【发布时间】：2020-02-17 16:05:00
【问题描述】：
有人可以就我如何修改 SQL 代码以显示超过 3 个工作日的工作提供任何帮助吗？
我有以下代码：
select
central_enquiry.enquiry_number,
central_enquiry.enquiry_time,
central_enquiry.officer_code,
type_of_service.service_code,
type_of_service.service_name,
enquiry_subject. subject_code,
enquiry_subject.subject_name,
central_site.site_name,
central_enquiry.enquiry_address,
central_enquiry.enquiry_desc,
enquiry_status.enq_status_code,
enquiry_status.enq_status_name,
central_enquiry.log_effective_date,
central_enquiry.follow_up_date,
CASE
when round((SYSDATE - central_enquiry.enquiry_time),2) >=3 then 'Over 3 days'
else ''
end as Days_since_reported
from
central_enquiry
inner join enquiry_subject on enquiry_subject.subject_code = central_enquiry.subject_code
inner join type_of_service on type_of_service.service_code = enquiry_subject.service_code
inner join enquiry_status_log on central_enquiry.enquiry_number = enquiry_status_log.enquiry_number and central_enquiry.enquiry_log_number = enquiry_status_log.enquiry_log_number
inner join enquiry_status on enquiry_status.enq_status_code = enquiry_status_log.enq_status_code
inner join central_site on central_site.site_code = central_enquiry.site_code
where
type_of_service.service_code = 'ECPE' and
round((SYSDATE - central_enquiry.enquiry_time),2) >=3 and
central_enquiry.officer_code = 'BSO' and
central_enquiry.outstanding_flag = 'Y'
order by central_enquiry.enquiry_number
这显示了基于当前日期超过 3 天前记录的所有作业。
where round((SYSDATE - central_enquiry.enquiry_time),2) >=3
但是，我希望这仅显示已在 3 个工作日前记录的作业 - 因此，如果作业是在 2 月 13 日星期四记录的 - 它只会在超过同一时间时显示在报告中2 月 18 日星期二。
【问题讨论】：
-
轻松为您提供帮助：minimal reproducible example.