我正在尝试获取此格式的数据
Tname FY FPY Week
test1 87 78 1
test1 45 34 2
test1 34 89 3
test1 34 56 4
test1 56 78 5
test1 68 45 6
test2 45 89 1
test2 76 65 2
test2 45 54 3
test3 56 57.8 2
test4 34 55 4
test4 32 52.2 7
test3 78 49.4 3
test5 89 46.6 2
test5 90 43.8 1
test5 98 41 6
test5 91 38.2 7
test6 92 35.4 5
test6 93 32.6 6
显示结果
Tname week1 Week2 Week3 Week4 Week5 Week6 Week7
FY FPY FY FPY FY FPY FY FPY FY FPY FY FPY FY FPY
test1 87 78 45 34 34 89 34 56 56 78 68 45 NULL NULL
test2 45 89 76 65 45 54 NULL NULL NULL NULL NULL NULL NULL NULL
test3 NULL NULL 56 57.8 78 49.4 NULL NULL NULL NULL NULL NULL NULL NULL
test4 NULL NULL NULL NULL NULL NULL 34 55 NULL NULL NULL NULL 32 52.2
test5 90 43.8 89 46.6 NULL NULL NULL NULL NULL NULL 98 41 91 38.2
test6 NULL NULL NULL NULL NULL NULL NULL NULL 92 35.4 93 32.6 NULL NULL
我想使用存储过程或gridview实现此目的。请指教
答案 0 :(得分：1)
您没有具体说明您使用的是哪种RDBMS，但之前的问题是使用SQL Server标记的，因此这个答案假定您使用的是数据库。
有几种方法可以获得结果。一种简单的方法是使用具有类似于：
的CASE表达式的聚合函数select tname,
sum(case when week = 1 then FY end) Week1FY,
sum(case when week = 1 then FPY end) Week1FPY,
sum(case when week = 2 then FY end) Week2FY,
sum(case when week = 2 then FPY end) Week2FPY,
sum(case when week = 3 then FY end) Week3FY,
sum(case when week = 3 then FPY end) Week3FPY,
sum(case when week = 4 then FY end) Week4FY,
sum(case when week = 4 then FPY end) Week4FPY,
sum(case when week = 5 then FY end) Week5FY,
sum(case when week = 5 then FPY end) Week5FPY,
sum(case when week = 6 then FY end) Week6FY,
sum(case when week = 6 then FPY end) Week6FPY,
sum(case when week = 7 then FY end) Week7FY,
sum(case when week = 7 then FPY end) Week7FPY
from yourtable
group by tname;
由于您需要在两列上进行PIVOT，因此您将首先取消FY和FPY列中数据的取消操作，然后应用PIVOT函数。再次取决于您的RDBMS，有几种方法可以显示数据我将使用CROSS APPLY和UNION ALL显示如何执行此操作：
select tname,
col = 'week_'+cast(week as varchar(2))+'_'+col,
value
from yourtable
cross apply
(
select 'FY', FY union all
select 'FPY', FPY
) c (col, value)
unpivot进程将您的数据从多列转换为多行：
| TNAME | COL | VALUE |
|-------|------------|-------|
| test1 | week_1_FY | 87 |
| test1 | week_1_FPY | 78 |
| test1 | week_2_FY | 45 |
| test1 | week_2_FPY | 34 |
| test1 | week_3_FY | 34 |
然后你可以应用PIVOT功能：
select tname,
week_1_FY, week_1_FPY, week_2_FY, week_2_FPY,
week_3_FY, week_3_FPY, week_4_FY, week_4_FPY,
week_5_FY, week_5_FPY, week_6_FY, week_6_FPY,
week_7_FY, week_7_FPY
from
(
select tname,
col = 'week_'+cast(week as varchar(2))+'_'+col,
value
from yourtable
cross apply
(
select 'FY', FY union all
select 'FPY', FPY
) c (col, value)
) d
pivot
(
sum(value)
for col in (week_1_FY, week_1_FPY, week_2_FY, week_2_FPY,
week_3_FY, week_3_FPY, week_4_FY, week_4_FPY,
week_5_FY, week_5_FPY, week_6_FY, week_6_FPY,
week_7_FY, week_7_FPY)
) piv;
最后，如果周数不明，那么您可以使用动态SQL来获得结果：
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX),
@weekstart int = 1,
@weekend int = 4
select @cols = STUFF((SELECT ',' + QUOTENAME('week_'+cast(week as varchar(2))+'_'+col)
from yourtable
cross apply
(
select 'FY', 1 union all
select 'FPY', 2
) c (col, so)
where week >= @weekstart
and week <= @weekend
group by col, so, week
order by week, so
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT tname, ' + @cols + '
from
(
select tname,
col = ''week_''+cast(week as varchar(2))+''_''+col,
value
from yourtable
cross apply
(
select ''FY'', FY union all
select ''FPY'', FPY
) c (col, value)
where week >= '+cast(@weekstart as varchar(2))+'
and week <= '+cast(@weekend as varchar(2))+'
) x
pivot
(
sum(value)
for col in (' + @cols + ')
) p '
execute sp_executesql @query;
见SQL Fiddle with Demo。所有版本的结果都是：
| TNAME | WEEK_1_FY | WEEK_1_FPY | WEEK_2_FY | WEEK_2_FPY | WEEK_3_FY | WEEK_3_FPY | WEEK_4_FY | WEEK_4_FPY | WEEK_5_FY | WEEK_5_FPY | WEEK_6_FY | WEEK_6_FPY | WEEK_7_FY | WEEK_7_FPY |
|-------|-----------|------------|-----------|------------|-----------|------------|-----------|------------|-----------|------------|-----------|------------|-----------|------------|
| test1 | 87 | 78 | 45 | 34 | 34 | 89 | 34 | 56 | 56 | 78 | 68 | 45 | (null) | (null) |
| test2 | 45 | 89 | 76 | 65 | 45 | 54 | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
| test3 | (null) | (null) | 56 | 58 | 78 | 49 | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) |
| test4 | (null) | (null) | (null) | (null) | (null) | (null) | 34 | 55 | (null) | (null) | (null) | (null) | 32 | 52 |
| test5 | 90 | 44 | 89 | 47 | (null) | (null) | (null) | (null) | (null) | (null) | 98 | 41 | 91 | 38 |
| test6 | (null) | (null) | (null) | (null) | (null) | (null) | (null) | (null) | 92 | 35 | 93 | 33 | (null) | (null) |
答案 1 :(得分：0)
尝试类似
的内容SELECT *
FROM (
SELECT TName, [1] as Week1FY, [2] as Week2FY, [3] as Week3FY,
[4] as Week4FY, [5] as Week5FY, [6] as Week6FY, [7] as Week7FY
FROM( SELECT t.TName, t.FY, t.Week FROM Table t
PIVOT
( MAX(FY) FOR Week in ([1], [2], [3], [4], [5], [6], [7])
)) as P
JOIN
(
SELECT TName, [1] as Week1FPY, [2] as Week2FPY, [3] as Week3FPY,
[4] as Week4FPY, [5] as Week5FPY, [6] as Week6FPY, [7] as Week7FPY
FROM( SELECT t.TName, t.FPY, t.Week FROM Table t
PIVOT
( MAX(FPY) FOR Week in ([1], [2], [3], [4], [5], [6], [7])
)) as Q ON P.TName = Q.TName