Question

我想将JSON编码的列表转换为本地BigQuery数组，理想情况下，最终将是具有值，位置条目的元组或字典的列表。因此，对python枚举功能的引用。

即

[(idx, elem) for idx, elem in enumerate(json_list_string)]

[{'pos':idx, 'value':elem} for idx, elem in enumerate(json_list_string)]

将json转换为数组的第一部分，我已经使用此question

解决了

编辑：

WITH
  my_ids AS (
  SELECT 'xyz' as grp, '["7f9f98fh9g4ef393d3h5", "chg3g33f26949hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' as ids
  UNION ALL
  SELECT 'abc' as grp, '["7f9f98fh9g4ef393d3h5", "chg3g33fdsfsdfs49hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' as ids
  )

SELECT
  *
FROM my_ids

在理想的世界中，我将得到如下输出：

xyz, 7f9f98fh9g4ef393d3h5, 1
xyz, chg3g33f26949hg6067d, 2
...
abc, 946637g145h48322686f, 6

请注意，列表可能会很长（最多24个条目，我有点不想对所有路径进行硬编码）

Edit2 ：（可能的解决方案）

WITH
  my_ids AS (
  SELECT 'xyz' as grp, '["7f9f98fh9g4ef393d3h5", "chg3g33f26949hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' as ids
  UNION ALL
  SELECT 'abc' as grp, '["7f9f98fh9g4ef393d3h5", "chg3g33fdsfsdfs49hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' as ids
  ),

  as_list AS (SELECT
    *,
    SPLIT(REGEXP_REPLACE(JSON_EXTRACT(ids,'$'), r'[\[\]\"]', ''), ',') AS split_items,
    GENERATE_ARRAY(1, ARRAY_LENGTH(SPLIT(REGEXP_REPLACE(JSON_EXTRACT(ids,'$'), r'[\[\]\"]', ''), ','))) AS positions
  FROM my_ids)

SELECT grp, ids, positions[OFFSET(off)] as pos
FROM as_list, unnest(split_items) as ids WITH OFFSET off

Answer 1

以下是用于BigQuery标准SQL

#standardSQL
WITH `project.dataset.my_ids` AS (
  SELECT 'xyz' AS grp, '["7f9f98fh9g4ef393d3h5", "chg3g33f26949hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' AS ids UNION ALL
  SELECT 'abc' AS grp, '["7f9f98fh9g4ef393d3h5", "chg3g33fdsfsdfs49hg6067d", "g477e5973ec04g7c3232", "0de1ec83304d761he786", "3c1h1f153530g90g35c2", "946637g145h48322686f"]' AS ids
)
SELECT grp, id, ROW_NUMBER() OVER(PARTITION BY grp ORDER BY OFFSET) pos
FROM `project.dataset.my_ids`, 
UNNEST(SPLIT(REGEXP_REPLACE(JSON_EXTRACT(ids,'$'), r'[\[\]\"]', ''), ',')) id WITH OFFSET

JSON数组上的类似于BigQuery枚举的函数

1 个答案: