MySQL 如何获取一年中出席和缺席学生的数量

为此，您可以使用IF()以及聚合函数SUM()。让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)

使用插入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)

以下是获取一年中出席和缺席学生数量的查询

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)

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