在BQ中,我使用ARRAY_AGG(STRUCT(...
来重构一些平面数据,但希望更进一步:在记录数组中创建另一个记录数组。
尽管在PostgreSQL中不存在STRUCT,但我很感兴的是如何解决这个问题。
考虑平面数据:
WITH a AS (
SELECT 'ABC' company, 'adress1' address, 'name1' name, 'email1' email, 'work' ph_type, '+123' ph_nr
UNION ALL
SELECT 'ABC' company, 'adress1' address, 'name1' name, 'email1' email, 'cell' ph_type, '+987'
UNION ALL
SELECT 'DEF' company, 'adress2' address, 'name2' name, 'email2' email, 'work' ph_type, '+127'
UNION ALL
SELECT 'DEF' company, 'adress2' address, 'name2' name, 'email2' email, 'cell' ph_type, '+988'
UNION ALL
SELECT 'XYZ' company, 'adress3' address, 'name3' name, 'email3' email, 'work' ph_type, '+456'
)
我可以像这样嵌套contact
SELECT company, address, ARRAY_AGG(STRUCT(name, email, ph_type, ph_nr)) contact
FROM a
GROUP BY company, address
ORDER BY 1
但我如何在同一个选择语句,phones
中嵌套(contact
中的记录数组)?
对于第一次接触,JSON表示看起来像是:
[
{
"company": "ABC",
"address": "adress1",
"contact": [
{
"name": "name1",
"email": "email1",
"phone": [
{
"ph_type": "work",
"ph_nr": "+123"
},
{
"ph_type": "cell",
"ph_nr": "+987"
}
},
...
这可以通过WITH
子句或子选择来顺序处理聚合,但不确定这会表现良好(数据读取两次?)。
我每天都有600M的记录要解析,所以想知道最有效的方式。
编辑:更正了名称定义
答案 0 :(得分:2)
您的问题的答案是两个级别的聚合。
然而,问题本身让我感到困惑,因为查询使用name
,但数据中没有定义。
以下是该怎么做的例子:
SELECT company, address, ARRAY_AGG(STRUCT(email, phones)) as contact
FROM (SELECT company, name, address, email, ARRAY_AGG(STRUCT(ph_type, ph_nr)) as phones
FROM a
GROUP BY company, name, address, email
) a
GROUP BY company, address
ORDER BY 1