我试图获得一个旅行的实时温度，如果没有实时数据，则从天体数据得到平均温度。 我用这些表格制作了我的问题的简单版本：
旅行
id departure_time arrival_time location_id
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1
位置
id name
1 Location
天气
id temperature date location_id
1 20 2018-04-07 1
2 20 2018-04-08 1
3 20 2018-04-09 1
4 20 2018-04-10 1
5 20 2018-04-11 1
6 20 2018-04-12 1
7 20 2018-04-13 1
8 20 2018-04-14 1
9 15 2016-04-07 1
10 15 2016-04-08 1
11 15 2016-04-09 1
12 15 2016-04-10 1
13 15 2016-04-11 1
14 15 2016-04-12 1
15 15 2016-04-13 1
16 15 2016-04-14 1
17 19 2017-04-07 1
18 19 2017-04-08 1
19 19 2017-04-09 1
20 19 2017-04-10 1
21 19 2017-04-11 1
22 19 2017-04-12 1
23 19 2017-04-13 1
24 19 2017-04-14 1
25 15 2017-04-15 1
26 15 2017-04-16 1
27 15 2017-04-17 1
28 15 2017-04-18 1
29 15 2017-04-19 1
30 15 2017-04-20 1
31 15 2017-04-21 1
32 19 2016-04-15 1
33 19 2016-04-16 1
34 19 2016-04-17 1
35 19 2016-04-18 1
36 19 2016-04-19 1
37 19 2016-04-20 1
38 19 2016-04-21 1
我遇到的问题是，由于这些旅行是最后一分钟旅行，我有＆＃34;生活＆＃34;下周内出发的旅行数据。 所以我希望得到一个现场预测（如果有的话），否则就可以得到前几年的温度。
http://sqlfiddle.com/#!17/bce59/3
以下是我尝试解决问题的方法。
如果忘记了任何细节，请询问。
预期结果：
id departure_time arrival_time location_id temperature
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
1 2018-04-07 07:00:00 2018-04-14 17:00:00 1 20
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 20
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
2 2018-04-14 07:00:00 2018-04-21 17:00:00 1 17
答案 0 :(得分：1)
使用generate_series函数在子查询上从trip表创建日历。
然后在Left JOIN的子查询上dates，您可能会得到匹配weather，您可以获得温度。如果temperature上的w.temperature为空，则获取avg temperature
你可以试试这个。
SELECT t.id,
t.departure_time,
t.arrival_time,
l.id as "location_id",
coalesce(w.temperature,(select FLOOR(avg(temperature)) from weather)) as "temperature"
FROM
location l inner join
(
select id,
location_id,
departure_time,
arrival_time,
generate_series(departure_time :: timestamp,arrival_time::timestamp,'1 day'::interval) as dates
from trip
) t on t.location_id = l.id LEFT JOIN weather w on t.dates::date = w.date::date
sqlfiddle：http://sqlfiddle.com/#!17/bce59/48
修改
您可以使用 CTE 查询Avg获取year而不是select子句中coalesce函数中的子查询。
WITH weather_avg AS (
SELECT floor(avg(a)) avgTemp
from
(
SELECT
extract(YEAR from weather.date) AS YEAR,
floor(avg(weather.temperature)) a
FROM weather
group by extract(YEAR from weather.date)
) t
)
SELECT t.id,
t.departure_time,
t.arrival_time,
t.location_id as "location_id",
coalesce(w.temperature,(select avgTemp from weather_avg)) as "temperature"
FROM
(
select t.id,
t.location_id,
t.departure_time,
t.arrival_time,
generate_series(departure_time :: timestamp,arrival_time::timestamp,'1 day'::interval) as dates
from trip t inner join location l on t.location_id = l.id
) t LEFT JOIN weather w
on t.dates::date = w.date::date
sqlfiddle：http://sqlfiddle.com/#!17/bce59/76