Question

有人可以帮我设计查询以获得低于o / p。

我有数据：

Table X:
Count Name Amount Days 
5     ABC  500    Day1
10    ABC  1000   Day2
3     BCD  100    Day1
4     BDC  200    Day2

我需要的结果：

Name  Count AmountDay1 Count AmountDay2
ABC   5     500         10   1000
BCD   3     100         4    200

等

这可能吗？

我尝试了以下查询，但未获得所需的o / p

select * from X
pivot
(sum(amount) for days in ('Day1', 'Day2'))

请帮忙我正在使用Oracle 11 G

Answer 1

由于您希望在两列Count和Amount上进行数据透视，因此可能更容易使用带有CASE表达式的聚合函数来获取结果：

select name,
  sum(case when days = 'Day1' then "count" else 0 end) CountDay1,
  sum(case when days = 'Day1' then amount else 0 end) AmountDay1,
  sum(case when days = 'Day2' then "count" else 0 end) CountDay2,
  sum(case when days = 'Day2' then amount else 0 end) AmountDay2
from tableX
group by name;

请参阅SQL Fiddle with Demo

如果您想使用PIVOT功能，那么您首先需要取消Amount和Count这两列：

select name, CountDay1, AmountDay1, CountDay2, AmountDay2
from 
(
  select name, col||days as col, value
  from tableX
  unpivot
  (
    value
    for col in ("Count", Amount)
  ) u
) d 
pivot
(
  sum(value)
  for col in ('CountDay1' as CountDay1, 'AMOUNTDay1' as AmountDay1, 
              'CountDay2' as CountDay2, 'AMOUNTDay2' as AmountDay2)
) piv;

请参阅SQL Fiddle with Demo

Answer 2

你需要在两列数量和数量上进行调整：

从x中选择* 枢（分（计）作为计数，（'Day1'，'Day2'）中的天数（金额））

Oracle进入列

2 个答案: