Question

PostgreSQL的！

我有一张桌子＆＃34;会议＆＃34;

create table meeting (
id SERIAL, 
name varchar, 
date_start timestamp not null, 
period interval default '0:0:0');

和一些数据

select * from meeting;
+----+---------------+---------------------+--------+
| id |     name      |     date_start      | period |
+----+---------------+---------------------+--------+
|  1 | Måndagsmöte   | 2017-06-06 09:00:00 | 7 days |
|  2 | torsdagssoppa | 2017-06-01 00:00:00 | 7 days |
+----+---------------+---------------------+--------+

可以从中选择参加会议;

select id,name, date from meeting,
generate_series( 
(select date_start::timestamptz from meeting where id=1)::timestamp,
'2017-08-01 00:00:00', (select period from meeting where id=1)) as date
 where date > '2017-06-15 00:00:00' 
 and date < '2017-07-20 00:00:00' and id=1;

+----+-------------+---------------------+
| id |    name     |        date         |
+----+-------------+---------------------+
|  1 | Måndagsmöte | 2017-06-20 09:00:00 |
|  1 | Måndagsmöte | 2017-06-27 09:00:00 |
|  1 | Måndagsmöte | 2017-07-04 09:00:00 |
|  1 | Måndagsmöte | 2017-07-11 09:00:00 |
|  1 | Måndagsmöte | 2017-07-18 09:00:00 |
+----+-------------+---------------------+

是否有任何可能的放松＆＃34; id = 1＆＃34;并举行所有名字的会议？

/ GH

Answer 1

根据您的数据，您可以使用下一个查询：

select *
from (select id,
             name, 
             generate_series(date_start, '2017-08-01 00:00:00', period) date
      from   meeting) mt
where  mt.date > '2017-06-15 00:00:00' 
and    mt.date < '2017-07-20 00:00:00'

id | name          | date               
-: | :------------ | :------------------
 1 | Måndagsmöte | 2017-06-20 09:00:00
 1 | Måndagsmöte | 2017-06-27 09:00:00
 1 | Måndagsmöte | 2017-07-04 09:00:00
 1 | Måndagsmöte | 2017-07-11 09:00:00
 1 | Måndagsmöte | 2017-07-18 09:00:00
 2 | torsdagssoppa | 2017-06-22 00:00:00
 2 | torsdagssoppa | 2017-06-29 00:00:00
 2 | torsdagssoppa | 2017-07-06 00:00:00
 2 | torsdagssoppa | 2017-07-13 00:00:00

dbfiddle here

postgresql中的generate_series来自表中的字段

1 个答案: