'Monday' 的 postgresql 时隙列表 | 09:00:00 | 11:00:00
postgresql list of time slots from 'Monday' | 09:00:00 | 11:00:00
我正在构建一个预订系统，用户可以在其中设置他们的可用性，例如：我在周一上午 9 点到上午 11 点、周二上午 9 点到下午 5 点等有空……并且需要生成一个时间段列表，时间间隔为 15 分钟他们的可用性。
我有以下 table（但我可以灵活地更改它）：
availabilities(day_of_week text, start_time: time, end_time: time)
其中 returns 记录如下：
‘Monday’ | 09:00:00 | 11:00:00
‘Monday’ | 13:00:00 | 17:00:00
‘Tuesday’ | 08:00:00 | 17:00:00
所以我正在尝试构建一个存储过程来生成一个时隙列表，到目前为止我得到了这个：
create or replace function timeslots ()
return setof timeslots as $$
declare
rec record;
begin
for rec in select * from availabilities loop
/*
convert 'Monday' | 09:00:00 | 11:00:00 into:
2020-02-03 09:00:00
2020-02-03 09:15:00
2020-02-03 09:30:00
2020-02-03 09:45:00
2020-02-03 10:00:00
and so on...
*/
return next
end loop
$$ language plpgsql stable;
我 return 一组而不是 table 因为我正在使用 Hasura 并且它需要 return 一组所以我只创建一个空白 table。
我认为我在正确的轨道上，但目前卡在：
- 如何从 'Monday' 09:00:00 创建下周一的时间戳，因为我只关心从今天开始的时间段？
- 如何将
'Monday' | 09:00:00 | 11:00:00 转换为间隔 15 分钟的时隙列表？
how do I create a timestamp from 'Monday' 09:00:00 for the next monday
as I only care about timeslots from today onwards?
You can use date_trunc for this (see for more info):
SELECT date_trunc('week', current_date) + interval '1 week';
From the docs re week:
The number of the ISO 8601 week-numbering week of the year. By
definition, ISO weeks start on Mondays
So taking this value and adding a week gives next Monday (you may need to ammend this behaviour based upon what you want to do if today is monday!).
how do I convert 'Monday' | 09:00:00 | 11:00:00 into a list of time
slots 15 mins apart?
This is a little tricker; generate_series will give you the timeslots but the trick is getting it into a result set. The following should do the job (I have included your sample data; change the values bit to refer to your table) - dbfiddle :
with avail_times as (
select
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + start_time as start_time,
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + end_time as end_time
from
(
values
('Monday','09:00:00'::time,'11:00:00'::time),
('Monday','13:00:00'::time,'17:00:00'::time),
('Tuesday','08:00:00'::time,'17:00:00'::time)
) as availabilities (day_of_week,
start_time,
end_time) )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts);
A few notes:
- The CTE
avail_times is used to simplify things; it generates two columns (start_time and end_time) which are the full timestamps (so including the date). In this example the first row is "2020-02-03 09:00:00, 2020-02-03 11:00:00" (I'm 运行 this on 2020-02-02 so 2020-02-03 is next Monday).
- The way I'm converting 'monday' etc to a day of the week is a bit of a hack (and I have not bothered to do the full week); there is probably a better way but storing the day of week as an integer would make this simpler.
- I subtract 1ms from the end time because I'm assuming you dont want this in the result set.
- The main query is using a LATERAL Subquery. See
for more info.
Aditional Question
how to adjust this so I can pass in a start and end date so I can get
time slots for a particular period
You could do something like the following (just adjust the dates CTE to return whatever days you want to include; you could convert to a function or just pass the dates in as parameters).
Note that as @Belayer mentions my original solution did not cater for shifts over midnight so this addresses that too.
with dates as (
select
day
from
generate_series('2020-02-20'::date, '2020-03-10'::date, '1 day') as day ),
availabilities as (
select
*
from
(
values (1,'09:00:00'::time,'11:00:00'::time),
(1,'13:00:00'::time,'17:00:00'::time),
(2,'08:00:00'::time,'17:00:00'::time),
(3,'23:00:00'::time,'01:00:00'::time)
) as availabilities
(day_of_week, -- 1 = monday
start_time,
end_time) ) ,
avail_times as (
select
d.day + start_time as start_time,
case
end_time > start_time
when true then d.day
else d.day + interval '1 day' end + end_time as end_time
from
availabilities a
inner join dates d on extract(ISODOW from d.day) = a.day_of_week )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts)
order by
g.ts;
The following uses much of the techniques mentioned by @Brits. They present some very good information, so I'll not repeat but suggest you review it (and the links).
I do however take a slightly different approach. First a couple table changes. I use the ISO day of week 1-7 (Monday-Sunday) rather than the day name. The day name is easily extracted for the dater later.
