说我有以下表/数据：
create table #a(id int, name varchar(2), score int)
go
insert #a values(0, 'a1', 1)
insert #a values(1, 'b1', 0)
insert #a values(2, 'c1', 1)
insert #a values(3, 'd1', 0)
insert #a values(4, 'd2', 1)
insert #a values(5, 'e1', 0)
insert #a values(6, 'e2', 2)
insert #a values(7, 'e3', 1)
insert #a values(8, 'e4', 0)
我想选择这些行：
id name score
1 b1 0
2 c1 1
4 d2 1
6 e2 2
标准：
这就是我提出的：
select id, name, score
into #b
from #a
where id > 0
group by left(name, 1)
having score = max(score)
go
select f.*
from #b f
left join #b g on left(g.name, 1) = left(f.name, 1) and g.name > f.name
where g.name is null
order by f.name
在不使用临时表/两个查询/重复（所有这些left s）和效率方面，这可以做得更好吗？
答案 0 :(得分：2)
假设
name是唯一的为了测试决胜局逻辑，我们将添加另一个'e'行：
insert #a values (9,'e5',2) -- same score as the 6/e2/2 record
由于SAP（Sybase）ASE有很多限制......
rank()功能row_number()功能offset/limit条款top子句的有限支持order by子句......我们需要一点“创造性”（阅读：这会有点复杂）
我们要做的第一件事就是找到每个单个字符的最高分数，其中id＆gt; 0：
select left(name,1) as name1,
max(score) as mscore
from #a
where id > 0
group by left(name,1)
order by 1
go
name1 mscore
----- -----------
b 0
c 1
d 1
e 2
接下来，我们将此结果集与原始表连接，根据第一个字符匹配行并得分=最大值（得分）：
select a2.name1,
a1.name,
a2.mscore
from #a a1
join (select left(name,1) as name1,
max(score) as mscore
from #a
where id > 0
group by left(name,1)) a2
on left(a1.name,1) = a2.name1
and a1.score = a2.mscore
and a1.id > 0
order by 1,2
go
name1 name mscore
----- ---- -----------
b b1 0
c c1 1
d d2 1
e e2 2
e e5 2
接下来我们将讨论打破平局规则;我们可以通过将max()函数应用到a1.name列来确保添加相应的group by子句来处理此问题：
select a2.name1,
max(a1.name) as mname,
a2.mscore
from #a a1
join (select left(name,1) as name1,
max(score) as mscore
from #a
where id > 0
group by left(name,1)) a2
on left(a1.name,1) = a2.name1
and a1.score = a2.mscore
and a1.id > 0
group by a2.name1,
a2.mscore
order by 1,2
go
name1 mname mscore
----- ----- -----------
b b1 0
c c1 1
d d2 1
e e5 2
最后一部分是将最终结果与原始表格联系起来以获得id：
select a3.id,
a4.mname as 'name',
a4.mscore as 'score'
from #a a3
join (select a2.name1,
max(a1.name) as mname,
a2.mscore
from #a a1
join (select left(name,1) as name1,
max(score) as mscore
from #a
where id > 0
group by left(name,1)) a2
on left(a1.name,1) = a2.name1
and a1.score = a2.mscore
and a1.id > 0
group by a2.name1,
a2.mscore) a4
on a3.name = a4.mname
order by 1,2
go
id name score
----------- ---- -----------
1 b1 0
2 c1 1
4 d2 1
9 e5 2
注意：在SAP（Sybase）ASE 16.0 SP03 PL01上验证的上述查询/结果。
净结果......
虽然可以通过单个查询完成所需的操作，但编码有点复杂（并且可能更难维护）。
原始代码（2个查询和一个中间临时表）更容易理解（并且可能更容易维护）。
答案 1 :(得分：0)
使用row_number()：
select a.*
from (select a.*,
row_number() over (partition by left(name, 1)
order by score desc, name asc
) as seqnum
from #a a
) a
where seqnum = 1;
编辑：
某些版本的Sybase确实支持row_number()，但不是全部。你也可以这样做：
select a.*
from #a a
where a.id = (select top 1 a2.id
from #a a2
where left(a2.name, 1) = left(a.name, 1)
order by a2.score desc, name asc
);
编辑II：
这有用吗？
select a.*
from (select a.*,
(select top 1 a2.id
from #a a2
where left(a2.name, 1) = left(a.name, 1)
order by a2.score desc, name asc
) as comp_id
from #a a
) a
where id = comp_id;
即TOP？
SELECT