gpt4 book ai didi

MySQL 基于连接表限制结果

转载 作者:行者123 更新时间:2023-11-29 11:14:51 24 4
gpt4 key购买 nike

我有 2 个表,但以多对多关系链接,因此有 3 个表:

Table Author :
idAuthor,
Name

+----------+-------+
| idAuthor | Name |
+----------+-------+
| 1 | Renee |
| 2 | John |
| 3 | Bob |
| 4 | Bryan |
+----------+-------+
Table Publication:
idPublication,
Title,
Type,
Date,
Journal,
Conference

+---------------+--------------+------+-------------+------------+-----------+
| idPublication | Title | Date | Type | Conference | Journal |
+---------------+--------------+------+-------------+------------+-----------+
| 1 | Flower thing | 2008 | book | NULL | NULL |
| 2 | Bees | 2009 | article | NULL | Le Monde |
| 3 | Wasps | 2010 | inproceding | KDD | NULL |
| 4 | Whales | 2010 | inproceding | DPC | NULL |
| 5 | Lyon | 2011 | article | NULL | Le Figaro |
| 6 | Plants | 2012 | book | NULL | NULL |
+---------------+--------------+------+-------------+------------+-----------+
Table author_has_publication :
Author_idAuthor,
Publication_idPublication

+-----------------+---------------------------+
| Author_idAuthor | Publication_idPublication |
+-----------------+---------------------------+
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 4 |
| 1 | 5 |
| 2 | 5 |
| 3 | 5 |
| 3 | 6 |
+-----------------+---------------------------+

我想做的是让排名前 X 的作者拥有最多的出版物。

我使用以下请求获得了具有最多出​​版物的 idAuthor 的结果:SELECT Author_idAuthor, COUNT(*) as count FROM Author_has_publication GROUP BY Author_idAuthor ORDER BY count DESC;我得到了作者 ID 的列表,按出版物数量排序:

+-----------------+-------+
| Author_idAuthor | count |
+-----------------+-------+
| 3 | 3 |
| 2 | 2 |
| 1 | 2 |
| 4 | 1 |
+-----------------+-------+

但是当我尝试选择与上一个查询的结果集的前 X 相对应的作者时,出现错误我正在尝试此 SELECT TOP 2 FROM 作者 WHERE (SELECT Author_idAuthor, COUNT(*) as count FROM Author_has_publication GROUP BY Author_idAuthor ORDER BY count DESC)=idAuthor;我认为这可能是因为我的内部查询返回 2 行,并且我在这里做了一个简单的 SELECT ,或者我需要一个 JOIN ,但我不知道如何在这里使用它.

最佳答案

MySQL 没有 TOP 关键字。但它确实有一个 LIMIT 关键字。无论如何,您的查询都是无效的。

这里有几个选项。以下是相关子查询的示例:https://en.wikipedia.org/wiki/Correlated_subquery

SELECT 
a.idAuthor,
a.Name ,
(SELECT COUNT(*) from author_has_publication ahp WHERE
ahp.Author_idAuthor = a.idAuthor) AS publication_count
FROM
author a
ORDER BY
publication_count DESC
LIMIT 2

正如引用的文章所述,上述方法效率低下,因为需要对结果的每一行重新执行子查询。如果您实际上不需要结果集中的计数,那么下面的代码会更有效,因为子查询是不相关的并且仅执行一次。

SELECT 
a.idAuthor,
a.Name
FROM
author a
INNER JOIN
(select ahp.Author_idAuthor AS idAuthor, COUNT(*) as publication_count
FROM author_has_publication ahp GROUP BY ahp.Author_idAuthor LIMIT 2)
AS TEMP ON TEMP.idAuthor = a.idAuthor

关于MySQL 基于连接表限制结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39930487/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com