Substr date date

我与老师有一个争论，他说在oracle的日期格式栏中使用substring是不可接受的。因为substr应该在字符串上使用而不是在日期上使用。你对此有何看法？Substr date date

的TAST是：

Lists those guest who were in one of our apartment at his/her birthday

查询是这样的：

select * 
from ACCOMODATION.GUEST 
    JOIN ACCOMODATION.RESERVATION ON (USERNAME = GUEST_FK) 
where to_date(CONCAT(SUBSTR(ACCOMODATION.RESERVATION.ARRIVAL,0,2), substr(BDAY, 3, length(BDAY)))) 
     between ACCOMODATION.RESERVATION.ARRIVAL and ACCOMODATION.RESERVATION.LEAVE ;

ACCOMODATION.RESERVATION.ARRIVAL，ACCOMODATION.RESERVATION.LEAVE和BDAY是日期

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2017-03-22 Daniel

什么是你列的数据类型'ACCOMODATION.RESERVATION.ARRIVAL'和'ACCOMODATION.RESERVATION.LEAVE'？ – FDavidov

今天的提示：表别名。（使查询更容易编写和阅读。） – jarlh

顺便说一下，你的老师绝对完全100％正确！ – FDavidov

正如评论，尝试TO_CHAR的日期。如果已经存储为char，那么你已经有了更大的问题

select AG.*, AR.* 
from ACCOMODATION.GUEST AG 
INNER JOIN ACCOMODATION.RESERVATION AR 
    ON (USERNAME = GUEST_FK) 
where (to_char(AR.LEAVE, 'MMDD') > to_char(AR.ARRIVAL, 'MMDD') 
     and to_char(AG.BDAY, 'MMDD') 
      between to_char(AR.ARRIVAL, 'MMDD') 
        and to_char(AR.LEAVE, 'MMDD')) -- for those who did not stay over the new year 
or (to_char(AR.LEAVE, 'MMDD') < to_char(AR.ARRIVAL, 'MMDD') -- for those who stayed over new year 
    and (to_char(AG.BDAY, 'MMDD') >= to_char(AR.ARRIVAL, 'MMDD') 
     or to_char(AG.BDAY, 'MMDD') <= to_char(AR.LEAVE, 'MMDD')))

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2017-03-22 10:18:49 JohnHC

是的，我首先教了类似的东西，但我觉得有一种更简单的方法来解决它。但我错了。 – Daniel

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