Question

我在Postgres数据库中有一个JSONB列。我正在存储一组JSON对象，每个对象都有一个键值对。我确信我可以设计得更好，但是现在我已经坚持了。

id | reviews
------------------
 1 | [{"apple": "delicious"}, {"kiwi": "not-delicious"}]
 2 | [{"orange": "not-delicious"}, {"pair": "not-delicious"}]
 3 | [{"grapes": "delicious"}, {"strawberry": "not-delicious"}, {"carrot": "delicious"}]

假设此表名为tasks。虽然每个对象中的键都是不可预测的，但值是。对于每一行，我想知道reviews数组中“美味”和“不美味”值的数量。

编辑以澄清：

我正在寻找上表中每个id /行的美味/不美味的数量。样本所需的输出：

id | delicious | not_delicious
-------------------------------
 1 |         1 |             1
 2 |         0 |             2
 3 |         2 |             1

Answer 1

让我们说r是你的桌子：

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)

我使用ctid作为行标识符，因为我没有其他列而且不想要长reviews

显然每行美味聚合：

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)

更新帖子

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)

从Postgres JSONB对象数组中访问（并计数）对象值

1 个答案: