解决上一户和下一户问题
这两个函数，是偏移量函数，其用途是：可以查出同一字段下一个值或上一个值。
lead(col_name,num,flag)
col_name是列名；num是取向下第几个值；flag是一个标志，也就是如果向下第几个值是空值的话就取flag；
例如lead(login_time,1,null)这个是向下取一个值，如果这个值为空则按空算，当然也可以用其他值替换。
lag(col_name,num,flag)
和lead类似，col_name是列名；num是取向上第几个值；flag是一个标志，也就是如果向上第几个值是空值的话就取flag；
例如lag(login_time,1,null)这个是向上取一个值，如果这个值为空则按空算，当然也可以用其他值替换。
举个例子：有一个表tmp_test(u_id,login_time)查一下这个表中连续7天都有登录机器的人是谁？
造下数据：
create table tmp_test(u_id number,login_time date);
insert into tmp_test
select 1 rn,sysdate + rownum as login_time
from dual
connect by level <=8
union
select 2 rn,sysdate + rownum as login_time
from dual
connect by level <=3
union
select 3 rn,sysdate + rownum as login_time
from dual
connect by level <=2
union
select 2 rn,sysdate + rownum+4 as login_time
from dual
connect by level <=5;
commit;
然后造几条重复数据：
insert into tmp_test
select 1 rn,sysdate + rownum as login_time
from dual
connect by level <=3;
查下数据：
select * from tmp_test;
U_ID LOGIN_TIME
---------- -----------
1 2012/3/8 6:33:24
1 2012/3/9 6:33:24
1 2012/3/10 6:33:24
1 2012/3/11 6:33:24
1 2012/3/12 6:33:24
1 2012/3/13 6:33:24
1 2012/3/14 6:33:24
1 2012/3/15 6:33:24
2 2012/3/8 6:33:24
2 2012/3/9 6:33:24
2 2012/3/10 6:33:24
2 2012/3/12 6:33:24
2 2012/3/13 6:33:24
2 2012/3/14 6:33:24
2 2012/3/15 6:33:24
2 2012/3/16 6:33:24
3 2012/3/8 6:33:24
3 2012/3/9 6:33:24
1 2012/3/8 6:37:35
1 2012/3/9 6:37:35
1 2012/3/10 6:37:35
从上面数据看出其实只有u_id=1满足条件，那么怎么用sql实现呢？
SQL> select distinct u_id
2 from (select u_id,
3 login_time last_login_time,
4 lead(login_time, 6) over(partition by u_id order by u_id, login_time) next_login_time
5 from (select distinct u_id, trunc(login_time) login_time
6 from tmp_test))
7 where next_login_time - last_login_time = 6;
U_ID
----------
1
ok，就是这个结果。其实用lag也可以实现相同结果，写法如下：
select distinct u_id
from (select u_id,
login_time last_login_time,
lag(login_time, 6) over(partition by u_id order by u_id, login_time) next_login_time
from (select distinct u_id, trunc(login_time) login_time
from tmp_test))
where last_login_time - next_login_time = 6;
