为此,您可以使用聚合函数SUM()
。让我们首先创建一个表-
mysql> create table DemoTable636 ( StudentId int NOT NULL AUTO_INCREMENT PRIMARY KEY,StudentFirstName varchar(100) );
使用插入命令在表中插入一些记录-
mysql> insert into DemoTable636(StudentFirstName) values('John'); mysql> insert into DemoTable636(StudentFirstName) values('Robert'); mysql> insert into DemoTable636(StudentFirstName) values('Robert'); mysql> insert into DemoTable636(StudentFirstName) values('Sam'); mysql> insert into DemoTable636(StudentFirstName) values('Mike'); mysql> insert into DemoTable636(StudentFirstName) values('John'); mysql> insert into DemoTable636(StudentFirstName) values('Robert');
使用select语句显示表中的所有记录-
mysql> select *from DemoTable636;
这将产生以下输出-
+-----------+------------------+ | StudentId | StudentFirstName | +-----------+------------------+ | 1 | John | | 2 | Robert | | 3 | Robert | | 4 | Sam | | 5 | Mike | | 6 | John | | 7 | Robert | +-----------+------------------+ 7 rows in set (0.00 sec)
以下是对已知(或枚举)不同值的出现进行计数的查询-
mysql> select sum(StudentFirstName='John') AS JOHN_COUNT, sum(StudentFirstName='Robert') AS ROBERT_COUNT, sum(StudentFirstName='Sam') AS SAM_COUNT, sum(StudentFirstName='Mike') AS MIKE_COUNT from DemoTable636;
这将产生以下输出-
+------------+--------------+-----------+------------+ | JOHN_COUNT | ROBERT_COUNT | SAM_COUNT | MIKE_COUNT | +------------+--------------+-----------+------------+ | 2 | 3 | 1 | 1 | +------------+--------------+-----------+------------+ 1 row in set (0.00 sec)