Return 多个键值对的 json 而不是单键值对列表 jsons
Return a json of multiple key-value pairs instead of list of single-paired key-value jsons
目前我的查询如下所示,return结果如下:
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
结果如下所示:
61 [{"in31" : "0.0"}, {"in32" : "145.8"}, {"in34" : "134.0"}]
如何将上面的查询修改为 return 同一 JSON 对象(而不是 json 列表)中的所有密钥对值?预期输出:
61 {"in31" : "0.0", "in32" : "145.8", "in34" : "134.0"}
替换
json_agg(json_build_object(ds.statement_ref, value)) as financials
和
json_object_agg(ds.statement_ref, value) as financials
目前我的查询如下所示,return结果如下:
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
结果如下所示:
61 [{"in31" : "0.0"}, {"in32" : "145.8"}, {"in34" : "134.0"}]
如何将上面的查询修改为 return 同一 JSON 对象(而不是 json 列表)中的所有密钥对值?预期输出:
61 {"in31" : "0.0", "in32" : "145.8", "in34" : "134.0"}
替换
json_agg(json_build_object(ds.statement_ref, value)) as financials
和
json_object_agg(ds.statement_ref, value) as financials