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
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)
java
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.id = rs.getLong("author_id");
author.name = rs.getString("author_name");
author.age = rs.getShort("author_age");
author.books = 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_ratings
和 books_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_ratings
和 books_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)