Also I use interval instead to time for start and end times. ( A time data type works for most scenarios but there is one it doesn't (more later).
One thing your description does not make clear is whether the ending time is included it the available time or not. If included the last interval would be 11:00-11:15. If excluded the last interval is 10:45-11:00. I have assumed to excluded it. In the final results the end time is to be read as "up to but not including".
-- setup
create table availabilities (weekday integer, start_time interval, end_time interval);
insert into availabilities (weekday , start_time , end_time )
select wkday
, start_time
, end_time
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (4, '10:30'::interval, '12:45'::interval)
) as v(wkday,start_time,end_time)
) r ;
select * from availabilities;
The Query
It begins with a CTE (next_week) generates a entry for each day of the week beginning Monday and the appropriate ISO day number for it. The main query joins these with the availabilities table to pick up times for matching days. Finally that result is cross joined with a generated timestamp to get the 15 minute intervals.
-- Main
with next_week (wkday,tm) as
(SELECT n+1, date_trunc('week', current_date) + interval '1 week' + n*interval '1 day'
from generate_series (0, 6) n
)
select to_char(gdtm,'Day'), gdtm start_time, gdtm+interval '15 min' end_time
from ( select wkday, tm, start_time, end_time
from next_week nw
join availabilities av
on (av.weekday = nw.wkday)
) s
cross join lateral
generate_series(start_time+tm, end_time+tm- interval '1 sec', interval '15 min') gdtm ;
The outlier
As mentioned there is one scenario where a time data type does not work satisfactory, but you may not nee it. What happens when a shift worker says they available time is 23:00-01:30. Believe me when a shift worker goes to work at 22:00 of Friday, 01:30 is still Friday night, even though the calendar might not agree. (I worked that shift for many years.) The following using interval handles that issue. Loading the same data as prior with an addition for the this case.
insert into availabilities (weekday, start_time, end_time )
select wkday
, start_time
, end_time + case when end_time < start_time
then interval '1 day'
else interval '0 day'
end
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (5, '23:30'::interval, '02:30'::interval) -- Friday Night - Saturday Morning
) as v(wkday,start_time,end_time)
) r
;
select * from availabilities;
希望对您有所帮助。
我正在构建一个预订系统，用户可以在其中设置他们的可用性，例如：我在周一上午 9 点到上午 11 点、周二上午 9 点到下午 5 点等有空……并且需要生成一个时间段列表，时间间隔为 15 分钟他们的可用性。
我有以下 table（但我可以灵活地更改它）：
availabilities(day_of_week text, start_time: time, end_time: time)
其中 returns 记录如下：
‘Monday’ | 09:00:00 | 11:00:00
‘Monday’ | 13:00:00 | 17:00:00
‘Tuesday’ | 08:00:00 | 17:00:00
所以我正在尝试构建一个存储过程来生成一个时隙列表，到目前为止我得到了这个：
create or replace function timeslots ()
return setof timeslots as $$
declare
rec record;
begin
for rec in select * from availabilities loop
/*
convert 'Monday' | 09:00:00 | 11:00:00 into:
2020-02-03 09:00:00
2020-02-03 09:15:00
2020-02-03 09:30:00
2020-02-03 09:45:00
2020-02-03 10:00:00
and so on...
*/
return next
end loop
$$ language plpgsql stable;
我 return 一组而不是 table 因为我正在使用 Hasura 并且它需要 return 一组所以我只创建一个空白 table。
我认为我在正确的轨道上，但目前卡在：
- 如何从 'Monday' 09:00:00 创建下周一的时间戳，因为我只关心从今天开始的时间段？
- 如何将
'Monday' | 09:00:00 | 11:00:00转换为间隔 15 分钟的时隙列表？
how do I create a timestamp from 'Monday' 09:00:00 for the next monday as I only care about timeslots from today onwards?
You can use date_trunc for this (see
SELECT date_trunc('week', current_date) + interval '1 week';
From the docs re week:
The number of the ISO 8601 week-numbering week of the year. By definition, ISO weeks start on Mondays
So taking this value and adding a week gives next Monday (you may need to ammend this behaviour based upon what you want to do if today is monday!).
how do I convert 'Monday' | 09:00:00 | 11:00:00 into a list of time slots 15 mins apart?
