BigQuery：在标准 SQL 中使用重复/数组 STRUCT 字段加入？

Question

我基本上有两张桌子，Orders 和 Items。由于这些表是从 Google Cloud Datastore 备份文件导入的，因此引用不是通过简单的 ID 字段进行的，而是通过 <STRUCT> 进行一对一关系，其中其 id 字段表示我想要匹配的实际唯一 ID .对于一对多关系（REPEATED），架构使用<STRUCT> 的数组。

我可以使用 LEFT OUTER JOIN 查询一对一的关系，我也知道如何加入非重复结构和重复的字符串或 int，但我无法使用重复结构。

一个订单，一个项目：

#standardSQL
WITH Orders AS (
  SELECT 1 AS __oid__, STRUCT(STRUCT(2 AS id, "default" AS ns) AS key) AS item UNION ALL 
  SELECT 2 AS __oid__, STRUCT(STRUCT(4 AS id, "default" AS ns) AS key) AS item UNION ALL 
  SELECT 3 AS __oid__, STRUCT(STRUCT(6 AS id, "default" AS ns) AS key) AS item
),
Items AS (
  SELECT STRUCT(1 AS id, "default" AS ns) AS key, "#1.1" AS title UNION ALL
  SELECT STRUCT(2 AS id, "default" AS ns) AS key, "#1.2" AS title UNION ALL
  SELECT STRUCT(3 AS id, "default" AS ns) AS key, "#1.3" AS title UNION ALL
  SELECT STRUCT(4 AS id, "default" AS ns) AS key, "#1.4" AS title UNION ALL
  SELECT STRUCT(5 AS id, "default" AS ns) AS key, "#1.5" AS title UNION ALL
  SELECT STRUCT(6 AS id, "default" AS ns) AS key, "#1.6" AS title
)

SELECT
   __oid__
  ,Order_item AS item
FROM Orders  

LEFT OUTER JOIN(
  SELECT
     key
    ,title
  FROM Items
) Order_item
ON Order_item.key.id = item.key.id

结果（按预期工作）：

+-----+---------+--------------+-------------+------------+
| Row | __oid__ |  item.key.id | item.key.ns | item.title |
+-----+---------+--------------+-------------+------------+
|   1 |       1 |            2 |     default |       #1.2 |
+-----+---------+--------------+-------------+------------+
|   2 |       2 |            4 |     default |       #1.4 |
+-----+---------+--------------+-------------+------------+
|   3 |       3 |            6 |     default |       #1.6 |
+-----+---------+--------------+-------------+------------+

类似的查询，但这次是一个带有 许多 项的订单：

#standardSQL
WITH Orders AS (
  SELECT 1 AS __oid__, ARRAY[STRUCT(STRUCT(1 AS id, "default" AS ns) AS key), STRUCT(STRUCT(2 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 2 AS __oid__, ARRAY[STRUCT(STRUCT(3 AS id, "default" AS ns) AS key), STRUCT(STRUCT(4 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 3 AS __oid__, ARRAY[STRUCT(STRUCT(5 AS id, "default" AS ns) AS key), STRUCT(STRUCT(6 AS id, "default" AS ns) AS key)] AS items
),
Items AS (
  SELECT STRUCT(1 AS id, "default" AS ns) AS key, "#1.1" AS title UNION ALL
  SELECT STRUCT(2 AS id, "default" AS ns) AS key, "#1.2" AS title UNION ALL
  SELECT STRUCT(3 AS id, "default" AS ns) AS key, "#1.3" AS title UNION ALL
  SELECT STRUCT(4 AS id, "default" AS ns) AS key, "#1.4" AS title UNION ALL
  SELECT STRUCT(5 AS id, "default" AS ns) AS key, "#1.5" AS title UNION ALL
  SELECT STRUCT(6 AS id, "default" AS ns) AS key, "#1.6" AS title
)

SELECT
   __oid__
  ,Order_items AS items
FROM Orders  

LEFT OUTER JOIN(
  SELECT
     key
    ,title
  FROM Items
) Order_items
ON Order_items.key.id IN (SELECT item.key.id FROM UNNEST(items) AS item)

错误：在连接谓词中不支持 IN 子查询。

我其实预料到了这个结果：

+-----+---------+--------------+-------------+------------+
| Row | __oid__ |  item.key.id | item.key.ns | item.title |
+-----+---------+--------------+-------------+------------+
|   1 |       1 |            1 |     default |       #1.1 |
|     |         |            2 |     default |       #1.2 |
+-----+---------+--------------+-------------+------------+
|   2 |       2 |            3 |     default |       #1.3 |
|     |         |            4 |     default |       #1.4 |
+-----+---------+--------------+-------------+------------+
|   3 |       3 |            5 |     default |       #1.5 |
|     |         |            6 |     default |       #1.6 |
+-----+---------+--------------+-------------+------------+

如何更改第二个查询以获得预期结果？

Answer 1

另一种选择是使用 CROSS JOIN 而不是 LEFT JOIN

#standardSQL
WITH Orders AS (
  SELECT 1 AS __oid__, ARRAY[STRUCT(STRUCT(1 AS id, "default" AS ns) AS key), STRUCT(STRUCT(2 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 2 AS __oid__, ARRAY[STRUCT(STRUCT(3 AS id, "default" AS ns) AS key), STRUCT(STRUCT(4 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 3 AS __oid__, ARRAY[STRUCT(STRUCT(5 AS id, "default" AS ns) AS key), STRUCT(STRUCT(6 AS id, "default" AS ns) AS key)] AS items
),
Items AS (
  SELECT STRUCT(1 AS id, "default" AS ns) AS key, "#1.1" AS title UNION ALL
  SELECT STRUCT(2 AS id, "default" AS ns) AS key, "#1.2" AS title UNION ALL
  SELECT STRUCT(3 AS id, "default" AS ns) AS key, "#1.3" AS title UNION ALL
  SELECT STRUCT(4 AS id, "default" AS ns) AS key, "#1.4" AS title UNION ALL
  SELECT STRUCT(5 AS id, "default" AS ns) AS key, "#1.5" AS title UNION ALL
  SELECT STRUCT(6 AS id, "default" AS ns) AS key, "#1.6" AS title
)

SELECT
   __oid__
  ,ARRAY_AGG(Order_items) AS items
FROM Orders  

CROSS JOIN(
  SELECT
     key
    ,title
  FROM Items
) Order_items
WHERE Order_items.key.id IN (SELECT item.key.id FROM UNNEST(items) AS item)
GROUP BY __oid__

Answer 2

问题在于 BigQuery 无法从两侧对连接键进行哈希分区（因为连接表示为 IN 条件）。您可以通过展平左侧的数组然后从右侧聚合项目来完成这项工作：

#standardSQL
WITH Orders AS (
  SELECT 1 AS __oid__, ARRAY[STRUCT(STRUCT(1 AS id, "default" AS ns) AS key), STRUCT(STRUCT(2 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 2 AS __oid__, ARRAY[STRUCT(STRUCT(3 AS id, "default" AS ns) AS key), STRUCT(STRUCT(4 AS id, "default" AS ns) AS key)] AS items UNION ALL 
  SELECT 3 AS __oid__, ARRAY[STRUCT(STRUCT(5 AS id, "default" AS ns) AS key), STRUCT(STRUCT(6 AS id, "default" AS ns) AS key)] AS items
),
Items AS (
  SELECT STRUCT(1 AS id, "default" AS ns) AS key, "#1.1" AS title UNION ALL
  SELECT STRUCT(2 AS id, "default" AS ns) AS key, "#1.2" AS title UNION ALL
  SELECT STRUCT(3 AS id, "default" AS ns) AS key, "#1.3" AS title UNION ALL
  SELECT STRUCT(4 AS id, "default" AS ns) AS key, "#1.4" AS title UNION ALL
  SELECT STRUCT(5 AS id, "default" AS ns) AS key, "#1.5" AS title UNION ALL
  SELECT STRUCT(6 AS id, "default" AS ns) AS key, "#1.6" AS title
)

SELECT
   __oid__
  ,ARRAY_AGG(Order_items) AS items
FROM Orders,
UNNEST(items) AS item

LEFT OUTER JOIN(
  SELECT
     key
    ,title
  FROM Items
) Order_items
ON Order_items.key.id = item.key.id
GROUP BY __oid__

无论如何，这看起来都是您想要的，因为您的原始查询将 items 作为一个结构而不是一个结构数组。