create table store (id integer primary key, name text);
create table opening (store integer references store(id),
wday text, start integer, end integer);
insert into store (name) values ('foo'), ('bar');
insert into opening (store, wday, start, end)
values (1, 'mon', 0, 60),
(1, 'mon', 60, 120),
(1, 'tue', 180, 240),
(1, 'tue', 300, 360),
(2, 'wed', 0, 60),
(2, 'wed', 60, 120),
(2, 'thu', 180, 240);
我试图在一个查询中以 JSON 格式在工作日之前查询所有商店及其各自的营业时间。
{
"1": {
"name": "foo",
"openings": {
"mon": [ [ 0, 60 ], [ 60, 120 ] ],
"tue": [ [180, 240 ], [ 300, 360 ] ]
}
},
"2": {
"name": "bar",
"openings": {
"wed": [ [0,60], [60,120] ],
"thu": [ [180,240] ]
}
}
}
这是我尝试过的演变。我想我想念一种做多层次的方法json_group_object。
select * from opening;
store wday start end
---------- ---------- ---------- ----------
1 mon 0 60
1 mon 60 120
1 tue 180 240
1 tue 300 360
2 wed 0 60
2 wed 60 120
2 thu 180 240
select * from opening group by store;
store wday start end
---------- ---------- ---------- ----------
1 mon 0 60
2 wed 0 60
select json_group_object(store, wday) from opening group by store;
json_group_object(store, wday)
-----------------------------------------
{"1":"mon","1":"mon","1":"tue","1":"tue"}
{"2":"wed","2":"wed","2":"thu"}
select store, wday, json_group_array(json_array(start, end))
from opening group by store, wday;
store wday json_group_array(json_array(start, end))
---------- ---------- ----------------------------------------
1 mon [[0,60],[60,120]]
1 tue [[180,240],[300,360]]
2 thu [[180,240]]
2 wed [[0,60],[60,120]]
select json_object('id', store,
'openings', json_group_object(wday, json_group_array(json_array(start, end)))
) from opening group by store, wday;
Error: near line 17: misuse of aggregate function json_group_array()
select json_object('id', store,
'openings', json_object(wday, json_group_array(json_array(start, end)))
) from opening group by store, wday;
{"id":1,"openings":{"mon":[[0,60],[60,120]]}}
{"id":1,"openings":{"tue":[[180,240],[300,360]]}}
{"id":2,"openings":{"thu":[[180,240]]}}
{"id":2,"openings":{"wed":[[0,60],[60,120]]}}
我怎样才能在这里分组相同的ID？
将为与 a 对应的每个唯一值返回一行group by。因此，最外面的select必须有一个group by store.
select json_group_object(store, x)
from (
select
store,
json_object(
'id', store,
'openings', json_object(wday, json_group_array(json_array(start, end)))
) x
from opening group by store, wday
) group by store;
然而，这个内部查询返回文字 JSON。将内部 JSON 解码然后在最外面的查询中对其进行编码似乎很愚蠢。
{"1":"{\"id\":1,\"openings\":{\"mon\":[[0,60],[60,120]]}}","1":"{\"id\":1,\"openings\":{\"tue\":[[180,240],[300,360]]}}"}
{"2":"{\"id\":2,\"openings\":{\"thu\":[[180,240]]}}","2":"{\"id\":2,\"openings\":{\"wed\":[[0,60],[60,120]]}}"}
Postgres 中的 IIRC 这个返回 JSON 的内部查询不会返回文字 JSON，但无论哪种方式，我都对如何继续感到困惑。
谢谢你的帮助。