这是我的表
ID account Value date
0 Tom 10 6/7
1 Tom -9 6/8
2 Tom -5 6/9
3 Tom -4 6/10
4 Tom 20 6/11
5 Tom 30 6/12
6 Tom -4 6/13
7 Tom -5 6/14
8 Tom 7 6/15
9 Tom -5 6/16
10 Tom -5 6/16
11 Tom 10 6/16
12 Tom -2 6/17
我必须在正值之间添加所有负值，并使用之前的正值减去它们。以下是上面的示例表
的结果 ID account Value date
0 Tom 10 6/7
4 Tom 2 6/11
5 Tom 30 6/12
8 Tom -2 6/15
11 Tom 0 6/16
12 Tom -2 6/17
答案 0 :(得分：2)
尝试此解决方案（警告：Sql Server 2012）：http://www.sqlfiddle.com/#!6/f5ce0/7
with summary as
(
select
*,
-- if negative make it zero
-- if positive make it 1
grp =
sum((sign(value)+1) / 2)
over(order by id desc)
from tbl
)
select
id = max(id),
account = max(account),
value = sum(value),
date = max(date)
from summary
group by grp
order by grp desc
输出：
ID ACCOUNT VALUE DATE
0 Tom 10 2001-06-07
4 Tom 2 2001-06-11
5 Tom 30 2001-06-12
8 Tom -2 2001-06-15
11 Tom 0 2001-06-16
12 Tom -2 2001-06-17
第一步，为正数和负数创建一个标记，稍后我们将使用它来分隔组：
select
*,
-- if negative make it zero
-- if positive make it 1
(sign(value)+1) / 2 as rn
from tbl;
输出：
ID ACCOUNT VALUE DATE RN
0 Tom 10 2001-06-07 1
1 Tom -9 2001-06-08 0
2 Tom -5 2001-06-09 0
3 Tom -4 2001-06-10 0
4 Tom 20 2001-06-11 1
5 Tom 30 2001-06-12 1
6 Tom -4 2001-06-13 0
7 Tom -5 2001-06-14 0
8 Tom 7 2001-06-15 1
9 Tom -5 2001-06-16 0
10 Tom -5 2001-06-16 0
11 Tom 10 2001-06-16 1
12 Tom -2 2001-06-17 0
第二步，SUM超过那些1s 0来创建分组，根据OP的问题这是向后的：http://www.sqlfiddle.com/#!6/f5ce0/3
select
*,
-- if negative make it zero
-- if positive make it 1
grp =
sum((sign(value)+1) / 2)
over(order by id desc)
from tbl;
输出：
ID ACCOUNT VALUE DATE GRP
12 Tom -2 2001-06-17 0
11 Tom 10 2001-06-16 1
10 Tom -5 2001-06-16 1
9 Tom -5 2001-06-16 1
8 Tom 7 2001-06-15 2
7 Tom -5 2001-06-14 2
6 Tom -4 2001-06-13 2
5 Tom 30 2001-06-12 3
4 Tom 20 2001-06-11 4
3 Tom -4 2001-06-10 4
2 Tom -5 2001-06-09 4
1 Tom -9 2001-06-08 4
0 Tom 10 2001-06-07 5
最后一步，将这些相关数字分组：http://www.sqlfiddle.com/#!6/f5ce0/7
with summary as
(
select
*,
-- if negative make it zero
-- if positive make it 1
grp =
sum((sign(value)+1) / 2)
over(order by id desc)
from tbl
)
select
id = max(id),
account = max(account),
value = sum(value),
date = max(date)
from summary
group by grp
order by grp desc
输出：
ID ACCOUNT VALUE DATE
0 Tom 10 2001-06-07
4 Tom 2 2001-06-11
5 Tom 30 2001-06-12
8 Tom -2 2001-06-15
11 Tom 0 2001-06-16
12 Tom -2 2001-06-17
答案 1 :(得分：2)
适用于Sql Server 2008：http://www.sqlfiddle.com/#!3/4b7cb/2
with a as
(
select z.id,z.account,z.date, z.value, coalesce(x.id,-1) as grp
from tbl z
outer apply
(
select top 1 y.id
from tbl y where y.id < z.id and y.value > 0
order by y.id desc
) as x
)
select
id = max(id),
account = max(account),
value = sum(value),
date = max(date)
from a
group by grp
order by id
数据样本：
CREATE TABLE [tbl]
([ID] int, [account] varchar(3), [Value] int, [date] date);
INSERT INTO [tbl]
([ID], [account], [Value], [date])
VALUES
(0, 'Tom', 10, '2001-06-07 00:00:00'),
(1, 'Tom', -9, '2001-06-08 00:00:00'),
(2, 'Tom', -5, '2001-06-09 00:00:00'),
(3, 'Tom', -4, '2001-06-10 00:00:00'),
(4, 'Tom', 20, '2001-06-11 00:00:00'),
(5, 'Tom', 30, '2001-06-12 00:00:00'),
(6, 'Tom', -4, '2001-06-13 00:00:00'),
(7, 'Tom', -5, '2001-06-14 00:00:00'),
(8, 'Tom', 7, '2001-06-15 00:00:00'),
(9, 'Tom', -5, '2001-06-16 00:00:00'),
(10, 'Tom', -5, '2001-06-16 00:00:00'),
(11, 'Tom', 10, '2001-06-16 00:00:00'),
(12, 'Tom', -2, '2001-06-17 00:00:00');
输出：
ID ACCOUNT VALUE DATE
0 Tom 10 2001-06-07
4 Tom 2 2001-06-11
5 Tom 30 2001-06-12
8 Tom -2 2001-06-15
11 Tom 0 2001-06-16
12 Tom -2 2001-06-17
工作原理：
select *, coalesce(x.id,-1) as z_id
from tbl z
outer apply
(
select top 1 y.id
from tbl y where y.id < z.id and y.value > 0
order by y.id desc
) as x
order by z.id desc;
输出：
ID ACCOUNT VALUE DATE Z_ID
12 Tom -2 2001-06-17 11
11 Tom 10 2001-06-16 8
10 Tom -5 2001-06-16 8
9 Tom -5 2001-06-16 8
8 Tom 7 2001-06-15 5
7 Tom -5 2001-06-14 5
6 Tom -4 2001-06-13 5
5 Tom 30 2001-06-12 4
4 Tom 20 2001-06-11 0
3 Tom -4 2001-06-10 0
2 Tom -5 2001-06-09 0
1 Tom -9 2001-06-08 0
0 Tom 10 2001-06-07 -1
答案 2 :(得分：1)
使用MySQL：http://www.sqlfiddle.com/#!2/61773/18
这里的Mysqlism太多，所选字段不需要在GROUP BY上：
select
id, account, sum(value), date
from
(
select *, @gn := @gn + ( (sign(value)+1) / 2 ) as gn
from (tbl, (select @gn := 0)) vars
order by id desc
) as x
group by gn
order by id
最好像这样写，意图是明确的：
select
max(id) as id, max(account) as account, sum(value) as value, max(date) as date
from
(
select *, @gn := @gn + ( (sign(value)+1) / 2 ) as gn
from (tbl, (select @gn := 0) vars)
order by id desc
) as x
group by gn
order by gn desc
sign(value)的逻辑解释，请参阅：https://stackoverflow.com/a/10514961
答案 3 :(得分：0)
在复制我的答案时，我意识到你需要它用于sql-server或plsql。我的解决方案适用于MySQL。对于那个很抱歉。也许它无论如何都有帮助。
不使用cursors我认为没办法。
所以来了。
测试数据：
drop table if exists test;
create table test(id int, value int, c_date date);
insert into test values
( 0 , 10 ,'2012/6/7'),
( 1 , -9 ,'2012/6/8'),
( 2 , -5 ,'2012/6/9'),
( 3 , -4 ,'2012/6/10'),
( 4 , 20 ,'2012/6/11'),
( 5 , 30 ,'2012/6/12'),
( 6 , -4 ,'2012/6/13'),
( 7 , -5 ,'2012/6/14'),
( 8 , 7 ,'2012/6/15'),
( 9 , -5 ,'2012/6/16'),
( 10 , -5 ,'2012/6/16'),
( 11 , 10 ,'2012/6/16'),
( 12 , -2 ,'2012/6/17');
您需要的程序：
drop procedure if exists read_test;
create procedure read_test()
begin
declare done int default false;
declare v_id, v_value int;
declare v_date date;
declare adding_result int;
declare cur cursor for select id, value, c_date from test order by id;
declare continue handler for not found set done = true;
drop table if exists tmp_result;
create temporary table tmp_result as select * from test where 1 = 0;
set adding_result = 0;
open cur;
read_loop: LOOP
fetch cur into v_id, v_value, v_date;
if done then
leave read_loop;
end if;
if v_value >= 0 then
set adding_result = adding_result + v_value;
insert into tmp_result (id, value, c_date) values (v_id, adding_result, v_date);
set adding_result = 0;
else
set adding_result = adding_result + v_value;
end if;
end loop;
insert into tmp_result (id, value, c_date) values (v_id, adding_result, v_date);
close cur;
select * from tmp_result;
end;
最后
call read_test();
就是这样。输出：
id value c_date
0 10 2012-06-07
4 2 2012-06-11
5 30 2012-06-12
8 -2 2012-06-15
11 0 2012-06-16
12 -2 2012-06-17