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不使用MySQL COUNT()执行多次计数？

2024-05-17 20:35:08 340

要进行计数，可以将SUM()与CASE语句一起使用来确定条件。让我们首先创建一个表-

create table DemoTable1485
   -> (
   -> StudentId int NOT NULL AUTO_INCREMENT PRIMARY KEY,
   -> StudentName varchar(20),
   -> StudentSubject varchar(20)
   -> );

使用插入命令在表中插入一些记录-

insert into DemoTable1485(StudentName,StudentSubject) values('Chris','MySQL');
insert into DemoTable1485(StudentName,StudentSubject) values('Robert','MongoDB');
insert into DemoTable1485(StudentName,StudentSubject) values('Robert','MongoDB');
insert into DemoTable1485(StudentName,StudentSubject) values('Chris','Java');

使用select语句显示表中的所有记录-

select * from DemoTable1485;

这将产生以下输出-

+-----------+-------------+----------------+
| StudentId | StudentName | StudentSubject |
+-----------+-------------+----------------+
|         1 | Chris       |          MySQL |
|         2 | Robert      |        MongoDB |
|         3 | Robert      |        MongoDB |
|         4 | Chris       |           Java |
+-----------+-------------+----------------+
4 rows in set (0.00 sec)

这是不使用COUNT()方法即可执行多次计数的查询-

select StudentSubject,
   -> sum(case when StudentName = 'Chris' THEN 1 ELSE 0 END) Chris_Count,
   -> sum(case when StudentName = 'Robert' THEN 1 ELSE 0 END) Robert_Count
   -> from DemoTable1485
   -> group by StudentSubject;

这将产生以下输出-

+----------------+-------------+--------------+
| StudentSubject | Chris_Count | Robert_Count |
+----------------+-------------+--------------+
| MySQL          |           1 |            0 |
| MongoDB        |           0 |            2 |
| Java           |           1 |            0 |
+----------------+-------------+--------------+
3 rows in set (0.00 sec)

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不使用MySQL COUNT()执行多次计数？

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