如何在SQL的第一行和最后一个上选择一个值？

Question

我有这样一个表：

ID | persid | value1 | value2 
---+--------+---------+--------- 
1 | A  | string1 | string2 
2 | A  | string3 | string4 
3 | A  | string5 | string6 

我需要用GROUP BY返回是这样的：

Field1 | field2 | field3 
-------+---------+---------- 
A  | string1 | string6 

我使用的是SQL Server 2008中， 这可能吗？

Answer 1

我会用有条件聚集：

select persid, 
     max(case when seqnum_asc = 1 then value1 end) as value1, 
     max(case when seqnum_desc = 1 then value2 end) as value2 
from (select t.*, 
      row_number() over (partition by persid order by id) as seqnum_asc, 
      row_number() over (partition by persid order by id desc) as seqnum_desc 
     from t 
    ) t 
group by persid; 

Answer 2

你可以尝试这样的事情：

create table tbl (id int, persid varchar(10), value1 varchar(10), value2 varchar(10)); 

insert into tbl (id, persid, value1, value2) values (1, 'A', 'string1', 'string2'); 
insert into tbl (id, persid, value1, value2) values (2, 'A', 'string3', 'string4'); 
insert into tbl (id, persid, value1, value2) values (3, 'A', 'string5', 'string6'); 

select tbl.persid, tbl.value1, persmax.value2 
from tbl 
inner join tbl persmax on persmax.id = (select max(id) from tbl where persid='A') 
where tbl.persid='A' 
and tbl.id = (select min(id) from tbl where persid='A') 

Answer 3

我认为你可以使用这样的查询：

select top(1) yourTable.persid field1, yourTable.value1 field2, t.value2 field3 
from yourTable 
    cross join (
    select top(1) value2 
    from yourTable 
    order by ID desc 
) t  
order by ID; 

Answer 4

这个怎么样？

select distinct a, b, c from 
    (select mytbl.persid as a, mytbl.value1 as b from mytbl, 
     (select persid, min(id) as id from mytbl group by persid) as t1 
       where mytbl.id = t1.id and mytbl.persid = t1.persid) as t2, 
    (select mytbl.persid as d, mytbl.value2 as c from mytbl, 
     (select persid, max(id) as id from mytbl group by persid) as t3 
       where mytbl.id = t3.id and mytbl.persid = t3.persid) as t4 
where t2.a = t4.d