postgresql - 如何按各组总计对ROLLUP结果进行排序

Question

所以我有以下查询，会产生以下结果:

    actname    | year     | tickets
---------------+----------+---------
 Join Division | 2016     |       2
 Join Division | 2018     |       2
 Join Division | 2020     |       3
 Join Division | Total    |       7 <<<
 QLS           | 2018     |       2
 QLS           | 2019     |       1
 QLS           | Total    |       3 <<<
 Scalar Swift  | 2017     |       3
 Scalar Swift  | 2018     |       1
 Scalar Swift  | 2019     |       1
 Scalar Swift  | Total    |       5 <<<
 The Selecter  | 2017     |       4
 The Selecter  | 2018     |       4
 The Selecter  | Total    |       8 <<<
 The Where     | 2016     |       1
 The Where     | 2017     |       3
 The Where     | 2018     |       5
 The Where     | 2020     |       4
 The Where     | Total    |      13 <<<
 ViewBee 40    | 2017     |       3
 ViewBee 40    | 2018     |       1
 ViewBee 40    | Total    |       4 <<<

我遇到的问题是我想对结果重新排序，使总计最低的组首先出现，这样结果将如下所示:

    actname    | year     | tickets
---------------+----------+---------
 QLS           | 2018     |       2
 QLS           | 2019     |       1
 QLS           | Total    |       3 <<<
 ViewBee 40    | 2017     |       3
 ViewBee 40    | 2018     |       1
 ViewBee 40    | Total    |       4 <<<
 Scalar Swift  | 2017     |       3
 Scalar Swift  | 2018     |       1
 Scalar Swift  | 2019     |       1
 Scalar Swift  | Total    |       5 <<<
 Join Division | 2016     |       2
 Join Division | 2018     |       2
 Join Division | 2020     |       3
 Join Division | Total    |       7 <<<
 The Selecter  | 2017     |       4
 The Selecter  | 2018     |       4
 The Selecter  | Total    |       8 <<<
 The Where     | 2016     |       1
 The Where     | 2017     |       3
 The Where     | 2018     |       5
 The Where     | 2020     |       4
 The Where     | Total    |      13 <<<

我使用以下组获取结果:

GROUP BY actname, ROLLUP(year)

将同一 Actor 名称和年份的所有门票金额合并在一起。

如有必要，我可以提供完整的查询!

谢谢

Answer 1

使用窗口函数(在本例中为 sum())，您可以将值设置为组(组按 actname 列分区)，因此现在每个组都来自actname 列，与它自己的行具有相同的值，其中 year='Total'。

然后只需按新列排序，如下所示:

with t(actname, year, tickets) as (
VALUES
('Join Division','2016',2),
('Join Division','2018',2),
('Join Division','2020',3),
('Join Division','Total',7),
('QLS','2018',2),
('QLS','2019',1),
('QLS','Total',3 ),
('Scalar Swift','2017',3),
('Scalar Swift','2018',1),
('Scalar Swift','2019',1),
('Scalar Swift','Total',5 ),
('The Selecter','2017',4),
('The Selecter','2018',4),
('The Selecter','Total',8 ),
('The Where','2016',1),
('The Where','2017',3),
('The Where','2018',5),
('The Where','2020',4),
('The Where','Total',13 ),
('ViewBee 40','2017',3),
('ViewBee 40','2018',1),
('ViewBee 40','Total',4 )
)
SELECT * FROM (
    select *,  sum(case when year = 'Total' then tickets end) over(partition by actname) sm from t
) tt
ORDER BY sm, year