3

为什么这个SQL Fiddle不起作用？

复制了完整的脚本：

create table tbl (
  id int,
  month varchar(9),
  value float);
insert tbl values
(1,'Jan',0.12),
(1,'Feb',0.36),
(1,'Mar',0.72),
(2,'Mar',0.11),
(2,'Apr',0.12),
(2,'May',0.36);

declare @tbl table (
  id int,
  number int,
  month varchar(9),
  value float);
insert @tbl
select id.id, Months.Number, Months.Name, t.value
from (values(1,'Jan'),
            (2,'Feb'),
            (3,'Mar'),
            (4,'Apr'),
            (5,'May'),
            (6,'Jun')) Months(Number,Name)
cross join (select distinct id from tbl) id
left join tbl t on t.month = Months.name and t.id=id.id;

;with cte as (
  select id,Number,month,isnull(Value,0.0)value
  from @tbl
  where Number=1
  union all
  select cte.id,cte.Number+1,cte.month,isnull(t.value,cte.Value)
  from cte
  join @tbl t on t.id=cte.id and t.number=cte.number+1
)
/*update t
set value=cte.value
from @tbl t
join cte on t.id=cte.id and t.number=cte.number;*/

select id, Jan,Feb,Mar,Apr,May,Jun
from (select id,month,value from /*@tbl*/ cte) p
pivot (max(value) for month in (Jan,Feb,Mar,Apr,May,Jun)) v;

预期结果：

ID  JAN FEB MAR APR MAY JUN
1   0.12    0.36    0.72    0.72    0.72    0.72
2   0   0   0.11    0.12    0.36    0.36

实际结果：

ID  JAN FEB MAR APR MAY JUN
1   0.72    (null)  (null)  (null)  (null)  (null)
2   0.36    (null)  (null)  (null)  (null)  (null)

如果您取消注释已注释掉的代码，它就可以工作。但是，如果您直接从 CTE 中选择，它会显示与UPDATE 语句之后SELECT * FROM CTE相同的值。@tbl

前段时间我花时间分析CTE + ROW_NUMBER()但希望有人能解释一下。

4

1 回答 1

2

我没有从我得到的结果中得到相同的CTE结果@tbl。对于CTE所有的月份都是JAN。如果你用这个改变你的 CTE 定义：

;with cte as (
  select id,Number,month,isnull(Value,0.0)value
  from @tbl
  where Number=1
  union all
  select cte.id,cte.Number+1,t.month /*here there was cte.month*/,
         isnull(t.value,cte.Value) 
  from cte
  join @tbl t on t.id=cte.id and t.number=cte.number+1
)

然后我得到相同的结果。

于 2012-10-03T21:04:50.583 回答