MySQL 如何选择落在一周特定日期的行？

对于特定日期，使用DAYOFWEEK()。

让我们首先创建一张表 –

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)

使用插入命令将一些记录插入表中 –

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)

使用选择语句显示表中的所有记录 –

mysql> select *from DemoTable785;

这将生成以下输出 –

+------------+--------------+--------------+
| 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)

下面是选择落在特定一周的查询 –

mysql> select *from DemoTable785 where DAYOFWEEK(ShoppingDate)=2;

这将生成以下输出 –

+------------+--------------+--------------+
| CustomerId | CustomerName | ShoppingDate |
+------------+--------------+--------------+
| 2          | Robert       | 2019-07-01   |
+------------+--------------+--------------+
1 row in set (0.00 sec)

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