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数据透视表：

id Jan   Feb   March  April   May   June
=========================================
1  0.12  0.36   0.72   null    null  null
2  null  null   0.11   0.12   0.36   null

结果：

id Jan   Feb   March  April   May   June
=========================================
1  0.12  0.36    0.72   0.72   0.72   0.72    
2  0.00  0.00    0.11   0.12   0.36   0.36

在行/列中单元格的开头，如果没有值，则 null 应替换为 0，如果为 null，则该行中的前一个单元格应填充同一行中的下一个单元格。

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1 回答 1

1 
-- sample table for discussion
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);

-- beginning of script
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;

-- fully populate the table with all id/month combinations filled in
;with cte as (
  select id,Number,month,isnull(Value,0.0)value
  from @tbl
  where Number=1
  union all
  select cte.id,t.Number,t.month,isnull(t.value,cte.Value)
  from cte
  join @tbl t on t.id=cte.id and t.number=cte.number+1
)
-- this is what the original pivot would have looked like
select id, Jan,Feb,Mar,Apr,May,Jun
from (select id,month,value from cte) p
pivot (max(value) for month in (Jan,Feb,Mar,Apr,May,Jun)) v;
于 2012-10-03T20:47:52.127 回答