如何从 NESTED 字段中的 ARRAY 中删除 STRUCT

How to delete a STRUCT from an ARRAY in the NESTED field

有没有一种简单的方法可以从 BigQuery 中的嵌套和重复字段中删除 STRUCT(BQ table 列类型:RECORD,模式:REPEATED)。

假设我有以下 tables:

  1. 心愿单
name    toy.id  toy.priority
Alice   1       high
        2       medium
        3       low
Kazik   3       high
        1       medium
  1. 玩具
id  name    available
1   car     0
2   doll    1
3   bike    1

我想从愿望清单中删除不可用的玩具 (toys.available==0)。在这种情况下,它是 toy.id==1.

因此,愿望清单将如下所示:

name    toy.id  toy.priority
Alice   2       medium
        3       low
Kazik   3       high

我知道如何select它:

WITH `project.dataset.wishlist` AS
(
  SELECT 'Alice' name, [STRUCT<id INT64, priority STRING>(1, 'high'), (2, 'medium'), (3, 'low')] toy UNION ALL
  SELECT 'Kazik' name, [STRUCT<id INT64, priority STRING>(3, 'high'), (1, 'medium')]
), toys AS (
  SELECT 1 id, 'car' name, 0 available UNION ALL
  SELECT 2 id, 'doll' name, 1 available UNION ALL
  SELECT 3 id, 'bike' name, 1 available
)
SELECT wl.name, ARRAY_AGG(STRUCT(unnested_toy.id, unnested_toy.priority)) as toy
FROM `project.dataset.wishlist` wl, UNNEST (toy) as unnested_toy
LEFT JOIN toys t ON unnested_toy.id=t.id
WHERE t.available != 0
GROUP BY name

但是当 toys.available==0.

时,我不知道如何从愿望清单中删除结构

有非常相似的问题,例如 or How to Delete rows from Structure in bigquery,但我在删除方面不清楚答案,或者建议使用 selection 将整个心愿单复制到新的 table陈述。我的 'wishlist' 很大而且 'toys.availabililty' 经常变化。复制它在我看来效率很低。

能否建议一个符合 BQ 最佳实践的解决方案?

谢谢!

您可以 UNNEST() 并重新聚合:

SELECT wl.name, 
       (SELECT ARRAY_AGG(t)
        FROM UNNEST(wl.toy) t JOIN
             toys
             ON toys.id = t.id
        WHERE toys.available <> 0
       ) as available_toys
FROM `project.dataset.wishlist` wl;

... since row Deletion was implemented in BQ, I thought that STRUCT deletion inside a row is also possible.

您可以为此使用 UPDATE DML(不是 DELETE,因为它用于删除整行,而 UPDATE 可用于修改行)

update `project.dataset.wishlist` wl
set toy = ((
  select array_agg(struct(unnested_toy.id, unnested_toy.priority)) 
  from unnest(toy) as unnested_toy
  left join `project.dataset.toys` t on unnested_toy.id=t.id
  where t.available != 0
))
where true;