SQL语句积累1:自身查询、区间查询、同号排查

数据库表:t_persons

select * from t_persons

ID name date money
1 李天 2012-01-01 00:00:00.000 100
2 李忠 2012-02-01 00:00:00.000 200
3 李天 2012-01-01 00:00:00.000 100
4 李忠 2012-02-02 00:00:00.000 500
5 李敏 2011-01-01 00:00:00.000 600
6 李吧 2011-02-03 00:00:00.000 1000

1.

/*查询相同名字的最大时间的记录*/
select * from t_persons a where a.date=(select max(date) from t_persons b where b.name=a.name)

2.

/*区间数据： 前第4-5条记录*/
select top 2 * from( select top 5 * from t_persons order by ID asc ) a order by ID desc
select top 2 * from t_persons where ID not in (select top 3 ID from t_persons)

扩展：

以查询前20到30条为例,主键名为id

方法一: 先正查,再反查
select top 10 * from (select top 30 * from tablename order by id asc) A order by id desc

方法二: 使用left join
select top 10 A.* from tablename A
left outer join (select top 20 * from tablename order by id asc) B
on A.id = B.id
where B.id is null
order by A.id asc

方法三: 使用not exists
select top 10 * from tablename A
where id not exists
(select top 20 * from tablename B on A.id = B.id)

方法四: 使用not in
select top 10 * from tablename
where id not in
(select top 20 id from tablename order by id asc)
order by id asc

方法五: 使用rank()
select id from
(select rank() over(order by id asc) rk, id from tablename) T
where rk between 20 and 30

中第五种方法看上去好像没有问题,查了下文档,当over()用于rank/row_number时,整型列不能描述一个列,所以会产生非预期的效果. 待考虑下,有什么办法可以修改为想要的结果.

3.删除重复的数据，只留一条

select * from t_persons where ID=
(select max(ID) from t_persons where
name =(select name from t_persons group by name,date,money having count(*)>1)
and date=(select date from t_persons group by name,date,money having count(*)>1)
and money=(select money from t_persons group by name,date,money having count(*)>1)
)