处理UNPIVOT中的NULL值

Question

我可以将表格取消,但结果中不包含空值.

create table pivot_task
(
age int null,
[a] numeric(8,2),
[b] numeric(8,2),
[c] numeric(8,2),
[d] numeric(8,2),
[e] numeric(8,2)
);

select * from pivot_task;

insert into pivot_task values (18, 0.5, null, 0.6, 1.21, 1.52),
(19, 7.51, 6.51, 5.51, null, 3.53),
(20, 4.52, 4.52, 6.52, 3.53, null);


select age, [over], [av]
from pivot_task
unpivot
(
 [av]
 for [over] in ([a], [b], [c], [d], [e])
) a;

您可以在http://sqlfiddle.com/#!6/2ab59/1上看到18岁[over] b 的结果,并且缺少空值我想要为每次空碰到都包含null.

我发现用不同的值替换null然后替换所有那些常量不同的值方法对我的工作是不可行的.我想只包括在unpivot中.

Answer 1

这很丑陋，但并不依赖于寻找带外替代品NULL：

declare @pivot_task table
(
age int null,
[a] numeric(8,2),
[b] numeric(8,2),
[c] numeric(8,2),
[d] numeric(8,2),
[e] numeric(8,2)
);

insert into @pivot_task values (18, 0.5, null, 0.6, 1.21, 1.52),
(19, 7.51, 6.51, 5.51, null, 3.53),
(20, 4.52, 4.52, 6.52, 3.53, null);


select a.age, pmu.[over], [av]
from (select 'a' as [over] union all select 'b' union all select 'c'
        union all select 'd' union all select 'e') pmu
cross join (select age from @pivot_task) as a
left join
@pivot_task pt
unpivot
(
 [av]
 for [over] in ([a], [b], [c], [d], [e])
) ex
on pmu.[over] = ex.[over] and
   a.age = ex.age

结果：

age         over av
----------- ---- ---------------------------------------
18          a    0.50
18          b    NULL
18          c    0.60
18          d    1.21
18          e    1.52
19          a    7.51
19          b    6.51
19          c    5.51
19          d    NULL
19          e    3.53
20          a    4.52
20          b    4.52
20          c    6.52
20          d    3.53
20          e    NULL

但如果你沿着这条路走下去，你可以UNPIVOT完全消除：

select a.age, pmu.[over],
      CASE pmu.[over]
           WHEN 'a' THEN a.a
           WHEN 'b' THEN a.b
           WHEN 'c' THEN a.c
           WHEN 'd' THEN a.d
           WHEN 'e' THEN a.e
        END [av]
from (select 'a' as [over] union all select 'b' union all select 'c'
        union all select 'd' union all select 'e') pmu
cross join @pivot_task as a