Question

我需要以递归交叉连接方式连接名称。我不知道如何做到这一点，我使用WITH RECURSIVE尝试过CTE但没有成功。

我有一张这样的表：

group_id | name
---------------
13       | A
13       | B
19       | C
19       | D
31       | E 
31       | F 
31       | G

期望的输出：

combinations
------------
ACE
ACF
ACG
ADE
ADF
ADG
BCE
BCF
BCG
BDE
BDF
BDG

当然，如果我添加第4组（或更多组），结果应该会增加。

Answer 1

Native Postgresql语法：

的 SqlFiddleDemo

WITH RECURSIVE cte1 AS ( SELECT *, DENSE_RANK() OVER (ORDER BY group_id) AS rn FROM mytable ),cte2 AS ( SELECT CAST(name AS VARCHAR(4000)) AS name, rn FROM cte1 WHERE rn = 1 UNION ALL SELECT CAST(CONCAT(c2.name,c1.name) AS VARCHAR(4000)) AS name ,c1.rn FROM cte1 c1 JOIN cte2 c2 ON c1.rn = c2.rn + 1 ) SELECT name as combinations FROM cte2 WHERE LENGTH(name) = (SELECT MAX(rn) FROM cte1) ORDER BY name;

<强>之前：

我希望你不介意我使用SQL Server语法：

样品：

CREATE TABLE #mytable( ID INTEGER NOT NULL ,TYPE VARCHAR(MAX) NOT NULL ); INSERT INTO #mytable(ID,TYPE) VALUES (13,'A'); INSERT INTO #mytable(ID,TYPE) VALUES (13,'B'); INSERT INTO #mytable(ID,TYPE) VALUES (19,'C'); INSERT INTO #mytable(ID,TYPE) VALUES (19,'D'); INSERT INTO #mytable(ID,TYPE) VALUES (31,'E'); INSERT INTO #mytable(ID,TYPE) VALUES (31,'F'); INSERT INTO #mytable(ID,TYPE) VALUES (31,'G');

主要查询：

WITH cte1 AS ( SELECT *, rn = DENSE_RANK() OVER (ORDER BY ID) FROM #mytable ),cte2 AS ( SELECT TYPE = CAST(TYPE AS VARCHAR(MAX)), rn FROM cte1 WHERE rn = 1 UNION ALL SELECT [Type] = CAST(CONCAT(c2.TYPE,c1.TYPE) AS VARCHAR(MAX)) ,c1.rn FROM cte1 c1 JOIN cte2 c2 ON c1.rn = c2.rn + 1 ) SELECT * FROM cte2 WHERE LEN(Type) = (SELECT MAX(rn) FROM cte1) ORDER BY Type;

的 LiveDemo

我假设“交叉加入”的顺序取决于升序ID。

cte1生成DENSE_RANK()，因为您的ID包含空白

cte2 CONCAT
的递归部分
主要查询只过滤掉所需的长度和排序字符串

Answer 2

Postgres中的递归查询稍微简单一些：

WITH RECURSIVE t AS (  -- to produce gapless group numbers
   SELECT dense_rank() OVER (ORDER BY group_id) AS grp, name
   FROM   tbl
   )
, cte AS (
   SELECT grp, name
   FROM   t
   WHERE  grp = 1

   UNION ALL
   SELECT t.grp, c.name || t.name
   FROM   cte c
   JOIN   t ON t.grp = c.grp + 1
   )
SELECT name AS combi
FROM   cte
WHERE  grp = (SELECT max(grp) FROM t)
ORDER  BY 1;

基本逻辑与SQL Server version provided by @lad2025中的基本逻辑相同，我添加了一些小改进。

或如果您的最大群组数量不是太大，则可以使用简单版（ 不能 > 非常大，真的，因为结果集呈指数级增长）。最多5组：

WITH t AS (  -- to produce gapless group numbers
   SELECT dense_rank() OVER (ORDER BY group_id) AS grp, name AS n
   FROM   tbl
   )
SELECT concat(t1.n, t2.n, t3.n, t4.n, t5.n) AS combi
FROM        (SELECT n FROM t WHERE grp = 1) t1
LEFT   JOIN (SELECT n FROM t WHERE grp = 2) t2 ON true
LEFT   JOIN (SELECT n FROM t WHERE grp = 3) t3 ON true
LEFT   JOIN (SELECT n FROM t WHERE grp = 4) t4 ON true
LEFT   JOIN (SELECT n FROM t WHERE grp = 5) t5 ON true
ORDER  BY 1;

对于少数群体来说可能更快。即使缺少更高的级别，LEFT JOIN .. ON true也会使这项工作成功。 concat()忽略NULL值。使用EXPLAIN ANALYZE进行测试以确定。

SQL Fiddle显示两者。

连接递归交叉连接

2 个答案: