在 Teradata 中创建具有 "with recursive" 语句的递归视图
Create a recursive view that has a "with recursive" statement in Teradata
我想根据以下可重现的示例在 Teradata 中创建递归视图(即 CREATE RECURSIVE VIEW
):
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 ║
╚═════╩═════════╩═════╩══════╩════════╝
其中我最终想"save"作为一个观点。我试过 CREATE RECURSIVE VIEW
和几种变体,但我想我不明白如何绕过 WITH RECURSIVE cte
语句。
有关了解发生了什么的相关问题,请参阅
好吧,这实际上比我想象的要难:
create recursive view db.test_view (
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 < 5 THEN
t.bar + cte.rsum
ELSE t.bar
END
FROM
vt1 t
JOIN test_view cte
ON t.foo = cte.foo
AND t.counter = cte.counter + 1
)
不限定视图的递归联接。 IE,JOIN test_view
,不是 JOIN db.test_view
。
volatile 表存储在用户的假脱机 space 配额中,应使用您的用户名进行限定。
P.s。为什么首先要使用易变表?
我想根据以下可重现的示例在 Teradata 中创建递归视图(即 CREATE RECURSIVE VIEW
):
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 ║
╚═════╩═════════╩═════╩══════╩════════╝
其中我最终想"save"作为一个观点。我试过 CREATE RECURSIVE VIEW
和几种变体,但我想我不明白如何绕过 WITH RECURSIVE cte
语句。
有关了解发生了什么的相关问题,请参阅
好吧,这实际上比我想象的要难:
create recursive view db.test_view (
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 < 5 THEN
t.bar + cte.rsum
ELSE t.bar
END
FROM
vt1 t
JOIN test_view cte
ON t.foo = cte.foo
AND t.counter = cte.counter + 1
)
不限定视图的递归联接。 IE,JOIN test_view
,不是 JOIN db.test_view
。
volatile 表存储在用户的假脱机 space 配额中,应使用您的用户名进行限定。
P.s。为什么首先要使用易变表?