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);
我试图在一个工作日内以JSON的形式查询所有商店及其开店情况。
{
"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] ]
}
}
}
这是我尝试过的事情的演变。我想念一种做多层json_group_object的方法。
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]]}}
如何在这里对相同的ID进行分组？
将为与group by对应的每个唯一值返回一行。因此，最外面的select必须具有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;
此内部查询返回文字JSON。解码内部JSON，然后将其全部编码到最外部的查询中，这似乎很愚蠢。
{"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]]}}"}
在Postgres中，IIRC返回JSON的内部查询不会返回原义JSON，但是无论哪种方式，我都困惑如何继续。
感谢您的帮助。
答案 0 :(得分：4)
添加示例以供一般参考。 Shawn关于在外部选择中使用json(x)的观点是关键。这是一个具有多层嵌套数组的示例
示例数据：select * from tblSmall
region|subregion |postalcode|locality |lat |lng |
------|-------------|----------|-------------------------------|-------|-------|
Delhi |Central Delhi| 110001|Connaught Place |28.6431|77.2197|
Delhi |Central Delhi| 110001|Parliament House |28.6407|77.2154|
Delhi |Central Delhi| 110003|Pandara Road |28.6431|77.2197|
Delhi |Central Delhi| 110004|Rashtrapati Bhawan |28.6453|77.2128|
Delhi |Central Delhi| 110005|Karol Bagh |28.6514|77.1907|
Delhi |Central Delhi| 110005|Anand Parbat |28.6431|77.2197|
Delhi |North Delhi | 110054|Civil Lines (North Delhi) |28.6804|77.2263|
Delhi |North Delhi | 110084|Burari |28.7557|77.1994|
Delhi |North Delhi | 110084|Jagatpur |28.7414|77.2199|
Delhi |North Delhi | 110086|Kirari Suleman Nagar |28.7441|77.0732|
对于每个region具有多个subregion值，每个subregion具有多个postalcode值，并且每个postalcode具有多个locality值。
这是sql：
select json_object('region', A2.region, 'subregions', json_group_array(json(A2.json_obj2))) from
(select A1.region, json_object('subregion',
A1.subregion,
'postalCodes',
json_group_array(json(A1.json_obj1)) ) as json_obj2 from
(select region, subregion, json_object('postalCode',
postalcode,
'localities',
json_group_array(json_object('locality',
locality, 'latitude',
lat, 'longitude', lng) ) ) as json_obj1
from tblSmall where subregion in ('Central Delhi', 'North Delhi')
group by region, subregion, postalcode) as A1
group by A1.region, A1.subregion) as A2
group by A2.region
请注意json(A1.json_obj1)和json(A2.json_obj2)位来处理来自内部查询的json的解码/重新编码。
这是结果（由于打印精美而有点久）-有一个subregions数组，其中包含一个postalcodes数组，其中包含一个localities数组：
{
"region": "Delhi",
"subregions": [
{
"subregion": "Central Delhi",
"postalCodes": [
{
"postalCode": 110001,
"localities": [
{
"locality": "Connaught Place",
"latitude": 28.6431,
"longitude": 77.2197
},
{
"locality": "Parliament House",
"latitude": 28.6407,
"longitude": 77.2154
}
]
},
{
"postalCode": 110003,
"localities": [
{
"locality": "Pandara Road",
"latitude": 28.6431,
"longitude": 77.2197
}
]
},
{
"postalCode": 110004,
"localities": [
{
"locality": "Rashtrapati Bhawan",
"latitude": 28.6453,
"longitude": 77.2128
}
]
},
{
"postalCode": 110005,
"localities": [
{
"locality": "Karol Bagh",
"latitude": 28.6514,
"longitude": 77.1907
},
{
"locality": "Anand Parbat",
"latitude": 28.6431,
"longitude": 77.2197
}
]
},
{
"postalCode": 110060,
"localities": [
{
"locality": "Rajender Nagar",
"latitude": 28.5329,
"longitude": 77.2004
}
]
},
{
"postalCode": 110069,
"localities": [
{
"locality": "Union Public Service Commission",
"latitude": 28.5329,
"longitude": 77.2004
}
]
},
{
"postalCode": 110100,
"localities": [
{
"locality": "Foreign Post Delhi IBC",
"latitude": 28.6563,
"longitude": 77.1366
}
]
}
]
},
{
"subregion": "North Delhi",
"postalCodes": [
{
"postalCode": 110054,
"localities": [
{
"locality": "Timarpur",
"latitude": 28.7038,
"longitude": 77.2227
},
{
"locality": "Civil Lines (North Delhi)",
"latitude": 28.6804,
"longitude": 77.2263
}
]
},
{
"postalCode": 110084,
"localities": [
{
"locality": "Burari",
"latitude": 28.7557,
"longitude": 77.1994
},
{
"locality": "Jagatpur",
"latitude": 28.7414,
"longitude": 77.2199
}
]
},
{
"postalCode": 110086,
"localities": [
{
"locality": "Kirari Suleman Nagar",
"latitude": 28.7441,
"longitude": 77.0732
}
]
}
]
}
]
}