[英]Access (and count) just object values from Postgres JSONB array of objects
我在Postgres数据库中有一个JSONB列。 我存储了一个JSON对象数组,每个对象都有一个键值对。 我敢肯定我可以设计得更好,但是现在我仍然坚持下去。
id | reviews
------------------
1 | [{"apple": "delicious"}, {"kiwi": "not-delicious"}]
2 | [{"orange": "not-delicious"}, {"pair": "not-delicious"}]
3 | [{"grapes": "delicious"}, {"strawberry": "not-delicious"}, {"carrot": "delicious"}]
假设此表称为tasks
。 尽管每个对象中的键都是不可预测的,但值却是可预测的。 对于每一行,我想知道reviews
数组中“可口”的数量和“不可口”值的数量。
编辑以澄清:
我正在寻找上表中每个id
/行的美味/不美味计数。 所需输出样本:
id | delicious | not_delicious
-------------------------------
1 | 1 | 1
2 | 0 | 2
3 | 2 | 1
假设r是您的桌子:
so=# select * from r;
reviews
-------------------------------------------------------------------------------------
[{"apple": "delicious"}, {"kiwi": "not-delicious"}]
[{"orange": "not-delicious"}, {"pair": "not-delicious"}]
[{"grapes": "delicious"}, {"strawberry": "not-delicious"}, {"carrot": "delicious"}]
(3 rows)
然后:
so=# with j as (select jsonb_array_elements(reviews) a, r, ctid from r)
select jsonb_object_keys(a), a->>jsonb_object_keys(a),ctid from j;
jsonb_object_keys | ?column? | ctid
-------------------+---------------+-------
apple | delicious | (0,1)
kiwi | not-delicious | (0,1)
orange | not-delicious | (0,2)
pair | not-delicious | (0,2)
grapes | delicious | (0,3)
strawberry | not-delicious | (0,3)
carrot | delicious | (0,3)
(7 rows)
我用ctid作为行标识符,因为我没有其他列,也不想长时间reviews
并且显然每行美味的聚集:
so=# with j as (select jsonb_array_elements(reviews) a, r, ctid from r)
select ctid, a->>jsonb_object_keys(a), count(*) from j group by a->>jsonb_object_keys(a),ctid;
ctid | ?column? | count
-------+---------------+-------
(0,1) | delicious | 1
(0,3) | delicious | 2
(0,1) | not-delicious | 1
(0,2) | not-delicious | 2
(0,3) | not-delicious | 1
(5 rows)
对于更新的帖子
so=# with j as (select jsonb_array_elements(reviews) a, r, ctid from r)
, n as (
select ctid,a->>jsonb_object_keys(a) k from j
)
, ag as (
select ctid
, case when k = 'delicious' then 1 else 0 end deli
, case when k = 'not-delicious' then 1 else 0 end notdeli
from n
)
select ctid, sum(deli) deli, sum(notdeli) notdeli from ag group by ctid;
ctid | deli | notdeli
-------+------+---------
(0,1) | 1 | 1
(0,2) | 0 | 2
(0,3) | 2 | 1
(3 rows)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.