在 postgres 中,用连接 table 执行 AND 的最佳方法是什么?

In postgres, what is the best way to do an AND with a join table?

我有不同的用户类别,还有一个连接 table 允许用户属于多个类别。我的联接 table 称为 categories_users,它由一个 user_id 和一个 category_id 组成。

我想筛选同时属于类别 1 和类别 2 的用户。例如,我想找到所有对棒球和足球都感兴趣的人。

在 PostgreSQL 中执行此操作的最佳方法是什么?我有以下工作:

select * from users 
  where users.id IN 
    ( Select categories_users.user_id from categories_users 
      JOIN categories ON categories.id = categories_users.category_id 
      where categories.id = 1 OR categories.parent_id = 1) 
  AND users.id IN 
    (Select categories_users.user_id from categories_users 
    JOIN categories ON categories.id = categories_users.category_id 
    where categories.id = 2 OR categories.parent_id = 2)

然而这感觉很笨拙,我想知道是否有更好的方法来做到这一点。我尝试了各种连接,但最终总是在 categories_users table 中搜索 category_id 为 1 和 2 的行,这是不可能的。

编辑,我实际上还需要搜索父类别,所以我更改了上面的查询以包含 parent_id

只需使用相同的 table 加入两次(使用别名):

SELECT u.*
    FROM users u
    JOIN categories_users cu1 ON cu1.user_id = u.id
    JOIN categories_users cu2 ON cu2.user_id = u.id
    WHERE cu1.category_id = 1 AND cu2.category_id = 2
select u.*
from
    users u
    inner join (
        select user_id
        from categories_users
        group by user_id
        having
            bool_or(1 in (category_id, parent_id)) and
            bool_or(2 in (category_id, parent_id))
    ) s on s.user_id = u.id

您还可以对分区使用 COUNT(*),以查看用户在搜索类别集中有多少个类别。

我创建了以下示例,以了解如何定义和参数化它。 我创建了一个函数 test.find_users_in_categories(BIGINT[]),它接受我们需要用户列表的类别数组。 因此该函数将 return 所有给定类别中的所有用户。

解决方案 - 获取在所有给定类别中找到的用户

CREATE SCHEMA test;

CREATE TABLE test.categories_users (
  category_id BIGINT NOT NULL,
  user_id BIGINT NOT NULL
);

INSERT INTO test.categories_users
  (user_id, category_id)
  VALUES
  (33, 103),
  (34, 104),
  (35, 105),
  (37, 105),
  (35, 106),
  (37, 106);

CREATE OR REPLACE FUNCTION test.find_users_in_categories(BIGINT[])
  RETURNS TABLE (
    user_id BIGINT
  )
AS
$$
DECLARE
  categories ALIAS FOR ;
BEGIN
  RETURN QUERY
  SELECT t.user_id
    FROM
      (
        SELECT
          cu.user_id,
          cu.category_id,
          COUNT(*) OVER (PARTITION BY cu.user_id ) AS cnt
        FROM test.categories_users AS cu
        WHERE cu.category_id = ANY(categories)
      ) AS t
      WHERE t.cnt = array_length(categories, 1)
      GROUP BY t.user_id;
END;
$$
LANGUAGE plpgsql;

SELECT * FROM test.find_users_in_categories(ARRAY[105, 106]);

DROP SCHEMA test CASCADE;

EDIT - [递归解决]

解决方案 - 获取在所有给定类别和子类别中找到的用户

请参阅以下有关使用 JOIN + 递归 CTE 实现解决方案的代码。我使用了 JOIN 而不是 COUNT() 因为它看起来更适合这种情况。

CREATE SCHEMA test;

CREATE TABLE test.categories (
  category_id BIGINT PRIMARY KEY,
  parent_id BIGINT REFERENCES test.categories(category_id)
);

CREATE TABLE test.categories_users (
  category_id BIGINT NOT NULL REFERENCES test.categories(category_id),
  user_id BIGINT NOT NULL
);

INSERT INTO test.categories
  (category_id, parent_id)
  VALUES
  (100, NULL),
  (101, 100),
  (102, 100),
  (103, 101),
  (104, 101),
  (105, 101),
  (106, NULL);


INSERT INTO test.categories_users
  (user_id, category_id)
  VALUES
  (33, 103),
  (34, 104),
  (35, 105),
  (37, 105),
  (35, 106),
  (37, 106);


CREATE OR REPLACE FUNCTION test.find_users_in_categories(BIGINT[])
  RETURNS TABLE (
    user_id BIGINT
  )
AS
$$
DECLARE
  main_categories ALIAS FOR ;
BEGIN
  RETURN QUERY
  WITH
    -- get all main categories and subcategories
    RECURSIVE cte_categories (category_id, main_category_id) AS
    (
      SELECT cat.category_id, cat.category_id AS main_category_id
        FROM test.categories AS cat
        WHERE cat.category_id = ANY(main_categories)
      UNION ALL
      SELECT cat.category_id, cte.main_category_id
        FROM cte_categories AS cte
        INNER JOIN test.categories AS cat
          ON cte.category_id = cat.parent_id
    ),
    -- filter main categories that are found as children of other categories
    cte_categories_unique AS
    (
      SELECT cte.*
        FROM cte_categories AS cte
        LEFT JOIN
        (
          SELECT category_id
            FROM cte_categories
            WHERE category_id <> main_category_id
            GROUP BY category_id
        ) AS to_exclude
          ON cte.main_category_id = to_exclude.category_id
        WHERE to_exclude.category_id IS NULL
    ),
    -- compute the count of main categories
    cte_main_categories_count AS
    (
      SELECT COUNT(DISTINCT main_category_id) AS cnt
        FROM cte_categories_unique
    )
  SELECT t.user_id
    FROM
      (
        -- get the users which are found in each category/sub-category then group them under the main category
        SELECT
          cu.user_id,
          cte.main_category_id
        FROM test.categories_users AS cu
        INNER JOIN cte_categories_unique AS cte
          ON cu.category_id = cte.category_id
        GROUP BY cu.user_id, cte.main_category_id
      ) AS t
      GROUP BY t.user_id
      -- filter users that do not have a match on all main categories or their sub-categories
      HAVING COUNT(*) = (SELECT cnt FROM cte_main_categories_count);
END;
$$
LANGUAGE plpgsql;


SELECT * FROM test.find_users_in_categories(ARRAY[101, 106]);

DROP SCHEMA test CASCADE;