SQL 累计求和 - 码农教程

今天看到有群友在群里提了这个问题，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 。

我们可通过自关联来实现累计求和的结果，关联的条件这么写 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。