Substr date date

我與老師有一個爭論，他說在oracle的日期格式欄中使用substring是不可接受的。因爲substr應該在字符串上使用而不是在日期上使用。你對此有何看法？Substr date date

的TAST是：

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

查詢是這樣的：

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

來源

2017-03-22 Daniel

什麼是你列的數據類型'ACCOMODATION.RESERVATION.ARRIVAL'和'ACCOMODATION.RESERVATION.LEAVE'？ – FDavidov

今天的提示：表別名。（使查詢更容易編寫和閱讀。） – jarlh

順便說一下，你的老師絕對完全100％正確！ – FDavidov

正如評論，嘗試TO_CHAR的日期。如果已經存儲爲char，那麼你已經有了更大的問題

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')))

來源

2017-03-22 10:18:49 JohnHC

是的，我首先教了類似的東西，但我覺得有一種更簡單的方法來解決它。但我錯了。 – Daniel

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