bigquery中的多个array_agg

Posted 2023-03-24

【中文标题】bigquery中的多个array_agg

【英文标题】：Multiple array_agg in bigquery

【发布时间】：2021-08-26 19:53:32

【问题描述】：

我有一张这样的桌子：

我想根据前两个变量将信息分组到数组中，这就是我所做的

WITH sample as (
SELECT 1023 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'chop' as type,
'c1023' as id_cus, 'julian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1023 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'ext' as type,
'c1023' as id_cus, 'julian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1021 as id,10 as valuation,'inegi' as    origin, 2021-01-01 as date_lead,'ext' as type,
'in-2020' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as  mail
UNION ALL
SELECT 1021 as id,10 as valuation,'inegi' as    origin, 2021-01-01 as date_lead,'ext' as type,
'in-2020' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as  mail
UNION ALL
SELECT 1021 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'int' as type,
'c1021' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1021 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'int' as type,
'c1021' as id_cus, 'lucas' as name, '202342' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'type' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'chop' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'rad' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'uls' as type,
'c1040' as id_cus, 'julieta' as name, '123123' as phone, '[e-mail]' as mail
)
SELECT id,valuation,ARRAY_AGG(STRUCT(origin,date_lead,type,id_cus,name,phone,mail)) as lead 
FROM sample
GROUP BY id,valuation

这里的问题是我注意到我在最后三个变量（姓名、电话和邮件）中有很多重复值。 我也想将它们分组，但我不知道该怎么做。我注意到我无法在第一个数组中创建额外的 array_agg。

我希望得到这样的东西：

反正有这样的事情吗？你会怎么做？

谢谢。

【参考方案1】：

我会在原始查询周围包装另一个 SELECT ... GROUP BY，例如：

WITH sample as (
SELECT 1023 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'chop' as type,
'c1023' as id_cus, 'julian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1023 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'ext' as type,
'c1023' as id_cus, 'julian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1021 as id,10 as valuation,'inegi' as    origin, 2021-01-01 as date_lead,'ext' as type,
'in-2020' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as  mail
UNION ALL
SELECT 1021 as id,10 as valuation,'inegi' as    origin, 2021-01-01 as date_lead,'ext' as type,
'in-2020' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as  mail
UNION ALL
SELECT 1021 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'int' as type,
'c1021' as id_cus, 'lucian' as name, '12345' as phone, '[e-mail]' as    mail
UNION ALL
SELECT 1021 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'int' as type,
'c1021' as id_cus, 'lucas' as name, '202342' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'type' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'chop' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'rad' as type,
'c1040' as id_cus, 'julieta' as name, '202112' as phone, '[e-mail]' as mail
UNION ALL
SELECT 1040 as id,10 as valuation,'tlv' as  origin, 2021-01-01 as date_lead,'uls' as type,
'c1040' as id_cus, 'julieta' as name, '123123' as phone, '[e-mail]' as mail
)
SELECT id,valuation,ARRAY_AGG(STRUCT(lead,name,phone,mail)) FROM (
    SELECT id,valuation,name,phone,mail, ARRAY_AGG(STRUCT(origin,date_lead,type,id_cus)) as lead 
    FROM sample
    GROUP BY id,valuation,name,phone,mail
)
GROUP BY id,valuation