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)
)