在 MySQL 中，选择具有相同 ID 但不同对应记录的值(是、否)的数量?
mysqlmysqli database更新于 2024/3/7 6:49:00
为此，您可以将 SUM() 与 CASE 语句一起使用。让我们首先创建一个 −
mysql> create table DemoTable1430
-> (
-> EmployeeId int,
-> isMarried ENUM('YES','NO')
-> );
Query OK, 0 rows affected (0.60 sec)
使用 insert 在表中插入一些记录 −
mysql> insert into DemoTable1430 values(1001,'Yes'); Query OK, 1 row affected (0.19 sec) mysql> insert into DemoTable1430 values(1001,'No'); Query OK, 1 row affected (0.12 sec) mysql> insert into DemoTable1430 values(1001,'Yes'); Query OK, 1 row affected (0.09 sec) mysql> insert into DemoTable1430 values(1001,'Yes'); Query OK, 1 row affected (0.16 sec)
使用 select 语句显示表中的所有记录 −
mysql> select * from DemoTable1430;
这将产生以下输出 −
+------------+-----------+ | EmployeeId | isMarried | +------------+-----------+ | 1001 | YES | | 1001 | NO | | 1001 | YES | | 1001 | YES | +------------+-----------+ 4 rows in set (0.00 sec)
这是选择值计数的查询(Yes, No) −
mysql> select EmployeeId,sum(isMarried='Yes') as NumberOfMarried, -> sum(isMarried='No') as NumberOfUnMarried -> from DemoTable1430 -> group by EmployeeId;
这将产生以下输出 −
+------------+-----------------+-------------------+ | EmployeeId | NumberOfMarried | NumberOfUnMarried | +------------+-----------------+-------------------+ | 1001 | 3 | 1 | +------------+-----------------+-------------------+ 1 row in set (0.00 sec)
