LIKE 运算符无法正常使用日期

Posted 2023-05-09

【中文标题】LIKE 运算符无法正常使用日期

【英文标题】：LIKE operator not working properly with date

【发布时间】：2018-07-03 12:46:29

【问题描述】：

我正在尝试返回包含日期年份的记录，让我解释一下，这是我的查询：

$query = "SELECT coach.*
  FROM coach_career coach_cr
  LEFT JOIN coach coach ON coach.id = coach_cr.coach_id
  LEFT JOIN competition_seasons s ON s.id = :season_id
  WHERE coach_cr.team_id = 95 AND `start` LIKE '%2017/2018%'";

我需要选择coach_career上的所有coach字段，一个coach可以在不同的seasons中训练不同的团队，所以我需要接受具有season.name值2017/2018的教练，但 start 日期的日期时间格式如下：2017-01-01

我该如何处理这种情况？

样本数据

教练：

id | name | last_name
 1   foo      test

教练生涯：

coach_id | team_id | start       | end
   1         95       2017-01-01   NULL

competition_seasons

id | name | 
1    2017/2018

【问题讨论】：

样本数据和期望的结果真的很有帮助。

@GordonLinoff 检查我的更新

如果start 是日期，则使用start >= '2017-01-01' AND start < '2019-01-01'

@SalmanA 问题不是开始字段，问题是 s.name 不是日期并且包含 2017/2018

如果你使用'2017/2018' like '%' + convert(varchar(10), datepart(year, [start])) + '%'而不是[start] like '%2017/2018%'会解决你的问题吗？

【参考方案1】：

您可以使用BETWEEN 关键字。

$query = "SELECT coach.*
      FROM coach_career coach_cr
      LEFT JOIN coach coach ON coach.id = coach_cr.coach_id
      LEFT JOIN competition_seasons s ON s.id = :season_id
      WHERE coach_cr.team_id = 95 AND `start` BETWEEN '2017-01-01' and '2018-12-31'";

【讨论】：

s.name 的值是2017/2018%，正如我在查询中所写的那样，所以不是2017/2018% 认为那是s.name，我不能在两者之间使用..

你能用：` WHERE coach_cr.team_id = 95 AND (s.name like '%2017%' OR s.name like '%2018%')`

【参考方案2】：

抱歉，我的答案是针对 SQL Server。这是修正后的版本：

create table coach(id integer, name char(100));
insert into coach(id, name) values(1, 'Jack'), (2, 'Peter');

create table coach_career (coach_id integer, season_id integer, start date, end date);
insert into coach_career values (1, 95, '20170101', null), (2, 95, '20010101', null);

create table competition_seasons (id integer, name char(100));
insert into competition_seasons values (95, '2017/2018');

SELECT coach.*
FROM coach_career cc
LEFT JOIN coach ON coach.id = cc.coach_id
LEFT JOIN competition_seasons s ON s.id = cc.season_id
WHERE cc.season_id = 95 
      AND '2017/2018' like concat('%', cast(extract(year from cc.start) as char(100)), '%');

【讨论】：

感谢您的回复，很遗憾您的查询返回空结果...

这很奇怪，我在paiza.io/projects/SdtzcJ0SoUfg1QS7k3I8QQ?language=mysql 上测试过，它返回Jack

【参考方案3】：

试试这个

$query = "SELECT coach.*
    FROM coach_career coach_cr
    LEFT JOIN coach coach 
        ON coach.id = coach_cr.coach_id
    LEFT JOIN competition_seasons s
        ON s.id = :season_id
    WHERE coach_cr.team_id = 95 
        AND `start` LIKE '%2017%' 
        OR LIKE '%2018%'";