我有一个包含这些字段的表：
id | person_id | start_time | end_time | status
我想获得特定日期每小时存在的人数，如下面的伪代码：
select count(person) from table where dow of end_time=day and end_time >= hour and start_time < hour+1 for hour in working hours of organization and day is a certain day of week
如果有可能有一个小时的临时表可能是以下解决方案是一个解决方案：
select t.h, count(s.id)
from
session s cross join (temperoray table with one column of hours in a day as t)
where
s.start_time < (t.h + 1) and s.end_time > t.h
group by
t.h
但我不知道可以根据需要创建临时表的命令。
我发现这个question与我想要的非常相似，但它的所有解决方案都基于group by，我认为这在我的案例中没有任何意义，因为每个小组的部分都有共同点物品，例如，一个人可以在第11小时计算为人，在第12和第13小时计为人。
我希望我能找到一种能让我得到这样一张桌子的方法：
hour |number of persons
10 |2
11 |0
12 |3
13 |1
...
请注意，某些人可能会有零人数。
id | status | start_time | end_time | branch_id | person_id | session_type
------+--------+----------------------------------+----------------------------------+-----------+-----------+--------------
2675 | FI | 2018-04-23 10:30:50.939693+04:30 | 2018-04-23 12:31:39.340692+04:30 | 1 | 1085 | IN
2676 | FI | 2018-04-23 11:47:06.683374+04:30 | 2018-04-23 13:23:52.659714+04:30 | 1 | 2722 | IN
2677 | FI | 2018-04-23 11:47:59.341765+04:30 | 2018-04-23 13:25:46.339266+04:30 | 1 | 2721 | IN
2678 | FI | 2018-04-23 11:58:34.854222+04:30 | 2018-04-23 13:25:55.08795+04:30 | 1 | 2723 | IN
2679 | FI | 2018-04-23 12:27:58.817234+04:30 | 2018-04-23 13:12:28.278699+04:30 | 1 | 2724 | IN
2680 | FI | 2018-04-23 12:30:36.552407+04:30 | 2018-04-23 12:30:54.088159+04:30 | 1 | 2725 | IN
2681 | FI | 2018-04-23 14:55:50.886725+04:30 | 2018-04-23 16:08:27.076629+04:30 | 1 | 25 | IN
2682 | FI | 2018-04-23 15:06:30.443347+04:30 | 2018-04-23 15:52:20.128546+04:30 | 1 | 2653 | IN
2683 | FI | 2018-04-23 15:21:57.979387+04:30 | 2018-04-23 16:16:09.289267+04:30 | 1 | 2580 | IN
2684 | FI | 2018-04-23 15:26:18.057999+04:30 | 2018-04-23 16:02:44.704133+04:30 | 1 | 2726 | IN
2685 | FI | 2018-04-23 16:50:10.2957+04:30 | 2018-04-23 17:23:01.732404+04:30 | 1 | 2727 | IN
2686 | FI | 2018-04-23 16:52:28.474299+04:30 | 2018-04-23 17:23:51.013318+04:30 | 1 | 2728 | IN
2687 | FI | 2018-04-23 16:58:05.796563+04:30 | 2018-04-23 17:33:03.259335+04:30 | 1 | 1646 | IN
2688 | FI | 2018-04-23 17:50:02.738009+04:30 | 2018-04-23 18:43:27.152203+04:30 | 1 | 2729 | IN
2689 | FI | 2018-04-23 18:47:12.19468+04:30 | 2018-04-23 19:25:46.606731+04:30 | 1 | 2730 | IN
2690 | FI | 2018-04-23 19:18:32.922065+04:30 | 2018-04-23 20:11:26.703693+04:30 | 1 | 2408 | IN
2691 | FI | 2018-04-23 19:18:53.133712+04:30 | 2018-04-23 19:56:47.702305+04:30 | 1 | 2409 | IN
2692 | FI | 2018-04-23 19:21:00.348889+04:30 | 2018-04-23 20:24:25.882451+04:30 | 1 | 2731 | IN
2693 | FI | 2018-04-23 19:30:05.908247+04:30 | 2018-04-23 20:12:36.627888+04:30 | 1 | 2591 | IN
2694 | FI | 2018-04-23 19:36:02.700379+04:30 | 2018-04-23 20:13:35.146002+04:30 | 1 | 2732 | IN
2695 | FI | 2018-04-23 19:50:15.13214+04:30 | 2018-04-23 20:09:37.168147+04:30 | 1 | 2491 | IN
2696 | FI | 2018-04-23 19:51:54.754169+04:30 | 2018-04-23 20:09:59.029376+04:30 | 1 | 2733 | IN
2697 | FI | 2018-04-23 19:53:13.529475+04:30 | 2018-04-23 20:09:49.229139+04:30 | 1 | 2734 | IN
2698 | FI | 2018-04-23 19:59:27.70488+04:30 | 2018-04-23 20:21:47.862433+04:30 | 1 | 1762 | IN
2699 | FI | 2018-04-23 19:59:57.86605+04:30 | 2018-04-23 20:22:05.171377+04:30 | 1 | 1761 | IN
2700 | FI | 2018-04-23 20:24:21.212784+04:30 | 2018-04-23 20:47:31.854373+04:30 | 1 | 2735 | IN
2701 | FI | 2018-04-23 21:58:57.308547+04:30 | 2018-04-23 22:43:20.075321+04:30 | 1 | 1705 | IN
2702 | FI | 2018-04-23 21:59:44.974384+04:30 | 2018-04-23 22:43:45.946989+04:30 | 1 | 1704 | IN
2703 | FI | 2018-04-23 22:10:20.991216+04:30 | 2018-04-23 22:40:51.16409+04:30 | 1 | 2711 | IN
我今天得到的结果如下：
hour | number
10 | 1
11 | 4
12 | 6
13 | 4
14 | 1
15 | 4
16 | 6
17 | 4
18 | 2
19 | 11
20 | 10
21 | 2
22 | 3
答案 0 :(得分：2)
如果我理解正确
您需要使用generate_series函数创建24小时，然后left join。
您可以在Where子句中添加一些条件。
获取小时
SELECT gs.hours,
Sum(
CASE
WHEN start_time IS NOT NULL THEN 1
WHEN end_time IS NOT NULL THEN 1
ELSE 0
END ) AS "count"
FROM (
SELECT hours
FROM Generate_series(1,24) AS gs(hours) ) gs
LEFT JOIN
(
SELECT *,
Generate_series(start_time::timestamp, end_time::timestamp, '1 hours') invhour
FROM t )t
ON gs.hours = To_char(t.invhour,'HH24')::integer
GROUP BY gs.hours
获取小时和日期
SELECT To_char(t.invhour,'yyyy-MM-dd') AS "dates",
gs.hours,
Sum(
CASE
WHEN start_time IS NOT NULL THEN 1
WHEN end_time IS NOT NULL THEN 1
ELSE 0
END ) AS "count"
FROM (
SELECT hours
FROM Generate_series(1,24) AS gs(hours) ) gs
LEFT JOIN
(
SELECT *,
Generate_series(start_time::timestamp, end_time::timestamp, '1 hours') invhour
FROM t )t
ON gs.hours = To_char(t.invhour,'HH24')::integer
GROUP BY gs.hours,
to_char(t.invhour,'yyyy-MM-dd')
sqlfiddle：http://sqlfiddle.com/#!17/717fa/1
答案 1 :(得分：2)
您需要为24小时创建一个集合，然后按小时进行左连接和分组。
这里有人建议在子查询中使用generate_series，但我个人认为递归ctes对于创建范围更好一些。这样，您可以将系列集保留在主查询之外，使其更容易理解和维护。
; WITH RECURSIVE Hours AS
(
SELECT 1 AS hour
UNION ALL
SELECT hour + 1 FROM Hours WHERE Hour < 24
)
SELECT hour, COUNT(person_id)
FROM Hours
LEFT JOIN T on hour BETWEEN extract(hour from start_time) AND extract(hour from end_time)
GROUP BY hour
ORDER BY hour
答案 2 :(得分：1)
使用D-Shih's answer中的generate_series，我得到以下解决方案。
第一个解决方案显示至少有一个人在工作的所有时间。
select Hour, count(1) as "Users"
from generate_series(1,24) as gs (Hour)
join Log as l
on date_part('hour',l.TimeFrom) <= gs.Hour
and date_part('hour',l.TimeTo) >= gs.Hour
group by Hour
order by Hour;
如果您还需要无人工作的时间，请使用left join。
这需要更改count()以仅在找到记录时进行计数。
select Hour, count(case when l.UserId is not null then 1 end) as "Users"
from generate_series(1,24) as gs (Hour)
left join Log as l
on date_part('hour',l.TimeFrom) <= gs.Hour
and date_part('hour',l.TimeTo) >= gs.Hour
group by Hour
order by Hour;
有关示例数据和输出，请参阅this SQL Fiddle。
答案 3 :(得分：0)
看看这个article，它在PostgreSQL中创建一个Temp表（我是一个SQL服务器人）。
有一列是一天中的小时，我认为你将它加入你的桌子的方法是可行的。
它也取决于该表中的数据。您能提供前1行数据来了解它的外观吗？你可能甚至不需要临时表。