今天看到有群友在群里了这个问题，问题描述见下图。这种需求在做报表统计时经常会遇到，会的人觉得不难，没有接触过可能会被困住，所以我把它拿出来和大家分享。
图中已把问题描述清楚，再结合数据看就更清晰了。用算法来描述就是：给定一张表(假设表名叫作 t)，t 表有字段(oid,period,amount,balance),对同一时期(period 字段的值相等)的金额(amount)按 oid 的顺序做累加求和操作，累加的结果放到 balance 字段。
结合数据来看，在原始数据中，当 oid = 1 时，amount = 3500.00，由于此时只有一条记录，所以 balance = 3500.00 ；当 oid = 2 时，amount = 5100.00，balance = 3500.00 + 5100.00 = 8600.00；同理，当 oid = 3 时，balance = 3500.00 + 5100.00 + 10000.00 = 18600.00 。
我们通过自关联来实现累计求和的结果，关联的条件可以这么写 t as t1 INNER JOIN t as t2 ON t2.period = t1.period AND t2.oid <= t1.oid。其中，t1 是主表，用来限定 t2 可以累加的数据的范围。比如，当 t1.oid = 5 时，t2.oid 只能是(4,5)，对应的 balance 的计算过程就是 2560.00(t2.oid = 4 时的 amount) + 4700.00(t2.oid = 5 时的 amount) 。
完整的 SQL 就可以这么写：
WITH t AS
(SELECT
1 AS oid,
2009 AS period,
3500.00 AS amount,
0.00 AS balance
UNION
SELECT
2 AS oid,
2009 AS period,
5100.00 AS amount,
0.00 AS balance
UNION
SELECT
3 AS oid,
2009 AS period,
10000.00 AS amount,
0.00 AS balance
UNION
SELECT
4 AS oid,
2010 AS period,
2560.00 AS amount,
0.00 AS balance
UNION
SELECT
5 AS oid,
2010 AS period,
4700.00 AS amount,
0.00 AS balance)
# 上面的是造数据的SQL，下面这段才是核心实现SELECT
t1.oid,
t1.period,
t1.amount,
SUM(t2.amount) AS balance
FROM
t AS t1
LEFT JOIN t AS t2
ON t2.period = t1.period
AND t2.oid <= t1.oid
GROUP BY t1.oid,
t1.period,
t1.amount ;
看不惯自连接的写法可以换成标量子查询：
SELECT
oid,
period,
amount,
(SELECT
SUM(amount)
FROM
t
WHERE period = t1.period
AND oid <= t1.oid) AS balance
FROM
t AS t1
如果你的 MySQL 环境是 8.0 及其以上，可以尝试使用窗口函数。
SELECT
oid,
period,
amount,
SUM(amount) over (PARTITION BY period
ORDER BY oid) AS balance
FROM
t
实现累计求和差不多就是这些写法。如果你有不同的实现方式，欢迎在评论区留言，和大家分享你的思路。
来源：SQL实现
作者：zero
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