在大查询中从非结构化 JSON 中提取数据

Question

我在一个大查询字段中将 JSON 作为字符串：

[{"name":"user_group","value":"regular"},{"name":"checkout_version","value":"2.2"},{"name":"currency","value":"EUR"},{"name":"currency_exchange_rate","value":"1"},{"name":"currency_symbol","value":"€"},{"name":"variant","value":"default"},{"name":"snowplow_id","value":"XXXXXXX"},{"name":"ip_address","value":"XXXX"},{"name":"user_agent","value":"Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/102.0.0.0 Safari/537.36"},{"name":"is_test_order","value":"false"}]

我要疯了，试图从本节中提取值（“默认”）：

{"name":"variant","value":"default"}

我想要的部分将始终遵循"name":"variant","value":"并在末尾有一个" 。

我试过json_extract但regexp_extract似乎是最好的选择，我试过这个：（

select REGEXP_EXTRACT(json_string_field, r'/\{"value":"([^"]+)"/') as variant
from source_table

)

但我只是让空值回来......会很感激想法......

Answer 1

考虑以下查询

WITH json_data AS (
  SELECT '[{"name":"user_group","value":"regular"},{"name":"checkout_version","value":"2.2"},{"name":"currency","value":"EUR"},{"name":"currency_exchange_rate","value":"1"},{"name":"currency_symbol","value":"€"},{"name":"variant","value":"default"},{"name":"snowplow_id","value":"XXXXXXX"},{"name":"ip_address","value":"XXXX"},{"name":"user_agent","value":"Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/102.0.0.0 Safari/537.36"},{"name":"is_test_order","value":"false"}]' json
)
SELECT JSON_VALUE(kv, '$.value') AS value
  FROM json_data, UNNEST(JSON_QUERY_ARRAY(json)) kv
 WHERE JSON_VALUE(kv, '$.name') = 'variant';