双重职责问题
问题来自于《SQL puzzles and answers》一书的第36个Puzzle。问题的描述很简单，书中给出了很多种解答，我只能想到其中的1、2种，故在这里进行一下分享。有意思的是书中的解法1是无法通过SQL解析的，大家可以查看原书并进行尝试。本文中的解法在SQL Server 2008中测试通过，可能与原书有部分差异。
问题描述
我们有一张权责表：
| person | role |
| Smith | O |
| Smith | D |
| Jones | O |
| White | D |
| Brown | X |
表中包含2列，person列中存储人名，role列中存储职责代码，O代表Officer，D代表Director等等。创建表的脚本：
use
tempdb;
create
table
Roles
(
person
char
(
5
)
not
null
,
role
char
(
1
)
not
null
);
insert
into
Roles
(
person,
role
)
values
(
'
Smith
'
,
'
O
'
),
(
'
Smith
'
,
'
D
'
),
(
'
Jones
'
,
'
O
'
),
(
'
White
'
,
'
D
'
),
(
'
Brown
'
,
'
X
'
);
需要写一个查询，只关心role为O和D的人。若有人既有职责代码O，又有职责代码D，则合并显示为B。上表的查询结果应为：
| person | combined_role |
| Smith | B |
| Jones | O |
| White | D |
解决方案1
With
DirectorRole
as
(
select
person,
role
from
Roles
where
role
=
'
D
'
),
OfficerRole
as
(
select
person,
role
from
Roles
where
role
=
'
O
'
)
select
coalesce
(DirectorRole.person, OfficerRole.person)
as
person,
case
when
(DirectorRole.person
is
not
null
and
OfficerRole.person
is
not
null
)
then
'
B
'
else
coalesce
(DirectorRole.role, OfficerRole.role)
end
as
combined_role
from
DirectorRole
full
outer
join
OfficerRole
on
DirectorRole.person
=
OfficerRole.person;
解决方案2
select
person,
'
B
'
as
combined_role
from
Roles
where
role
in
(
'
O
'
,
'
D
'
)
group
by
person
having
COUNT
(
*
)
=
2
union
all
select
person,
max
(role)
as
combined_role
from
Roles
where
role
in
(
'
O
'
,
'
D
'
)
group
by
person
having
COUNT
(
*
)
=
1
解决方案3
select
distinct
R1.person,
case
when
exists
(
select
*
from
Roles
as
R2
where
R2.person
=
R1.person
and
R2.role
<>
R1.role
and
R2.role
in
(
'
D
'
,
'
O
'
))
then
'
B
'
else
R1.role
end
as
combined_role
from
Roles
as
R1
where
R1.role
in
(
'
O
'
,
'
D
'
);
解决方案4
select
person,
case
when
COUNT
(
*
)
=
1
then
max
(role)
else
'
B
'
end
as
combined_role
from
Roles
where
role
in
(
'
D
'
,
'
O
'
)
group
by
person;
解决方案5
select
person,
case when MIN (role) <> MAX (role)
then ' B '
else MIN (role)
end as combined_role
from
Roles
where
role in ( ' D ' , ' O ' )
group by
person;
解决方案6
select
person,
SUBSTRING ( ' ODB ' , sum ( charindex (role, ' OD ' , 1 )), 1 ) as combined_role
from
Roles
where
role in ( ' D ' , ' O ' )
group by
person;
总结
从这个例子里可以充分感受到SQL的丰富性，大家可以进一步比较各方法的性能。第6个方案相对前几个方案难理解一些，我在这里多说两句。此查询首先按照人名来分组（这点很重要）。若某人只有一个O职责，charindex函数将返回1；若某人只有一个D职责，charindex函数将返回2；若某人既有O又有D职责，charindex函数将返回2条记录值分别为1和2，经外层聚合函数sum求和后为3。之后substring再将sum的结果值作为索引取‘ODB’中的一个字母。