Question

user_id | date                 | distance
1       | 2019-04-09 00:00:00  | 2
1       | 2019-04-09 00:00:30  | 5
1       | 2019-04-09 00:01:00  | 3
1       | 2019-04-09 00:01:45  | 7
1       | 2019-04-09 00:02:30  | 6
1       | 2019-04-09 00:03:00  | 1

如何求和下一行的总和，直到达到阈值点，然后再次重置计数器。

例如，如果阈值为10，我正在尝试获得以下输出：

1       | 2019-04-09 00:00:00  | 2
1       | 2019-04-09 00:00:30  | 7            (2 + 5)
1       | 2019-04-09 00:01:00  | 10           ( 7 + 3 )
1       | 2019-04-09 00:01:45  | 7            RESET
1       | 2019-04-09 00:02:30  | 13           (7 + 6 )
1       | 2019-04-09 00:03:00  | 1            RESET

但是我所能实现的是通过以下查询获得累积距离：

SELECT *, sum(distance) over (order by date asc) as running_distance FROM table;

我正在使用PostgreSQL。

Answer 1

使用user-defined aggregate

实时测试：http://sqlfiddle.com/#!17/16716/2

SELECT *, sum_with_reset(distance, 10) over (order by date asc) as running_distance 
FROM tbl;

用户定义的汇总sum_with_reset定义：

create or replace function sum_reset_accum(
    _accumulated numeric, _current numeric, _threshold numeric
)
returns numeric as
$$
    select case when _accumulated >= _threshold then
        _current
    else
        _current + _accumulated
    end    
$$ language sql;


create aggregate sum_with_reset(numeric, numeric)
(
    sfunc = sum_reset_accum,
    stype = numeric,
    initcond = 0
);

数据

CREATE TABLE tbl
    ("user_id" int, "date" timestamp, "distance" int)
;

INSERT INTO tbl
    ("user_id", "date", "distance")
VALUES
    (1, '2019-04-09 00:00:00', 2),
    (1, '2019-04-09 00:00:30', 5),
    (1, '2019-04-09 00:01:00', 3),
    (1, '2019-04-09 00:01:45', 7),
    (1, '2019-04-09 00:02:30', 6),
    (1, '2019-04-09 00:03:00', 1)
;

输出：

| user_id |                 date | distance | running_distance |
|---------|----------------------|----------|------------------|
|       1 | 2019-04-09T00:00:00Z |        2 |                2 |
|       1 | 2019-04-09T00:00:30Z |        5 |                7 |
|       1 | 2019-04-09T00:01:00Z |        3 |               10 |
|       1 | 2019-04-09T00:01:45Z |        7 |                7 |
|       1 | 2019-04-09T00:02:30Z |        6 |               13 |
|       1 | 2019-04-09T00:03:00Z |        1 |                1 |

单线：

create or replace function sum_reset_accum(
    _accumulated numeric, _current numeric, _threshold numeric
)
returns numeric as
$$
    select _current + _accumulated * (_accumulated < _threshold)::int
$$ language 'sql';

通过使用强制转换运算符::int，Postgres布尔值可以将true转换为1，将false转换为0。

您也可以使用plpgsql语言：

create or replace function sum_reset_accum(
    _accumulated numeric, _current numeric, _threshold numeric
)
returns numeric as
$$begin
    return _current + _accumulated * (_accumulated < _threshold)::int;
end$$ language 'plpgsql';

请注意，您无法在sqlfiddle.com上创建plpgsql函数，因此无法在sqlfiddle.com上测试该plpgsql代码。您可以，但是在您的计算机上。

Answer 2

作为变体，您可以使用RECURSIVE CTE来获取

测试表：

CREATE TABLE public.table_b (
    user_id int4 NULL,
    "date" timestamp NULL,
    distance int4 NULL
);

INSERT INTO public.table_b (user_id,"date",distance) VALUES 
(1,'2019-04-09 00:00:00.000',2)
,(1,'2019-04-09 00:00:30.000',5)
,(1,'2019-04-09 00:01:00.000',3)
,(1,'2019-04-09 00:01:45.000',7)
,(1,'2019-04-09 00:02:30.000',6)
,(1,'2019-04-09 00:03:00.000',1)
,(2,'2019-04-09 00:00:00.000',2)
,(2,'2019-04-09 00:00:30.000',5)
,(2,'2019-04-09 00:01:00.000',3)
,(2,'2019-04-09 00:01:45.000',7)
,(2,'2019-04-09 00:02:30.000',6)
,(2,'2019-04-09 00:03:00.000',1);

查询：

WITH RECURSIVE cte1 AS(
    SELECT *,ROW_NUMBER()OVER(PARTITION BY user_id ORDER BY date) n
    FROM table_b
),
cte2 AS(
    SELECT user_id,date,distance,distance running_distance,n
    FROM cte1
    WHERE n=1

    UNION ALL

    SELECT c1.user_id,c1.date,c1.distance,CASE WHEN c2.running_distance<10 THEN c2.running_distance ELSE 0 END+c1.distance running_distance,c1.n
    FROM cte1 c1
    JOIN cte2 c2 ON c2.user_id=c1.user_id AND c2.n+1=c1.n
)
SELECT user_id,date,distance,running_distance
FROM cte2
ORDER BY user_id,date

求和，直到达到阈值，然后重置计数器

2 个答案: