1

我认为ARRAY_AGGBigQuery 中的函数似乎在ORDER BY. 这里有一些 SQL 来解释：

#standardSQL
WITH t1 AS (
  SELECT *
  FROM UNNEST ( [
    STRUCT(1 AS user_id, 1 AS team_id, "2018-07-17" AS date_str),
    (  2, 1, "2018-07-17" ),
    (  3, 1, "2018-07-17" ),
    (  4, 1, "2018-07-17" ),
    (  5, 1, "2018-07-17" ),
    (  6, 1, "2018-07-17" ),
    (  7, 1, "2018-07-17" ),
    (  8, 2, "2018-07-17" ),
    (  9, 2, "2018-07-17" ),
    ( 10, 2, "2018-07-17" ),
    ( 11, 2, "2018-07-17" ),
    ( 14, 3, "2018-07-17" ),
    ( 15, 3, "2018-07-17" ),
    ( 16, 3, "2018-07-17" ),
    ( 17, 3, "2018-07-17" ),
    (  1, 1, "2018-07-18" ),
    (  4, 1, "2018-07-18" ),
    (  5, 1, "2018-07-18" ),
    (  6, 1, "2018-07-18" ),
    (  7, 1, "2018-07-18" ),
    (  8, 2, "2018-07-18" ),
    (  9, 2, "2018-07-18" ),
    ( 10, 2, "2018-07-18" ),
    ( 11, 2, "2018-07-18" ),
    ( 12, 2, "2018-07-18" ),
    ( 13, 2, "2018-07-18" ),
    ( 14, 3, "2018-07-18" ),
    ( 15, 3, "2018-07-18" ),
    ( 16, 3, "2018-07-18" ),
    ( 17, 3, "2018-07-18" ),
    ( 18, 3, "2018-07-18" ) ] ) )

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM (
  SELECT
    date_str,
    users,
    COUNT(*) AS teams
  FROM (
    SELECT
      date_str,
      team_id,
      COUNT(*) AS users
    FROM t1
    GROUP BY date_str, team_id
  )
  GROUP BY date_str, users
)
GROUP BY date_str
ORDER BY date_str;

此查询返回；

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  1 |  4 |  1 |  4 |        2 |        4 |
|     |            |  2 |  7 |  2 |  7 |        1 |        7 |
|   2 | 2018-07-18 |  1 |  5 |  1 |  5 |        2 |        5 |
|     |            |  2 |  6 |  2 |  6 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

但我希望是；

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  2 |  4 |  1 |  7 |        2 |        4 |
|     |            |  1 |  7 |  2 |  4 |        1 |        7 |
|   2 | 2018-07-18 |  2 |  5 |  1 |  6 |        2 |        5 |
|     |            |  1 |  6 |  2 |  5 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

似乎函数中的ORDER BY子句ARRAY_AGG不能正常工作，因为a1并且a4顺序错误。

此外，当我用or替换两个COUNT(*)部分中的任何一个时，很难理解查询完全按预期工作，这意味着；COUNT(user_id)COUNT(team_id)

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM (
  SELECT
    date_str,
    users,
    COUNT(*) AS teams
  FROM (
    SELECT
      date_str,
      team_id,
      COUNT(user_id) AS users
    FROM t1
    GROUP BY date_str, team_id
  )
  GROUP BY date_str, users
)
GROUP BY date_str
ORDER BY date_str;

或者

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM (
  SELECT
    date_str,
    users,
    COUNT(team_id) AS teams
  FROM (
    SELECT
      date_str,
      team_id,
      COUNT(*) AS users
    FROM t1
    GROUP BY date_str, team_id
  )
  GROUP BY date_str, users
)
GROUP BY date_str
ORDER BY date_str;

据我了解，在这种情况下，这些查询必须返回与原始查询相同的结果。这对我来说很困惑。可能是错误或我误解的东西？

一些额外的信息。

内部子查询；

SELECT
  date_str,
  users,
  COUNT(*) AS teams
FROM (
  SELECT
    date_str,
    team_id,
    COUNT(*) AS users
  FROM t1
  GROUP BY date_str, team_id
)
GROUP BY date_str, users

这返回；

+-----+------------+-------+-------+
| Row | date_str   | users | teams |
+-----+------------+-------+-------+
|   1 | 2018-07-18 |     5 |     2 |
|   2 | 2018-07-17 |     7 |     1 |
|   3 | 2018-07-18 |     6 |     1 |
|   4 | 2018-07-17 |     4 |     2 |
+-----+------------+-------+-------+

因此，直接通过 with 子句创建这些数据并运行相同的聚合查询；

#standardSQL
With t2 AS (
  SELECT *
  FROM UNNEST ( [
    STRUCT("2018-07-18" AS date_str, 5 AS users, 2 AS teams),
    (  "2018-07-17", 7, 1 ),
    (  "2018-07-18", 6, 1 ),
    (  "2018-07-17", 4, 2 ) ] )
)

SELECT
  date_str,
  ARRAY_AGG(teams ORDER BY users) AS a1,
  ARRAY_AGG(users ORDER BY users) AS a2,
  ARRAY_AGG(teams ORDER BY teams) AS a3,
  ARRAY_AGG(users ORDER BY teams) AS a4,
  ARRAY_AGG(STRUCT(teams, users) ORDER BY users) AS a5
FROM t2
GROUP BY date_str
ORDER BY date_str;

结果变成了我想要的；

+-----+------------+----+----+----+----+----------+----------+
| Row | date_str   | a1 | a2 | a3 | a4 | a5.teams | a5.users |
+-----+------------+----+----+----+----+----------+----------+
|   1 | 2018-07-17 |  2 |  4 |  1 |  7 |        2 |        4 |
|     |            |  1 |  7 |  2 |  4 |        1 |        7 |
|   2 | 2018-07-18 |  2 |  5 |  1 |  6 |        2 |        5 |
|     |            |  1 |  6 |  2 |  5 |        1 |        6 |
+-----+------------+----+----+----+----+----------+----------+

我不明白发生这种情况的原因。我完全不解。任何想法或建议表示赞赏。

4

1 回答 1

1

对不起，如果我误解了，但默认顺序是升序，所以它排序正确？

ARRAY_AGG(teams ORDER BY users desc) AS a1,
ARRAY_AGG(users ORDER BY users) AS a2,
ARRAY_AGG(teams ORDER BY teams) AS a3,
ARRAY_AGG(users ORDER BY teams desc) AS a4,

如果我将它们更改为降序排序，我会得到想要的结果

于 2018-09-14T08:50:19.913 回答