Question

以下代码应为列lag和lag2生成相同的值：

CREATE TABLE bug1 (
  id INT,
  value INT
);

INSERT INTO bug1 VALUES (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3);

SELECT
  id,
  value,
         (LAG(value) OVER (PARTITION BY id ORDER BY value))     lag,
  NULLIF((LAG(value) OVER (PARTITION by id ORDER BY value)), 0) lag2
FROM bug1
ORDER BY id, value;

实际上，在PostgreSQL 9上运行它会产生预期的输出：

id  v  lag  lag2
1   1       
1   2   1   1
1   3   2   2
2   1       
2   2   1   1
2   3   2   2

但是，在Redshift上运行它会产生错误的结果：

id  v  lag  lag2
1   1       2
1   2   1   3
1   3   2   
2   1       2
2   2   1   3
2   3   2

这是预期的吗？我http://docs.spring.io/spring/docs/current/spring-framework-reference/html/integration-testing.html#testing-ctx-management但没有回复。

Answer 1

这不是预期的;对我来说看起来像个bug。如果打算这样做，那么使用LEAD()代替并反转ORDER BY子句，我们期望与原始LAG()查询中的结果完全相同：

SELECT
  id,
  value,
         (LEAD(value) OVER (PARTITION BY id ORDER BY value desc))     lead,
  NULLIF((LEAD(value) OVER (PARTITION by id ORDER BY value desc)), 0) lead2
FROM bug1
ORDER BY id, value;

但这是我们得到的：

 id | value | lead | lead2
----+-------+------+-------
  1 |     1 |      |
  1 |     2 |    1 |     1
  1 |     3 |    2 |     2
  2 |     1 |      |
  2 |     2 |    1 |     1
  2 |     3 |    2 |     2

NULLIF在Amazon Redshift中产生错误的输出？

1 个答案: