【发布时间】:2017-04-17 20:32:42
【问题描述】:
我想通过以下可重现的示例在 Teradata(即CREATE RECURSIVE VIEW)中创建递归视图:
CREATE VOLATILE TABLE vt1
(
foo VARCHAR(10)
, counter INTEGER
, bar INTEGER
)
ON COMMIT PRESERVE ROWS;
INSERT INTO vt1 VALUES ('a', 1, '1');
INSERT INTO vt1 VALUES ('a', 2, '2');
INSERT INTO vt1 VALUES ('a', 3, '2');
INSERT INTO vt1 VALUES ('a', 4, '4');
INSERT INTO vt1 VALUES ('a', 5, '1');
INSERT INTO vt1 VALUES ('b', 1, '3');
INSERT INTO vt1 VALUES ('b', 2, '1');
INSERT INTO vt1 VALUES ('b', 3, '1');
INSERT INTO vt1 VALUES ('b', 4, '2');
WITH RECURSIVE cte (foo, counter, bar, rsum) AS
(
SELECT
foo
, counter
, bar
, bar AS rsum
FROM
vt1
QUALIFY ROW_NUMBER() OVER (PARTITION BY foo ORDER BY counter) = 1
UNION ALL
SELECT
t.foo
, t.counter
, t.bar
, CASE WHEN cte.rsum < 3 THEN t.bar + cte.rsum ELSE t.bar END
FROM
vt1 t JOIN cte ON t.foo = cte.foo AND t.counter = cte.counter + 1
)
SELECT
cte.*
, CASE WHEN rsum < 5 THEN 0 ELSE 1 END AS tester
FROM
cte
ORDER BY
foo
, counter
;
这将创建此输出:
╔═════╦═════════╦═════╦══════╦════════╗
║ foo ║ counter ║ bar ║ rsum ║ tester ║
╠═════╬═════════╬═════╬══════╬════════╣
║ a ║ 1 ║ 1 ║ 1 ║ 0 ║
║ a ║ 2 ║ 2 ║ 3 ║ 0 ║
║ a ║ 3 ║ 2 ║ 5 ║ 1 ║
║ a ║ 4 ║ 4 ║ 4 ║ 0 ║
║ a ║ 5 ║ 1 ║ 5 ║ 1 ║
║ b ║ 1 ║ 3 ║ 3 ║ 0 ║
║ b ║ 2 ║ 1 ║ 4 ║ 0 ║
║ b ║ 3 ║ 1 ║ 5 ║ 1 ║
║ b ║ 4 ║ 2 ║ 2 ║ 0 ║
╚═════╩═════════╩═════╩══════╩════════╝
我最终希望将其“保存”为视图。我试过CREATE RECURSIVE VIEW 和几个变种,但我想我不明白如何绕过WITH RECURSIVE cte 声明。
有关了解发生了什么的相关问题,请参阅this question
【问题讨论】:
-
语法为
create recursive view ....。这是文档的链接:info.teradata.com/HTMLPubs/DB_TTU_13_10/index.html#page/…
标签: sql recursion view teradata recursive-query