本地Postgresql语法:
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;
翻译后:
希望您不介意我使用SQL Server语法:
示例:
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;
实时演示
我假设“cross join”的顺序取决于升序ID。
- cte1 生成
DENSE_RANK()
,因为您的ID包含间隙
- cte2 递归部分使用
CONCAT
- 主查询只是过滤所需长度并对字符串进行排序