公共表表达式 (CTE)

由于业务的客观复杂性,有时候会写出长达 2000 行的单条 SQL 语句,其中包含大量的聚合和多层子查询嵌套,维护此类 SQL 堪称开发人员的噩梦。

在前面的小节当中已经介绍了如何使用视图简化查询,也介绍了如何使用临时表来缓存中间查询结果。

在这一小节当中,将介绍 TiDB 当中的公共表表达式(CTE)语法,它是一种更加便捷的复用查询结果的方法。

TiDB 从 5.1 版本开始支持 ANSI SQL 99 标准的 CTE 及其递归的写法,极大提升开发人员和 DBA 编写复杂业务逻辑 SQL 的效率,增强代码的可维护性。

基本使用

公共表表达式 (CTE) 是一个临时的中间结果集,能够在 SQL 语句中引用多次,提高 SQL 语句的可读性与执行效率。在 TiDB 中可以通过 WITH 语句使用公共表表达式。

公共表表达式可以分为非递归和递归两种类型。

非递归的 CTE

非递归的 CTE 使用如下语法进行定义:

WITH <query_name> AS ( <query_definition> ) SELECT ... FROM <query_name>;

例如,假设还想知道最年长的 50 位作家分别编写过多少书籍。

  • SQL
  • Java

在 SQL 中,可以将临时表小节当中的例子改为以下 SQL 语句:

WITH top_50_eldest_authors_cte AS ( SELECT a.id, a.name, (IFNULL(a.death_year, YEAR(NOW())) - a.birth_year) AS age FROM authors a ORDER BY age DESC LIMIT 50 ) SELECT ANY_VALUE(ta.id) AS author_id, ANY_VALUE(ta.age) AS author_age, ANY_VALUE(ta.name) AS author_name, COUNT(*) AS books FROM top_50_eldest_authors_cte ta LEFT JOIN book_authors ba ON ta.id = ba.author_id GROUP BY ta.id;

查询结果如下:

+------------+------------+---------------------+-------+ | author_id | author_age | author_name | books | +------------+------------+---------------------+-------+ | 1238393239 | 80 | Araceli Purdy | 1 | | 817764631 | 80 | Ivory Davis | 3 | | 3093759193 | 80 | Lysanne Harris | 1 | | 2299112019 | 80 | Ray Macejkovic | 4 | ... +------------+------------+---------------------+-------+ 50 rows in set (0.01 sec)

在 Java 中的示例如下:

public List<Author> getTop50EldestAuthorInfoByCTE() throws SQLException { List<Author> authors = new ArrayList<>(); try (Connection conn = ds.getConnection()) { Statement stmt = conn.createStatement(); ResultSet rs = stmt.executeQuery(""" WITH top_50_eldest_authors_cte AS ( SELECT a.id, a.name, (IFNULL(a.death_year, YEAR(NOW())) - a.birth_year) AS age FROM authors a ORDER BY age DESC LIMIT 50 ) SELECT ANY_VALUE(ta.id) AS author_id, ANY_VALUE(ta.name) AS author_name, ANY_VALUE(ta.age) AS author_age, COUNT(*) AS books FROM top_50_eldest_authors_cte ta LEFT JOIN book_authors ba ON ta.id = ba.author_id GROUP BY ta.id; """); while (rs.next()) { Author author = new Author(); author.setId(rs.getLong("author_id")); author.setName(rs.getString("author_name")); author.setAge(rs.getShort("author_age")); author.setBooks(rs.getInt("books")); authors.add(author); } } return authors; }

这时,可以发现名为 “Ray Macejkovic” 的作者写了 4 本书,继续通过 CTE 查询来了解这 4 本书的销量和评分:

WITH books_authored_by_rm AS ( SELECT * FROM books b LEFT JOIN book_authors ba ON b.id = ba.book_id WHERE author_id = 2299112019 ), books_with_average_ratings AS ( SELECT b.id AS book_id, AVG(r.score) AS average_rating FROM books_authored_by_rm b LEFT JOIN ratings r ON b.id = r.book_id GROUP BY b.id ), books_with_orders AS ( SELECT b.id AS book_id, COUNT(*) AS orders FROM books_authored_by_rm b LEFT JOIN orders o ON b.id = o.book_id GROUP BY b.id ) SELECT b.id AS `book_id`, b.title AS `book_title`, br.average_rating AS `average_rating`, bo.orders AS `orders` FROM books_authored_by_rm b LEFT JOIN books_with_average_ratings br ON b.id = br.book_id LEFT JOIN books_with_orders bo ON b.id = bo.book_id ;

查询结果如下:

+------------+-------------------------+----------------+--------+ | book_id | book_title | average_rating | orders | +------------+-------------------------+----------------+--------+ | 481008467 | The Documentary of goat | 2.0000 | 16 | | 2224531102 | Brandt Skiles | 2.7143 | 17 | | 2641301356 | Sheridan Bashirian | 2.4211 | 12 | | 4154439164 | Karson Streich | 2.5833 | 19 | +------------+-------------------------+----------------+--------+ 4 rows in set (0.06 sec)

在这个 SQL 语句,定义了三个 CTE 块,CTE 块之间使用 , 进行分隔。

先在 CTE 块 books_authored_by_rm 当中将该作者(作者 ID 为 2299112019)所编写的书查出来,然后在 books_with_average_ratingsbooks_with_orders 中分别查出这些书的平均评分和订单数,最后通过 JOIN 语句进行汇总。

值得注意的是,books_authored_by_rm 中的查询只会执行一次,TiDB 会开辟一块临时空间对查询的结果进行缓存,当 books_with_average_ratingsbooks_with_orders 引用时会直接从该临时空间当中获取数据。

递归的 CTE

递归的公共表表达式可以使用如下语法进行定义:

WITH RECURSIVE <query_name> AS ( <query_definition> ) SELECT ... FROM <query_name>;

比较经典的例子是通过递归的 CTE 生成一组斐波那契数

WITH RECURSIVE fibonacci (n, fib_n, next_fib_n) AS ( SELECT 1, 0, 1 UNION ALL SELECT n + 1, next_fib_n, fib_n + next_fib_n FROM fibonacci WHERE n < 10 ) SELECT * FROM fibonacci;

查询结果如下:

+------+-------+------------+ | n | fib_n | next_fib_n | +------+-------+------------+ | 1 | 0 | 1 | | 2 | 1 | 1 | | 3 | 1 | 2 | | 4 | 2 | 3 | | 5 | 3 | 5 | | 6 | 5 | 8 | | 7 | 8 | 13 | | 8 | 13 | 21 | | 9 | 21 | 34 | | 10 | 34 | 55 | +------+-------+------------+ 10 rows in set (0.00 sec)

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