Question

我的桌子;

Date       |  Cost

01.01.2010 |  100

02.01.2010 |  200

03.01.2010 |  300

04.01.2010 |  400

10.01.2010 |  800

11.01.2010 |  800

12.01.2010 |  800

25.01.2010 |  500

26.01.2010 |  500

05.02.2010 |  600

13.02.2010 |  700

15.02.2010 |  700

ı想在'01 .01.2010'和'28 .02.2010'之间制作“日期”每周视图

Week 1 | Week 2 | week 3 | week . .. . 

1000   | 2400   |    0   | 32432.... . .

如何帮助谢谢你？

Answer 1

SQL> create table mytable (the_date,cost)
  2  as
  3  select date '2010-01-01', 100 from dual union all
  4  select date '2010-01-02', 200 from dual union all
  5  select date '2010-01-03', 300 from dual union all
  6  select date '2010-01-04', 400 from dual union all
  7  select date '2010-01-10', 800 from dual union all
  8  select date '2010-01-11', 800 from dual union all
  9  select date '2010-01-12', 800 from dual union all
 10  select date '2010-01-25', 500 from dual union all
 11  select date '2010-01-26', 500 from dual union all
 12  select date '2010-02-05', 600 from dual union all
 13  select date '2010-02-13', 700 from dual union all
 14  select date '2010-02-15', 700 from dual
 15  /

Table created.

此查询使用MAX-DECODE作为标准数据透视技术。如果您使用的是版本11，则还可以使用PIVOT运算符。以下版本适用于任何版本。

SQL> select nvl(max(decode(the_week,'01',cost)),0) "Week 1"
  2       , nvl(max(decode(the_week,'02',cost)),0) "Week 2"
  3       , nvl(max(decode(the_week,'03',cost)),0) "Week 3"
  4       , nvl(max(decode(the_week,'04',cost)),0) "Week 4"
  5       , nvl(max(decode(the_week,'05',cost)),0) "Week 5"
  6       , nvl(max(decode(the_week,'06',cost)),0) "Week 6"
  7       , nvl(max(decode(the_week,'07',cost)),0) "Week 7"
  8       , nvl(max(decode(the_week,'08',cost)),0) "Week 8"
  9       , nvl(max(decode(the_week,'09',cost)),0) "Week 9"
 10    from ( select to_char(the_date,'ww') the_week
 11                , sum(cost) cost
 12             from mytable
 13            where the_date between date '2010-01-01' and date '2010-02-28'
 14            group by to_char(the_date,'ww')
 15         )
 16  /

    Week 1     Week 2     Week 3     Week 4     Week 5     Week 6     Week 7     Week 8     Week 9
---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ----------
      1000       2400          0       1000          0        600       1400          0          0

1 row selected.

此致罗布。

Answer 2

select to_char(date, 'ww'), sum(cost)
from table
group by to_char(date, 'ww');

或者这些行中的某些内容应该按周计算结果中的周数。 Link到Oracle 11g to_char语法，link用于格式化值。如果没有这个并且您不需要周数截码（日期，'DAY'）可能就是您要找的。

Answer 3

不是优雅的解决方案，但它的作品......

   SELECT SUM(Week1) Week1, SUM(Week2) Week2 ... SUM(Week36) Week36,
   SUM(Week36) Week37
   FROM   (SELECT DECODE(WeekNo, 1, Cost, 0) Week1,
            DECODE(WeekNo, 2, Cost, 0) Week2,
            ...
            DECODE(WeekNo, 36, Cost, 0) Week36,
            DECODE(WeekNo, 37, Cost, 0) Week37
     FROM   (SELECT to_char(DateFrom, 'IW') WeekNo, SUM(cost) Cost
              FROM   (SELECT trunc(SYSDATE) + LEVEL - 1 DateFrom,
                              LEVEL * 100 Cost
                       FROM   dual
                       CONNECT BY LEVEL < 40)
              GROUP  BY to_char(DateFrom, 'IW')))

如何从Oracle中的单个列中选择数据并在多个列中显示结果？

3 个答案: