Question

我的DW中有一个TIME维度，我想添加一个阵营来填补今年的赛季。我想做这样的事情：

create or replace
PROCEDURE "PROC_UPDATE_SEASON" 

IS 
CURSOR curs IS   
SELECT * FROM indw.time FOR UPDATE;  
 cur_d NUMBER;   
 cur_m NUMBER;   
 start_d NUMBER;   
 start_m NUMBER;   
 end_d NUMBER;   
 end_m NUMBER;
 cur_Date DATE;     

 BEGIN 
 FOR cs IN curs 
 LOOP 
     cur_d := to_number(to_char(cs.time_DAY,'dd')); 
     cur_m := to_number(to_char(cs.time_DAY,'mm')); 
     cur_date:= cs.time_DAY; -- this is my date dd-mm-yyyy or dd/mm/yyyy


     if ((cur_date in format dd/mm is >= 21/12 and <=31/12) OR cur_date>=01/01 and <=20/03) 
     UPDATE time t set t.time_SEASON = 'Winter' WHERE CURRENT OF curs; 
     else if cur_date_in format dd/mm is between a date and another then it's Spring)
       -- and so on ... 
     END LOOP; 
 END;

有没有办法用to_char和to_date来实现这个目的？我正在尝试，但总是得到错误:(分别比较月份和日期将是一个巨大的，IMO，愚蠢的工作。

你能给我一些提示吗？

我真的只使用像dd / mm＆lt; = another_date_with_only_day_and_month

这样的东西

Answer 1

无需游标和循环：

update indw.time
set t.time_SEASON = case
                      when to_char(time_DAY,'MMDD') between '1221' and '1231' 
                        or to_char(time_DAY,'MMDD') between '0101' and '0320' 
                         then 'Winter'
                      when ...
                    end

更好的是，在Oracle 11G中，您可以根据该CASE表达式向表中添加一个虚拟列，以便每行自动拥有正确的time_season而无需更新。

Answer 2

我无法访问Oracle以便为您播放选项。但我确实有一条建议：
- 如果此代码经常运行，请谨慎使用字符串转换和字符串比较，与数字处理相比，它们通常非常慢 - 但是，如果只是为了偶尔使用，请选择最简单的方法，并使用螺杆性能:)

考虑到这一点，我会使用EXTRACT ...... （注意，正如我所说，我无法测试这个）

UPDATE indw.time
SET    t.time_SEASON = CASE
                         WHEN EXTRACT(MONTH, time_DAY) < 3 THEN 'Winter'
                         WHEN EXTRACT(MONTH, time_DAY) = 3 THEN
                           CASE WHEN EXTRACT(DAY, time_DAY) <= 20 THEN 'Winter' ELSE 'Spring' END
                         WHEN EXTRACT(MONTH, time_DAY) < 6 THEN 'Spring'
                         WHEN EXTRACT(MONTH, time_DAY) = 6 THEN
                           CASE WHEN EXTRACT(DAY, time_DAY) <= 20 THEN 'Spring' ELSE 'Summer' END
                         WHEN EXTRACT(MONTH, time_DAY) < 9 THEN 'Summer'
                         WHEN EXTRACT(MONTH, time_DAY) = 9 THEN
                           CASE WHEN EXTRACT(DAY, time_DAY) <= 20 THEN 'Summer' ELSE 'Autumn' END
                         WHEN EXTRACT(MONTH, time_DAY) < 12 THEN 'Autumn'
                         WHEN EXTRACT(MONTH, time_DAY) = 12 THEN
                           CASE WHEN EXTRACT(DAY, time_DAY) <= 20 THEN 'Autumn' ELSE 'Winter' END
                       END

Answer 3

我会这样从日期获取季节：

CASE
    WHEN EXTRACT(MONTH FROM time_DAY) IN (12,1,2) THEN 'winter'
    WHEN EXTRACT(MONTH FROM time_DAY) IN (3,4,5) THEN 'spring'
    WHEN EXTRACT(MONTH FROM time_DAY) IN (6,7,8) THEN 'summer'
    WHEN EXTRACT(MONTH FROM time_DAY) IN (9,10,11) THEN 'fall'
END season

在SQL中将日期与一年中的季节分开的问题

3 个答案: