这很难看,但不依赖于必须找到NULL的带外替换：
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,你可以完全取消UNPIVOT：
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
