如果 SQL sum() 达到类型容量（溢出）会怎样？

Question

我知道这个问题取决于供应商，但我应该担心像 SUM 这样的聚合函数是否在小型类型上运行？

例如 MariaDB 使用 4 个字节来表示类型 INT。开发者可能会假设每笔交易的金额不超过几千。

但是，如果我们尝试为所有部门获取一整年的收入，会发生什么？例如：

-- CREATE TABLE income (dt DATETIME, department INT, amount INT);
SELECT SUM(amount) FROM income WHERE dt BETWEEN '2014-01-01' and '2014-12-31'

增加存储大小只是为了修复聚合函数SUM的溢出问题看起来很愚蠢。

我应该担心什么？ SQL 92/99/2008 标准是否有任何保证或澄清？

JDBC 驱动程序有什么特殊支持吗？

我应该重写表单中的选择吗：

SELECT SUM(CAST(amount AS BIGINT)) FROM income
  WHERE dt BETWEEN '2014-01-01' and '2014-12-31'

Answer 1

在 mysql 上测试相当容易：

32位溢出：

mysql> select sum(x) from (
    select pow(2,31) as x
    union all
    select pow(2,31)
    union all
    select pow(2,31)
) as bignums;
+------------+
| sum(x)     |
+------------+
| 6442450944 | // returned as a "bigint"
+------------+
1 row in set (0.00 sec)

64位：

mysql> select sum(x) from (
    select pow(2,63) as x
    union all
    select pow(2,63)
    union all
    select pow(2,63)
) as bignums;
+-----------------------+
| sum(x)                |
+-----------------------+
| 2.7670116110564327e19 | // returned as float
+-----------------------+
1 row in set (0.00 sec)

双：

mysql> select sum(x) from (
    select 1.7e+308 as x
    union all
    select 1.7e+308
    union all
    select 1.7e+308
) as bignums;
+--------+
| sum(x) |
+--------+
|      0 |
+--------+

在 mysql 上测试相当容易：

32位溢出：

mysql> select sum(x) from (
    select pow(2,31) as x
    union all
    select pow(2,31)
    union all
    select pow(2,31)
) as bignums;
+------------+
| sum(x)     |
+------------+
| 6442450944 | // returned as a "bigint"
+------------+
1 row in set (0.00 sec)

64位：

mysql> select sum(x) from (
    select pow(2,63) as x
    union all
    select pow(2,63)
    union all
    select pow(2,63)
) as bignums;
+-----------------------+
| sum(x)                |
+-----------------------+
| 2.7670116110564327e19 | // returned as float
+-----------------------+
1 row in set (0.00 sec)

双：

mysql> select sum(x) from (
    select 1.7e+308 as x
    union all
    select 1.7e+308
    union all
    select 1.7e+308
) as bignums;
+--------+
| sum(x) |
+--------+
|      0 |
+--------+

---

评论跟进：

mysql> describe overflow
    -> ;
+-------+------------+------+-----+---------+-------+
| Field | Type       | Null | Key | Default | Extra |
+-------+------------+------+-----+---------+-------+
| x     | int(11)    | YES  |     | NULL    |       |
| y     | bigint(20) | YES  |     | NULL    |       |
| z     | double     | YES  |     | NULL    |       |
+-------+------------+------+-----+---------+-------+
3 rows in set (0.00 sec)

mysql> select * from overflow;
+------------+---------------------+---------+
| x          | y                   | z       |
+------------+---------------------+---------+
| 2147483647 | 9223372036854775807 | 1.7e308 |
| 2147483647 | 9223372036854775807 | 1.7e308 |
| 2147483647 | 9223372036854775807 | 1.7e308 |
+------------+---------------------+---------+
3 rows in set (0.00 sec)

mysql> select sum(x), sum(y), sum(z) from overflow;
+------------+----------------------+--------+
| sum(x)     | sum(y)               | sum(z) |
+------------+----------------------+--------+
| 6442450941 | 27670116110564327421 |      0 |
+------------+----------------------+--------+
1 row in set (0.00 sec)

Answer 2

Postgres 处理这个没有溢出或截断：

来自手册：

sum(expression)，返回类型：bigint 表示 smallint 或 int 参数，numeric 表示 bigint 参数，否则与参数数据类型相同

http://www.postgresql.org/docs/current/static/functions-aggregate.html

快速测试证明：

psql (9.4.5)
Type "help" for help.

postgres=> create table x (amount int);
CREATE TABLE
postgres=>
postgres=> insert into x values (2147483647), (2147483647);
INSERT 0 2
postgres=> select sum(amount)
wbtest-> from x;
    sum
------------
 4294967294
(1 row)

postgres=>

---

有趣的是，SQL 标准要求语句在这种情况下失败：

如果在计算 AF 结果期间，中间结果不能以包含该中间结果的站点的声明类型表示，则
...
否则，会引发异常情况：数据异常 — 数值超出范围。

^{（AF = 聚合函数）}

Answer 3

当我理解你的正确时，你问的是如果溢出会发生什么。

至少对于 SQL Server，请查阅此文档：

https://msdn.microsoft.com/de-de/library/ms187810%28v=sql.120%29.aspx

这里说明sum() 的返回类型对于特定输入类型是什么：

Expression result               Return type
------------------------------------------------
tinyint                         int
smallint                        int
int                             int
bigint                          bigint
decimal category (p, s)         decimal(38, s)
money and smallmoney category   money
float and real category         float 

这意味着，可能确实存在溢出。所以我建议你使用float 或money 来表示薪水，而不是int。