如何在 MySQL 中计算另一列中每个不重复值的总和?
借助聚合函数 SUM() 和 GROUP BY 命令,可以获取另一列中每个不重复值的总和。为理解上述概念,让我们创建一个表。创建表的查询如下
mysql> create table SumOfEveryDistinct -> ( -> Id int not null, -> Amount int -> ); Query OK, 0 rows affected (0.59 sec)
使用 insert 命令在表中插入一些记录。查询如下
mysql> insert into SumOfEveryDistinct values(10,100); Query OK, 1 row affected (0.19 sec) mysql> insert into SumOfEveryDistinct values(11,200); Query OK, 1 row affected (0.20 sec) mysql> insert into SumOfEveryDistinct values(12,300); Query OK, 1 row affected (0.14 sec) mysql> insert into SumOfEveryDistinct values(10,400); Query OK, 1 row affected (0.20 sec) mysql> insert into SumOfEveryDistinct values(11,500); Query OK, 1 row affected (0.10 sec) mysql> insert into SumOfEveryDistinct values(12,600); Query OK, 1 row affected (0.13 sec) mysql> insert into SumOfEveryDistinct values(10,700); Query OK, 1 row affected (0.10 sec) mysql> insert into SumOfEveryDistinct values(11,800); Query OK, 1 row affected (0.18 sec) mysql> insert into SumOfEveryDistinct values(12,900); Query OK, 1 row affected (0.19 sec)
使用 select 语句显示表中的所有记录。查询如下
mysql> select *from SumOfEveryDistinct;
输出如下
+----+--------+ | Id | Amount | +----+--------+ | 10 | 100 | | 11 | 200 | | 12 | 300 | | 10 | 400 | | 11 | 500 | | 12 | 600 | | 10 | 700 | | 11 | 800 | | 12 | 900 | +----+--------+ 9 rows in set (0.00 sec)
以下是计算另一列中每个不重复值的总和的查询
mysql> select Id, sum(Amount) as TotalSum from SumOfEveryDistinct -> group by Id;
输出如下
+----+----------+ | Id | TotalSum | +----+----------+ | 10 | 1200 | | 11 | 1500 | | 12 | 1800 | +----+----------+ 3 rows in set (0.00 sec)
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