我想跨行连接数组，然后进行不同的计数。理想情况下，这会起作用:

WITH test AS
(
  SELECT
  DATE('2018-01-01') as date,
  2 as value,
  [1,2,3] as key
  UNION ALL
  SELECT
  DATE('2018-01-02') as date,
  3 as value,
  [1,4,5] as key
)
SELECT
  SUM(value) as total_value,
  ARRAY_LENGTH(ARRAY_CONCAT_AGG(DISTINCT key)) as unique_key_count
FROM test

很遗憾，ARRAY_CONCAT_AGG 函数不支持 DISTINCT 运算符。我可以取消嵌套数组，但随后我得到一个扇出并且值列的总和是错误的:

WITH test AS
(
  SELECT
  DATE('2018-01-01') as date,
  2 as value,
  [1,2,3] as key
  UNION ALL
  SELECT
  DATE('2018-01-02') as date,
  3 as value,
  [1,4,5] as key
)

SELECT
  SUM(value) as total_value,
  COUNT(DISTINCT k) as unique_key_count

FROM test
  CROSS JOIN UNNEST(key) k

是否有任何我遗漏的东西可以让我避免加入未嵌套的数组？

最佳答案

这里有一个替代方案:

CREATE TEMP FUNCTION DistinctCount(arr ANY TYPE) AS (
  (SELECT COUNT(DISTINCT x) FROM UNNEST(arr) AS x)
);

WITH test AS
(
  SELECT
  DATE('2018-01-01') as date,
  2 as value,
  [1,2,3] as key
  UNION ALL
  SELECT
  DATE('2018-01-02') as date,
  3 as value,
  [1,4,5] as key
)

SELECT
  SUM(value) as total_value,
  DistinctCount(ARRAY_CONCAT_AGG(key)) as unique_key_count
FROM test

这避免了子查询或需要将数组与表连接(导致求和中的重复值)。

https://stackoverflow.com/questions/52485871/