这个问题太不可抗拒了,所以我只是猜测了我们正在处理的数据结构。问题中没有具体说明该技术。这是在 Transact-SQL 中。
create table student
(
id int not null primary key identity,
name nvarchar(100) not null default '',
graduation_date date not null default getdate(),
)
go
create table school
(
id int not null primary key identity,
name nvarchar(100) not null default ''
)
go
create table student_school_asc
(
student_id int not null foreign key references student (id),
school_id int not null foreign key references school (id),
primary key (student_id, school_id)
)
go
insert into student (name, graduation_date) values ('john', '2001-11-11')
insert into student (name, graduation_date) values ('paul', '2011-11-11')
insert into school (name) values ('nyu')
insert into school (name) values ('ucla')
insert into school (name) values ('uft')
insert into school (name) values ('mit')
insert into student_school_asc (student_id, school_id) values (1,1)
insert into student_school_asc (student_id, school_id) values (1,2)
insert into student_school_asc (student_id, school_id) values (2,3)
insert into student_school_asc (student_id, school_id) values (2,4)
select
s.id,
s.name,
s.graduation_date as [date],
(select max(name) from
(select name,
RANK() over (order by name) as rank_num
from school sc
inner join student_school_asc ssa on ssa.school_id = sc.id
where ssa.student_id = s.id) s1 where s1.rank_num = 1) as school,
(select max(name) from
(select name,
RANK() over (order by name) as rank_num
from school sc
inner join student_school_asc ssa on ssa.school_id = sc.id
where ssa.student_id = s.id) s2 where s2.rank_num = 2) as school1
from
student s
结果:
id name date school school1
--- ----- ---------- ------- --------
1 john 2001-11-11 nyu ucla
2 paul 2011-11-11 mit uft