显示超过3个工作日的职位

时间:2020-02-17 16:05:00

标签: sql oracle days

有人可以提供任何关于如何修改SQL代码以显示超过3个工作日的工作的帮助吗?

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

这会向我显示基于当前日期超过3天前已记录的所有作业。

where  round((SYSDATE - central_enquiry.enquiry_time),2) >=3 

但是,我只想让我看到在3个工作日前已记录的作业-因此,如果作业是在2月13日(星期四)记录的,则该作业只会在过去的同一时间显示在报告中2月18日,星期二。

1 个答案:

答案 0 :(得分:1)

您需要按以下方式操作您的状况:

where round((SYSDATE - central_enquiry.enquiry_time),2) 
- case when to_char(SYSDATE,'dy') in ('mon','tue') then 2 else 0 end >=3

干杯!

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