按照我上一个问题的回答对工作进行健全性检查。我意识到我没有为我的目标找到正确的逻辑。我在表格trajectories中有用户的时间戳 GPS 序列：
CREATE TABLE trajectories
(
user_id integer,
session_id bigint NOT NULL,
"timestamp" timestamp with time zone NOT NULL,
lat double precision NOT NULL,
lon double precision NOT NULL,
alt double precision,
PRIMARY KEY (session_id, "timestamp")
);
例如：
INSERT INTO trajectories (user_id,session_id,timestamp,lat,lon,alt)
VALUES (11,1010,'2007-06-26 11:32:29+01',37.780927,113.677553,2160),
(11,1010,'2007-06-26 11:32:30+01',37.78093,113.677627,2160),
(11,1010,'2007-06-26 11:32:33+01',37.780932,113.677698,2160),
(11,1010,'2007-06-26 11:32:36+01',37.780938,113.677772,2159),
(11,1010,'2007-06-26 11:32:37+01',37.780945,113.677845,2159),
(11,1010,'2007-06-26 11:32:39+01',37.780952,113.677918,2159),
(11,1010,'2007-06-26 11:32:47+01',37.780962,113.67799,2159),
(11,1010,'2007-06-26 11:32:51+01',37.780973,113.67806,2159),
(11,1010,'2007-06-26 11:32:55+01',37.78098,113.678128,2159),
(11,1010,'2007-06-26 11:32:59+01',37.780992,113.678192,2157)
在我的第二张表labels中是用户在旅行期间使用的交通方式，如下所示：
CREATE TABLE labels
(
user_id integer NOT NULL,
session_id bigint,
start_timestamp timestamp with time zone NOT NULL,
end_timestamp timestamp with time zone NOT NULL,
travelmode text COLLATE pg_catalog."default",
PRIMARY KEY (user_id, start_timestamp, end_timestamp)
)
INSERT INTO labels (user_id,session_id,start_timestamp,end_timestamp,travelmode)
VALUES (11,1010,'2007-06-26 11:32:29+01','2007-06-26 11:40:29+01','bus'),
(11,1010,'2008-03-28 14:52:54+00','2008-03-28 15:59:59+00','train'),
(11,1010,'2008-03-28 16:00:00+00','2008-03-28 22:02:00+00','train'),
(11,1010,'2008-03-29 01:27:50+00','2008-03-29 15:59:59+00','train'),
(11,1010,'2008-03-29 16:00:00+00','2008-03-30 15:59:59+01','train')
出于统计目的，我想知道每种模式的 GPS 点数。在上表中，我使用以下模式获得总点数bus：
SELECT COUNT(*)
FROM trajectories t
JOIN labels l
ON t.user_id = l.user_id
WHERE travelmode='bus' AND t.timestamp BETWEEN l.start_timestamp AND l.end_timestamp;
count
---------
10
(1 row)
GPS点的采样率是不均匀的，所以我想得到bus区间内模式的点数：1 sec, 2-5sec, 6-10sec, above 10sec.
我的查询：
SELECT COUNT(*) count_all,
COUNT(*) FILTER (WHERE end_timestamp >= start_timestamp AND end_timestamp < start_timestamp + interval '1 second') one_sec_inv,
COUNT(*) FILTER (WHERE end_timestamp >= start_timestamp AND end_timestamp < start_timestamp + interval '5 second') two_to_5,
COUNT(*) FILTER (WHERE end_timestamp >= start_timestamp + interval '5 second' AND end_timestamp < start_timestamp + interval '10 second') five_to_10,
COUNT(*) FILTER (WHERE end_timestamp >= start_timestamp + interval '10 second' AND end_timestamp > start_timestamp + interval '15 second') ten_to_15
FROM trajectories t JOIN
labels l
ON t.user_id = l.user_id
WHERE travelmode='bus' AND t.timestamp >= l.start_timestamp AND t.timestamp <= l.end_timestamp;
count_all | one_sec_inv | two_to_5 | five_to_10 | ten_to_15
-----------+-------------+----------+------------+-----------
10 | 0 | 0 | 0 | 10
(1 row)
承认我在这个查询中缺少正确的逻辑。显然，分别有2 pointsin 区间one_sec_inv、5 pointsintwo_to_5区间和1 pointin 区间five_to_10。我该如何解决它以实现目标？
我在这个db<>fiddle中展示了这些表。
编辑
SELECT * FROM trajectories
+---------+------------+------------------------+-----------+------------+------+
| user_id | session_id | timestamp | lat | lon | alt |
+---------+------------+------------------------+-----------+------------+------+
| 11 | 1010 | 2007-06-26 11:32:29+01 | 37.780927 | 113.677553 | 2160 |
| 11 | 1010 | 2007-06-26 11:32:30+01 | 37.78093 | 113.677627 | 2160 |
| 11 | 1010 | 2007-06-26 11:32:33+01 | 37.780932 | 113.677698 | 2160 |
| 11 | 1010 | 2007-06-26 11:32:36+01 | 37.780938 | 113.677772 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:37+01 | 37.780945 | 113.677845 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:39+01 | 37.780952 | 113.677918 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:47+01 | 37.780962 | 113.67799 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:51+01 | 37.780973 | 113.67806 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:55+01 | 37.78098 | 113.678128 | 2159 |
| 11 | 1010 | 2007-06-26 11:32:59+01 | 37.780992 | 113.678192 | 2157 |
+---------+------------+------------------------+-----------+------------+------+
预期输出：
count_all | one_sec_inv | two_to_5 | five_to_10 | ten_to_15
-----------+-------------+----------+------------+-----------
10 | 2 | 6 | 1 | 0
编辑-2
@MikeOrganek 的这个答案没有产生预期的结果：
SELECT COUNT(*) count_all,
COUNT(*) FILTER (WHERE timestamp <= start_timestamp + interval '1 second') one_sec_inv,
COUNT(*) FILTER (WHERE timestamp > start_timestamp + interval '1 seconds'
AND timestamp <= start_timestamp + interval '5 seconds') two_to_5,
COUNT(*) FILTER (WHERE timestamp > start_timestamp + interval '5 seconds'
AND timestamp <= start_timestamp + interval '10 seconds') five_to_10,
COUNT(*) FILTER (WHERE timestamp > start_timestamp + interval '10 second'
AND timestamp <= start_timestamp + interval '15 second') ten_to_15
FROM trajectories t
JOIN labels l
ON t.user_id = l.user_id
WHERE travelmode='bus'
AND t.timestamp >= l.start_timestamp
AND t.timestamp <= l.end_timestamp;
+-----------+-------------+----------+------------+-----------+
| count_all | one_sec_inv | two_to_5 | five_to_10 | ten_to_15 |
+-----------+-------------+----------+------------+-----------+
| 10 | 2 | 1 | 3 | 0 |
+-----------+-------------+----------+------------+-----------+