目前我的查询如下所示,并返回以下结果:
select
c.id as company_id,
json_agg(json_build_object(ds.statement_ref, value)) as financials
from
st.data_statements ds
join st.company_data cd on ds.company_datum_id = cd.id
join st.companies c on cd.company_id = c.id
where
c.id = 61
group by
c.id
结果如下:
61 [{"in31" : "0.0"}, {"in32" : "145.8"}, {"in34" : "134.0"}]
如何修改上面的查询以返回同一JSON对象中的所有密钥对值(而不是jsons列表)?预期产出:
61 {"in31" : "0.0", "in32" : "145.8", "in34" : "134.0"}
答案 0 :(得分:1)
替换
json_agg(json_build_object(ds.statement_ref, value)) as financials
与
json_object_agg(ds.statement_ref, value) as financials