如何在 MySQL 中获取一年中在校和缺课学生的数量?
mysqlmysqli database
为此，您可以将 IF() 与聚合函数 SUM() 一起使用。我们首先创建一个表。 −
mysql> create table DemoTable1617 -> ( -> Attendance varchar(20), -> CurrentYear int -> ); Query OK, 0 rows affected (0.48 sec)
使用 insert 命令在表中插入一些记录 −
mysql> insert into DemoTable1617 values('Present',2019);
Query OK, 1 row affected (0.15 sec)
mysql> insert into DemoTable1617 values('Absent',2019);
Query OK, 1 row affected (0.20 sec)
mysql> insert into DemoTable1617 values('Absent',2017);
Query OK, 1 row affected (0.13 sec)
mysql> insert into DemoTable1617 values('Present',2019);
Query OK, 1 row affected (0.15 sec)
mysql> insert into DemoTable1617 values('Present',2018);
Query OK, 1 row affected (0.10 sec)
mysql> insert into DemoTable1617 values('Present',2019);
Query OK, 1 row affected (0.14 sec)
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)
以下查询获取一年内在校和缺勤学生的数量
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)
