为此,您可以将其IF()
与聚合函数一起使用SUM()
。让我们首先创建一个表-
mysql> create table DemoTable1617 -> ( -> Attendance varchar(20), -> CurrentYear int -> );
使用插入命令在表中插入一些记录-
mysql> insert into DemoTable1617 values('Present',2019); mysql> insert into DemoTable1617 values('Absent',2019); mysql> insert into DemoTable1617 values('Absent',2017); mysql> insert into DemoTable1617 values('Present',2019); mysql> insert into DemoTable1617 values('Present',2018); mysql> insert into DemoTable1617 values('Present',2019);
使用select语句显示表中的所有记录
mysql> select * from DemoTable1617;
这将产生以下输出-
+------------+-------------+ | Attendance | CurrentYear | +------------+-------------+ | Present | 2019 | | Absent | 2019 | | Absent | 2017 | | Present | 2019 | | Present | 2018 | | Present | 2019 | +------------+-------------+ 6 rows in set (0.00 sec)
以下是查询以获取当前和缺勤学生一年的数量的查询
mysql> select sum(if(Attendance='Present',1,0)) as Present , -> sum(if(Attendance='Absent',1,0)) as Absent -> from DemoTable1617 -> where CurrentYear LIKE '2019%';
这将产生以下输出-
+---------+--------+ | Present | Absent | +---------+--------+ | 3 | 1 | +---------+--------+ 1 row in set (0.00 sec)