This is a little tricker; generate_series will give you the timeslots but the trick is getting it into a result set. The following should do the job (I have included your sample data; change the values bit to refer to your table) - dbfiddle :
with avail_times as (
select
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + start_time as start_time,
date_trunc('week', current_date) + interval '1 week' + case day_of_week when 'Monday' then interval '0 day' when 'Tuesday' then interval '1 day' end + end_time as end_time
from
(
values
('Monday','09:00:00'::time,'11:00:00'::time),
('Monday','13:00:00'::time,'17:00:00'::time),
('Tuesday','08:00:00'::time,'17:00:00'::time)
) as availabilities (day_of_week,
start_time,
end_time) )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts);
A few notes:
- The CTE
avail_timesis used to simplify things; it generates two columns (start_timeandend_time) which are the full timestamps (so including the date). In this example the first row is "2020-02-03 09:00:00, 2020-02-03 11:00:00" (I'm 运行 this on 2020-02-02 so 2020-02-03 is next Monday). - The way I'm converting 'monday' etc to a day of the week is a bit of a hack (and I have not bothered to do the full week); there is probably a better way but storing the day of week as an integer would make this simpler.
- I subtract 1ms from the end time because I'm assuming you dont want this in the result set.
- The main query is using a LATERAL Subquery. See
Aditional Question
how to adjust this so I can pass in a start and end date so I can get time slots for a particular period
You could do something like the following (just adjust the dates CTE to return whatever days you want to include; you could convert to a function or just pass the dates in as parameters).
Note that as @Belayer mentions my original solution did not cater for shifts over midnight so this addresses that too.
with dates as (
select
day
from
generate_series('2020-02-20'::date, '2020-03-10'::date, '1 day') as day ),
availabilities as (
select
*
from
(
values (1,'09:00:00'::time,'11:00:00'::time),
(1,'13:00:00'::time,'17:00:00'::time),
(2,'08:00:00'::time,'17:00:00'::time),
(3,'23:00:00'::time,'01:00:00'::time)
) as availabilities
(day_of_week, -- 1 = monday
start_time,
end_time) ) ,
avail_times as (
select
d.day + start_time as start_time,
case
end_time > start_time
when true then d.day
else d.day + interval '1 day' end + end_time as end_time
from
availabilities a
inner join dates d on extract(ISODOW from d.day) = a.day_of_week )
select
g.ts
from
(
select
start_time,
end_time
from
avail_times) avail,
generate_series(avail.start_time, avail.end_time - interval '1ms', '15 minutes') g(ts)
order by
g.ts;
The following uses much of the techniques mentioned by @Brits. They present some very good information, so I'll not repeat but suggest you review it (and the links).
I do however take a slightly different approach. First a couple table changes. I use the ISO day of week 1-7 (Monday-Sunday) rather than the day name. The day name is easily extracted for the dater later.
Also I use interval instead to time for start and end times. ( A time data type works for most scenarios but there is one it doesn't (more later).
One thing your description does not make clear is whether the ending time is included it the available time or not. If included the last interval would be 11:00-11:15. If excluded the last interval is 10:45-11:00. I have assumed to excluded it. In the final results the end time is to be read as "up to but not including".
-- setup
create table availabilities (weekday integer, start_time interval, end_time interval);
insert into availabilities (weekday , start_time , end_time )
select wkday
, start_time
, end_time
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (4, '10:30'::interval, '12:45'::interval)
) as v(wkday,start_time,end_time)
) r ;
select * from availabilities;
The Query
It begins with a CTE (next_week) generates a entry for each day of the week beginning Monday and the appropriate ISO day number for it. The main query joins these with the availabilities table to pick up times for matching days. Finally that result is cross joined with a generated timestamp to get the 15 minute intervals.
-- Main
with next_week (wkday,tm) as
(SELECT n+1, date_trunc('week', current_date) + interval '1 week' + n*interval '1 day'
from generate_series (0, 6) n
)
select to_char(gdtm,'Day'), gdtm start_time, gdtm+interval '15 min' end_time
from ( select wkday, tm, start_time, end_time
from next_week nw
join availabilities av
on (av.weekday = nw.wkday)
) s
cross join lateral
generate_series(start_time+tm, end_time+tm- interval '1 sec', interval '15 min') gdtm ;
The outlier
As mentioned there is one scenario where a time data type does not work satisfactory, but you may not nee it. What happens when a shift worker says they available time is 23:00-01:30. Believe me when a shift worker goes to work at 22:00 of Friday, 01:30 is still Friday night, even though the calendar might not agree. (I worked that shift for many years.) The following using interval handles that issue. Loading the same data as prior with an addition for the this case.
insert into availabilities (weekday, start_time, end_time )
select wkday
, start_time
, end_time + case when end_time < start_time
then interval '1 day'
else interval '0 day'
end
from (select *
from (values (1, '09:00'::interval, '11:00'::interval)
, (1, '13:00'::interval, '17:00'::interval)
, (2, '08:00'::interval, '17:00'::interval)
, (3, '08:30'::interval, '10:45'::interval)
, (5, '23:30'::interval, '02:30'::interval) -- Friday Night - Saturday Morning
) as v(wkday,start_time,end_time)
) r
;
select * from availabilities;
希望对您有所帮助。