Question

我有一些代码

select 
'WEEK '|| row_number() over(order by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW')) WEEK,
count(*) 
from opencall
where trunc(to_date(substr(LOGDATE,1,10), 'DD-MM-YY')) BETWEEN '01-JAN-14' AND '31-DEC-14'
group by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW')   
order by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW');

另一个stackoverflow chap为我编写了一个很棒的代码，用于从数据库中提取调用并按周计算/分组它们，如果每周都有调用，它就能很好地工作。

但是我尝试将此代码重新用于其他用途，如果在给定的一周内没有结果，我会得到更少的分组，例如，可能只返回37周而不是当前的50周。

由于这会消除没有电话的周数，因此周数不等于一年中的实际周数。所以我实际上并不知道从几个星期开始按顺序返回来的几周，而不是在没有电话的几周内零。

如果给定的一周没有返回结果，有没有办法返回零值？我想这可能会使代码变得更加复杂。

如果我在一周内使用此代码

e.g：

WEEK 1  231
WEEK 2  657
WEEK 3  529
WEEK 4  606
WEEK 5  594
WEEK 6  517
WEEK 7  683
WEEK 8  664
WEEK 9  583
WEEK 10 740

Answer 1

如果你想知道你在哪个星期展示，那么为什么用row_number（）大惊小怪？

这为您提供以周为单位的年份，从第1周开始的前七天开始，依此类推：

select to_char(mydate, 'ww'), count(*)
from 
(
  select to_date(substr(logdate,1,10), 'dd-mm-yy') as mydate
  from opencall
)
where extract(year from mydate) = 2014
group by to_char(mydate, 'ww')
order by to_char(mydate, 'ww');

（为什么LOGDATE是一个字符串？为什么它的前10个字符的格式为'DD-MM-YY'，长度为8个字符？）

Answer 2

select 
'WEEK '|| row_number() over(order by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW')) WEEK,
row_number() over(order by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW')) mWeek, 
count(*) 
from (SELECT to_char(SYSDATE - (LEVEL * 7),'IW') TheDate
      FROM dual
      CONNECT BY LEVEL <= 52) WoY
LEFT JOIN opencall
  on Woy.TheDate = OpenCall.mWeek
where trunc(to_date(substr(LOGDATE,1,10), 'DD-MM-YY')) BETWEEN '01-JAN-14' AND '31-DEC-14'
group by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW')   
order by trunc(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'SATURDAY'), 'IW');

这样的事可能有用。

Answer 3

我用测试数据尝试了类似的东西并且有效。我希望这也适合你：

select 
'WEEK '|| row_number() over(order by nvl(to_char(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY') ,'SATURDAY'), 'IW'),week)) WEEK
,count(wk) 
from (select t.*,to_char(to_date(substr(LOGDATE,1,10), 'DD-MM-YY'),'IW') wk,w.week from
      table1 t right join (SELECT lpad(level,2,0) week
      FROM dual
      CONNECT BY LEVEL <= 52 ) w on (to_char(to_date(substr(t.LOGDATE,1,10), 'DD-MM-YY'),'IW') = w.week)
      order by w.week)
where trunc(to_date(substr(LOGDATE,1,10), 'DD-MM-YY')) BETWEEN '01-JAN-14' AND '31-DEC-14'
group by nvl(to_char(NEXT_DAY(to_date(substr(LOGDATE,1,10), 'DD-MM-YY') ,'SATURDAY'), 'IW'),week)   
;

Answer 4

- 在架构控制器下创建一个测试表，并在周数倍数为4的所有日子上插入数据。

create table alber.week_table (mydate date, mydescription varchar2(100));
declare
cur_date date :=  to_date('01/01/'||to_char(sysdate,'yyyy'),'dd/mm/yyyy') ;
week number := 01;
begin
  EXECUTE IMMEDIATE 'TRUNCATE TABLE alber.week_table';
  while (cur_date < sysdate) loop
    if mod(week,4) = 0 then
      insert into alber.week_table values(cur_date, to_char(sysdate, 'day dd month yyyy hh24:mi:ss'))  ;
    end if; 
    cur_date := cur_date + 1;
    week := to_number(to_char(cur_date,'IW'));
  end loop;  
end;

- 然后一个可能的解决方案是：

select theweek "Week", sum( decode( nvl(mydescription,0),'0',0,1) ) "Num of Records" 
FROM 
    (
      SELECT to_char(SYSDATE - (LEVEL * 7),'IW') theweek
      FROM dual
      CONNECT BY LEVEL <= 52
     ) w
LEFT JOIN  week_table ON w.theweek =  to_char(mydate,'IW')   
group by theweek
order by 1;

结果

Week Num of Records         
---- ---------------------- 
01   0                      
02   0                      
03   0                      
04   7                      
05   0                      
06   0                      
07   0                      
08   7                      
09   0                      
10   0                      
11   0                      
12   7                      
13   0                      
14   0                      
15   0                      
16   7                      
17   0                      
18   0        
.
.
.

选择后按周分组时返回零 -

4 个答案: