如何在 MySQL 中选择特定星期几的行?
mysqlmysqli database更新于 2025/11/14 21:00:17
对于特定星期几，请使用 DAYOFWEEK()。
首先我们创建一个表 −
mysql> create table DemoTable785 ( CustomerId int NOT NULL AUTO_INCREMENT PRIMARY KEY, CustomerName varchar(100), ShoppingDate date ); Query OK, 0 rows affected (0.61 sec)
使用 insert 命令在表中插入一些记录 −
mysql> insert into DemoTable785(CustomerName,ShoppingDate) values('Chris','2019-07-03');
Query OK, 1 row affected (0.20 sec)
mysql> insert into DemoTable785(CustomerName,ShoppingDate) values('Robert','2019-07-01');
Query OK, 1 row affected (0.13 sec)
mysql> insert into DemoTable785(CustomerName,ShoppingDate) values('David','2019-07-06');
Query OK, 1 row affected (0.13 sec)
mysql> insert into DemoTable785(CustomerName,ShoppingDate) values('Carol','2019-07-19');
Query OK, 1 row affected (0.19 sec)
使用 select 语句显示表中的所有记录 −
mysql> select *from DemoTable785;
这将产生以下输出 -
+------------+--------------+--------------+ | CustomerId | CustomerName | ShoppingDate | +------------+--------------+--------------+ | 1 | Chris | 2019-07-03 | | 2 | Robert | 2019-07-01 | | 3 | David | 2019-07-06 | | 4 | Carol | 2019-07-19 | +------------+--------------+--------------+ 4 rows in set (0.00 sec)
以下查询用于选择位于特定星期几的行 −
mysql> select *from DemoTable785 where DAYOFWEEK(ShoppingDate)=2;
这将产生以下输出 -
+------------+--------------+--------------+ | CustomerId | CustomerName | ShoppingDate | +------------+--------------+--------------+ | 2 | Robert | 2019-07-01 | +------------+--------------+--------------+ 1 row in set (0.00 sec)
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