mysql的语句如下

/*下拉框显示与实际值*/

select 

concat('2022年第',help_topic_id,'周(',date_add(DATE_FORMAT(subdate('2022-01-01',date_format('2022-01-01','%w')-1),'%Y-%m-%d'), INTERVAL ((help_topic_id-1)*7) DAY ) 

,'至',date_add(DATE_FORMAT(subdate('2022-01-01',date_format('2022-01-01','%w')-7),'%Y-%m-%d'),INTERVAL ((help_topic_id-1)*7) DAY ),')') as 显示值,

concat(date_add(DATE_FORMAT(subdate('2022-01-01',date_format('2022-01-01','%w')-1),'%Y-%m-%d'), INTERVAL ((help_topic_id-1)*7) DAY ),'至',

date_add(DATE_FORMAT(subdate('2022-01-01',date_format('2022-01-01','%w')-7),'%Y-%m-%d'),INTERVAL ((help_topic_id-1)*7) DAY )) as 实际值

 from mysql.help_topic where help_topic_id

-----------------------显示-------------------------

你的另外数据集中获取实际值就类似这样

where 1=1

and mytime >='${left(周次,10)}'

and mytime <='${mid(周次,12,10)}'

WEEK最后一天不就获取到有几周了

---------------

第一个星期一（有了周一周末就简单了瑟）

SWITCH(WEEKDAY(DATEINYEAR(today(),1)),1,DATEINYEAR(today(),1),

2,DATEDELTA(DATEINYEAR(today(),1),-1),

3,DATEDELTA(DATEINYEAR(today(),1),-2),

4,DATEDELTA(DATEINYEAR(today(),1),-3),

5,DATEDELTA(DATEINYEAR(today(),1),-4),

6,DATEDELTA(DATEINYEAR(today(),1),-5),

0,DATEDELTA(DATEINYEAR(today(),1),-6)

)

oracle :

select * from (

SELECT LEVEL 周次,(TRUNC(sysdate,'YYYY')-7) + (7-TO_CHAR(TRUNC(sysdate,'YYYY'),'D')+1)+(LEVEL-1)*7-1 当周第一天,  

                  (TRUNC(sysdate,'YYYY')-7) + (7-TO_CHAR(TRUNC(sysdate,'YYYY'),'D')+1)+(LEVEL-1)*7+5 当周最后一天  

FROM DUAL CONNECT BY LEVEL<=53 )