获得非付费会员

发布于 2024-11-28 23:19:51 字数 544 浏览 2 评论 0原文

我有一张会员表，

 member_Id
 member_Firstname

另一张表 Paymentschedules

  paymentschedule_Id
  member_Id
  paymentschedule_amountdue values(100 rs,200rs)
  paymentschedule_amonutpaid values (30 rs,40 rs)
  paymentschedule_ActualDatetobepaid  values(2011-02-03,2011-02-01,2011-03-01)
  paymentschedule_ datepaid  values like(2011-09-12,2011-08-10,2011-11-12,2010-08-01)

我的问题是，考虑到这些值“ActualDatetobepaid”和“amonutpaid”以及“amountdue”和“datepaid”，我可以获取截至“今天”逾期的会员名字吗

？任何人都可以提供帮助吗？一

原文

I have a member table

 member_Id
 member_Firstname

another table Paymentschedules

  paymentschedule_Id
  member_Id
  paymentschedule_amountdue values(100 rs,200rs)
  paymentschedule_amonutpaid values (30 rs,40 rs)
  paymentschedule_ActualDatetobepaid  values(2011-02-03,2011-02-01,2011-03-01)
  paymentschedule_ datepaid  values like(2011-09-12,2011-08-10,2011-11-12,2010-08-01)

My problem is can i get the members firstname who are overdue upto "today" taking into account these values "ActualDatetobepaid" and "amonutpaid " and "amountdue" and "datepaid"

Can any one help on this one

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放我走吧 2024-12-05 23:19:51

您的最后一个请求“添加 7 天”通过在 J0HN 的解决方案中添加 7 天来解决：

SELECT DISTINCT member_Firstname  
FROM member m 
  JOIN Paymentscedules p ON p.member_id = m.member_id 
WHERE DATE_ADD(paymentschedule_ActualDatetobepaid, INTERVAL 7 DAY)< NOW() 
  AND paymentschedule_amountdue > paymentschedule_amonutpaid

Your last request, "Add 7 days", is solved by adding 7 days to J0HN's solution:

SELECT DISTINCT member_Firstname  
FROM member m 
  JOIN Paymentscedules p ON p.member_id = m.member_id 
WHERE DATE_ADD(paymentschedule_ActualDatetobepaid, INTERVAL 7 DAY)< NOW() 
  AND paymentschedule_amountdue > paymentschedule_amonutpaid

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只是我以为 2024-12-05 23:19:51

不确定我理解列的含义，但试试这个：

select distinct member_Firstname 
from member m
join Paymentscedules p on p.member_id = m.member_id
where paymentschedule_ActualDatetobepaid<now() and paymentschedule_amountdue > paymentschedule_amonutpaid

如果您可以提供该列的数据类型用户和列含义，我可以提出更精确的解决方案。

Not sure I understand columns' meanings, but try this:

select distinct member_Firstname 
from member m
join Paymentscedules p on p.member_id = m.member_id
where paymentschedule_ActualDatetobepaid<now() and paymentschedule_amountdue > paymentschedule_amonutpaid

If you could provide data types user for that columns and column meanings I could come up with more precise solution.

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