Bri*_*n H 6 sqlite json sqlite-json1
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);
我正在尝试以 JSON 形式按工作日获取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。我想我缺少一种进行多层次的方法json_group_object。
{
"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] ]
}
}
}
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 * from opening;
json_group_object(store, wday)
-----------------------------------------
{"1":"mon","1":"mon","1":"tue","1":"tue"}
{"2":"wed","2":"wed","2":"thu"}
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
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 * from opening group by store;
Error: near line 17: misuse of aggregate function json_group_array()
store wday start end
---------- ---------- ---------- ----------
1 mon 0 60
2 wed 0 60
{"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]]}}
我如何在这里对相同的 id 进行分组？
对于与 对应的每个唯一值，将返回一行group by。因此，最外层select必须有一个group by store.
select json_group_object(store, wday) from opening group by store;
然而，这个内部查询返回文本 JSON。解码内部 JSON 然后将其全部编码到最外层查询中似乎很愚蠢。
{"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，我都很困惑如何继续。
谢谢你的帮助。
小智 6
添加一个示例以供一般参考。肖恩关于使用的观点json(x)在外部选择中使用的观点是关键。这是一个具有多层嵌套数组的示例
样本数据：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都有多个subregion值，每个subregion都有多个postalcode值，每个postalcode都有多个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（由于漂亮的打印而有点长） - 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
}
]
}
]
}
]
}
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