental_date) y, EXTRACT (MONTH FROM rental_date) M, NULL::double precision d, COUNT (rental_id) AS count
FROM rental
GROUP BY 1, 2
UNION ALL
SELECT EXTRACT (YEAR FROM rental_date) y, NULL::double precision M, NULL::double precision d, COUNT (rental_id) AS count
FROM rental
GROUP BY 1
;

Postgresql 8.0中的汇总数据层次结构？

1 个答案: