我有以下查询：
SELECT
usersq1.id AS user_id, name, completed_at,
COUNT(usersq1.id) AS trips,
SUM(cost_amount_cents) AS daily_cost_amount_cents
FROM usersq1
LEFT OUTER JOIN tripsq1
ON usersq1.id = user_id
GROUP by usersq1.id, name, completed_at
ORDER by user_id, name, completed_at;
返回以下内容：
user_id | name | completed_at | trips | daily_cost_amount_cents
---------+---------------------+--------------+-------+-------------------------
1001 | Makeda Mosser | 2017-06-01 | 2 | 125
1001 | Makeda Mosser | 2017-06-02 | 1 | 125
1001 | Makeda Mosser | 2017-06-03 | 2 | 350
1001 | Makeda Mosser | 2017-06-04 | 2 | 200
1001 | Makeda Mosser | 2017-06-06 | 1 | 100
1001 | Makeda Mosser | 2017-06-07 | 1 | 125
1001 | Makeda Mosser | 2017-06-08 | 1 | 150
1002 | Libbie Luby | 2017-06-02 | 2 | 125
1002 | Libbie Luby | 2017-06-09 | 1 | 175
1003 | Linn Loughran | 2017-06-03 | 1 | 75
1004 | Natacha Ned | 2017-06-04 | 1 | 100
1005 | Lorrine Lunt | 2017-06-05 | 1 | 125
1006 | Tami Tineo | 2017-10-06 | 1 | 150
1007 | Delisa Deen | 2017-10-07 | 1 | 175
1008 | Mimi Miltenberger | 2017-10-08 | 1 | 200
1009 | Seth Sneller | 2017-10-09 | 1 | 25
1010 | Rickie Rossi | 2017-10-10 | 1 | 50
1011 | Jenise Jeanbaptiste | 2017-06-01 | 1 | 200
1011 | Jenise Jeanbaptiste | 2017-07-01 | 1 | 75
1012 | Genia Glatz | 2017-06-02 | 1 | 25
1012 | Genia Glatz | 2017-07-02 | 1 | 50
1013 | Onita Oddo | 2017-06-03 | 1 | 50
1014 | Dario Dreyer | 2017-06-04 | 1 | 75
1014 | Dario Dreyer | 2017-06-24 | 5 | 750
1015 | Toby Trent | | 1 |
我想生成另一个累积和列，每个用户保持每天运行总计daily_cost_amount_cents。我希望的预期展望是这样的：
+---------+---------------------+------------+-------+-------------------------+-----------+
| user_id | name | created_at | trips | daily_cost_amount_cents | cum_cents |
+---------+---------------------+------------+-------+-------------------------+-----------+
| 1001 | Makeda Mosser | 6/1/17 | 2 | 125 | 125 |
| 1001 | Makeda Mosser | 6/2/17 | 1 | 125 | 250 |
| 1001 | Makeda Mosser | 6/3/17 | 2 | 350 | 600 |
| 1001 | Makeda Mosser | 6/4/17 | 2 | 200 | 800 |
| 1001 | Makeda Mosser | 6/6/17 | 1 | 100 | 900 |
| 1001 | Makeda Mosser | 6/7/17 | 1 | 125 | 1025 |
| 1001 | Makeda Mosser | 6/8/17 | 1 | 150 | 1175 |
| 1002 | Libbie Luby | 6/2/17 | 2 | 125 | 125 |
| 1002 | Libbie Luby | 6/9/17 | 1 | 175 | 300 |
| 1003 | Linn Loughran | 6/3/17 | 1 | 75 | 75 |
| 1004 | Natacha Ned | 6/4/17 | 1 | 100 | 100 |
| 1005 | Lorrine Lunt | 6/5/17 | 1 | 125 | 125 |
| 1006 | Tami Tineo | 10/6/17 | 1 | 150 | 150 |
| 1007 | Delisa Deen | 10/7/17 | 1 | 175 | 175 |
| 1008 | Mimi Miltenberger | 10/8/17 | 1 | 200 | 200 |
| 1009 | Seth Sneller | 10/9/17 | 1 | 25 | 25 |
| 1010 | Rickie Rossi | 10/10/17 | 1 | 50 | 50 |
| 1011 | Jenise Jeanbaptiste | 6/1/17 | 1 | 200 | 200 |
| 1011 | Jenise Jeanbaptiste | 7/1/17 | 1 | 75 | 275 |
| 1012 | Genia Glatz | 6/2/17 | 1 | 25 | 25 |
| 1012 | Genia Glatz | 7/2/17 | 1 | 50 | 75 |
| 1013 | Onita Oddo | 6/3/17 | 1 | 50 | 50 |
| 1014 | Dario Dreyer | 6/4/17 | 1 | 75 | 75 |
| 1014 | Dario Dreyer | 6/24/17 | 5 | 750 | 750 |
| 1015 | Toby Trent | | 0 | | |
+---------+---------------------+------------+-------+-------------------------+-----------+
我很确定我需要使用窗口函数来执行此操作，但似乎无法通过user_id和created_by保留分组
答案 0 :(得分：1)
问题在于GROUP BY条款存在window functions iterate over each group rather than multiple grouped rows。将您的查询放入WITH子句中，您可以轻松地进行所需的窗口化：
WITH t AS (
SELECT usersq1.id AS user_id,
name,
completed_at,
COUNT(completed_at) AS trips, -- To correctly handle 0 trips
SUM(cost_amount_cents) AS daily_cost_amount_cents
FROM usersq1
LEFT OUTER JOIN tripsq1 ON usersq1.id = user_id
GROUP BY usersq1.id, name, completed_at
ORDER BY user_id, name, completed_at
) SELECT user_id,
name,
completed_at AS created_at,
trips,
daily_cost_amount_cents,
SUM(daily_cost_amount_cents) OVER (PARTITION BY user_id
ORDER BY user_id, completed_at)
FROM t;