我们能在 MySQL 中通过 DATEPART() 选择行吗?可能吗?

mysqlmysqli database更新于 2025/1/31 21:32:17

MySQL 中没有 DATEPART() 函数,您需要使用 MONTH() 函数从日期列中提取月份名称。语法如下:

SELECT *FROM yourTableName WHERE MONTH(yourDateColumnName)=yourValue;

为了理解上述语法,让我们创建一个表。创建表的查询如下:

mysql> create table SelectRowFromDatePart
   -> (
   -> Id int NOT NULL AUTO_INCREMENT,  
   -> LoginDate date,
   -> PRIMARY KEY(Id)
   -> );
Query OK, 0 rows affected (2.42 sec)

现在您可以使用 insert 命令在表中插入一些记录。查询如下:

mysql> insert into SelectRowFromDatePart(LoginDate) values(curdate());
Query OK, 1 row affected (0.21 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values(date_add(curdate(),interval -3 year));
Query OK, 1 row affected (0.19 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2014-05-21');
Query OK, 1 row affected (0.21 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2015-08-25');
Query OK, 1 row affected (0.23 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2018-12-31');
Query OK, 1 row affected (0.18 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2011-12-21');
Query OK, 1 row affected (0.21 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2013-10-02');
Query OK, 1 row affected (0.12 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2006-12-11');
Query OK, 1 row affected (0.17 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2010-03-09');
Query OK, 1 row affected (0.40 sec)
mysql> insert into SelectRowFromDatePart(LoginDate) values('2012-12-12');
Query OK, 1 row affected (0.70 sec)

使用 select 语句显示表中的所有记录。查询如下:

mysql> select *from SelectRowFromDatePart;

输出结果如下:

+----+------------+
| Id | LoginDate  |
+----+------------+
|  1 | 2019-01-31 |
|  2 | 2016-01-31 |
|  3 | 2014-05-21 |
|  4 | 2015-08-25 |
|  5 | 2018-12-31 |
|  6 | 2011-12-21 |
|  7 | 2013-10-02 |
|  8 | 2006-12-11 |
|  9 | 2010-03-09 |
| 10 | 2012-12-12 |
+----+------------+
10 rows in set (0.00 sec)

这是按日期部分选择行的查询,即 MONTH():

mysql> select *from SelectRowFromDatePart where MONTH(LoginDate)=12;

输出结果如下:

+----+------------+
| Id | LoginDate  |
+----+------------+
|  5 | 2018-12-31 |
|  6 | 2011-12-21 |
|  8 | 2006-12-11 |
| 10 | 2012-12-12 |
+----+------------+
4 rows in set (0.00 sec)


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