Question

给出SQL表

id     date       employee_type  employee_level
1   10/01/2015        other            2
1   09/13/2011      full-time          1
1   09/25/2010       intern            1
2   09/25/2013      full-time          3
2   09/25/2011      full-time          2
2   09/25/2008      full-time          1
3   09/23/2015      full-time          5
3   09/23/2013      full-time          4

是否可以使用employee_type＆＃34来搜索具有employee_type＆＃34; intern＆＃34;的一行的ID以及表格中的上一行（相同的ID以及更晚的日期）;全职＆＃ 34 ;.

在这种情况下，id 1符合我的要求。

非常感谢！

Answer 1

假设您的意思是与前一个日期相同的id，那么您可以使用lag()，这是大多数数据库支持的ANSI标准函数：

select t.*
from table t
where t.id in (select id
               from (select t.*,
                            lag(employee_type) over (partition by id order by date) as prev_et
                     from table t
                    ) tt
               where tt.employee_type = 'intern' and tt.prev_et = 'full-time'
              );

如果您的数据库不支持lag()，您可以使用相关子查询执行类似操作。

Answer 2

我认为请求不是问题所描述的;相反，你似乎想要的是为实习生列出所有行。

SELECT
      t1.*
FROM yourtable AS t1
      INNER JOIN (
            SELECT DISTINCT
                  id
            FROM yourtable
            WHERE employee_type = 'intern'
      ) AS t2 ON t1.id = t2.id
;

或者你可能只想要那些既是'实习'又是'全职'的人，在这种情况下你可以使用下面使用HAVING子句的查询：

SELECT
      t1.*
FROM yourtable AS t1
      INNER JOIN (
            SELECT id
            FROM yourtable
            WHERE employee_type = 'intern'
                  OR employee_type = 'full-time'
            GROUP BY id
            HAVING COUNT(DISTINCT employee_type) > 1
      ) AS t2 ON t1.id = t2.id
;

如何在SQL中连续搜索两行？

2 个答案: