如何查找重复记录，而不是使用聚合函数和分组依据？

Question

db里有singers和songs表，我们要通过不使用聚合函数和关键字分组的方式来找出谁拥有最多的歌曲数。

tables in turkish

Table 翻译

song(sarkino, name, tour, duration, composerno, author no)
singer(singer, name, type, birthDate, birthPlace)
album(albumno, name, year, price, singer, stock quantity)
Song in the album(albumno, song, order)
composer(composcino, name, tour)
author(author, name)

问题原文：
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问题已翻译：

Solve the author numbers of the authors who wrote the most lyrics, as we did in the classroom (in the example in the presentation) without using the aggregate function or group by.

Answer 1

正如 Ergest 提到的，非常标准的方法是使用 GROUP BY 和 COUNT。不过好吧，今天你的老师喜欢挑战。

GarethD 的替代解决方案（完全符合要求）

我想在我的回答中回顾一下 GarethD 评论，功劳全归于他。

If the instructor considers COUNT(*) an aggregate function even when used in a windowed function, this wouldn't be a viable solution. If the only requirement is to get the singer with the most songs, then an alternative is to use ROW_NUMBER() which is almost certainly not an aggregate function by anyone's definition

SELECT s.name, 
       ROW_NUMBER() OVER (PARTITION BY s.id ORDER BY Songs.id) AS number_of_songs
FROM singers AS s
    JOIN Songs
        ON Songs.singer_id = s.id
ORDER BY number_of_songs DESC  
LIMIT 1

name	number_of_songs
Linkin Park	5

DBFiddle of GarethD

我的原始解决方案使用 COUNT OVER PARTITION BY

您需要使用 COUNT 的高级版本，结合 OVER (PARTITION BY singer_id) ，与 COUNT GROUP BY singer_id 具有相同的效果。这将统计指定分区 singer_id 上的所有歌曲记录，即总共 singer_id.

的歌曲记录

由于您添加了一堆不同 SQL 数据库类型的标签，我将选择 Postgres 来编写示例演示片段。不要担心，因为 COUNT(*) OVER PARTITION 语法对于每个 SQL 数据库

可能都是相同的

WITH singers(id, name) AS (
    VALUES(1, 'Lady Gaga'),
    (2, 'Elton John'),
    (3, 'Linkin Park')
),
songs(id, name, singer_id)
AS (
    VALUES(1, 'Born This Way', 1),
    (2, 'Rocket Man', 2),
    (3, 'Numb', 3),
    (4, 'In The End', 3),
    (5, 'Papercut', 3),
    (6, 'Can You Feel The Love Tonight?', 2),
    (7, 'Tiny Dancer', 2),
    (8, 'Your Song', 2),
    (9, 'Waiting For The End', 3),
    (10, 'Forgotten', 3)
)

SELECT singers.*, song_count.number_of_songs
FROM singers
JOIN
(SELECT DISTINCT singer_id, COUNT(*) OVER (PARTITION BY singer_id) as number_of_songs
FROM songs) song_count
ON singers.id = song_count.singer_id
ORDER BY song_count.number_of_songs DESC;

id	name	number_of_songs
3	Linkin Park	5
2	Elton John	4
1	Lady Gaga	1

SELECT子句中的DISTINCT用于删除singer_idnumber_of_songsselect

的重复记录

SELECT singer_id, COUNT(*) OVER (PARTITION BY singer_id) as number_of_songs
FROM songs;

singer_id	number_of_songs
1	1
2	4
2	4
2	4
2	4
3	5
3	5
3	5
3	5
3	5

看看这个 db fiddle here

参考 Postgresql Window Function Tutorial