--NBA连冠题
create table t_nba(
team_name varchar2(100),
champion_year number
);
insert into t_nba (team_name, champion_year) values ('活塞', 1990);
insert into t_nba (team_name, champion_year) values ('公牛', 1991);
insert into t_nba (team_name, champion_year) values ('公牛', 1992);
insert into t_nba (team_name, champion_year) values ('公牛', 1993);
insert into t_nba (team_name, champion_year) values ('火箭', 1994);
insert into t_nba (team_name, champion_year) values ('火箭', 1995);
insert into t_nba (team_name, champion_year) values ('公牛', 1996);
insert into t_nba (team_name, champion_year) values ('公牛', 1997);
insert into t_nba (team_name, champion_year) values ('公牛', 1998);
insert into t_nba (team_name, champion_year) values ('马刺', 1999);
insert into t_nba (team_name, champion_year) values ('湖人', 2000);
insert into t_nba (team_name, champion_year) values ('湖人', 2001);
insert into t_nba (team_name, champion_year) values ('湖人', 2002);
insert into t_nba (team_name, champion_year) values ('马刺', 2003);
insert into t_nba (team_name, champion_year) values ('活塞', 2004);
insert into t_nba (team_name, champion_year) values ('马刺', 2005);
insert into t_nba (team_name, champion_year) values ('热火', 2006);
insert into t_nba (team_name, champion_year) values ('马刺', 2007);
insert into t_nba (team_name, champion_year) values ('凯尔特人', 2008);
insert into t_nba (team_name, champion_year) values ('湖人', 2009);
insert into t_nba (team_name, champion_year) values ('湖人', 2010);
commit;
select * from t_nba;
--请写出一条SQL语句，查询出在此期间连续获得冠军的队伍有哪些，其连续的年份的起止时间是多少？
/*
例如：
team_name begin_date end_date
公牛 1991 1993
火箭 1994 1995
公牛 1996 1998
湖人 2000 2002
湖人 2009 2010
*/
--方法一：
select max(team_name) team_name,
min(champion_year) begin_date,
max(champion_year) end_date
from (select team_name, champion_year
from (select team_name,
champion_year,
lag(champion_year, 1) over(partition by team_name order by champion_year) before, --上一个
lead(champion_year, 1) over(partition by team_name order by champion_year) after --下一个
from t_nba) a
where a.champion_year = a.after - 1
or a.champion_year = a.before + 1) b
group by b.champion_year - rownum
order by begin_date;
--方法二：
with tmp_a as(
select t.team_name,
t.champion_year,
row_number() over(partition by t.team_name order by t.champion_year) rn
from t_nba t)
select t.team_name,
min(t.champion_year) begin_date,
max(t.champion_year) end_date
from tmp_a t
group by t.team_name,t.champion_year - rn
having count(t.champion_year - rn) > 1
order by begin_date;
--连续登陆题
create table t_user_login(
user_id number,
login_date date
);
insert into t_user_login(user_id, login_date) values(1,to_date('20200808','yyyy-mm-dd') - 3);
insert into t_user_login(user_id, login_date) values(1,to_date('20200808','yyyy-mm-dd') - 2);
insert into t_user_login(user_id, login_date) values(1,to_date('20200808','yyyy-mm-dd') - 1);
insert into t_user_login(user_id, login_date) values(1,to_date('20200808','yyyy-mm-dd'));
insert into t_user_login(user_id, login_date) values(2,to_date('20200808','yyyy-mm-dd') - 2);
insert into t_user_login(user_id, login_date) values(2,to_date('20200808','yyyy-mm-dd') - 1);
insert into t_user_login(user_id, login_date) values(2,to_date('20200808','yyyy-mm-dd'));
insert into t_user_login(user_id, login_date) values(3,to_date('20200808','yyyy-mm-dd') - 5);
insert into t_user_login(user_id, login_date) values(3,to_date('20200808','yyyy-mm-dd') - 4);
insert into t_user_login(user_id, login_date) values(3,to_date('20200808','yyyy-mm-dd') - 3);
insert into t_user_login(user_id, login_date) values(3,to_date('20200808','yyyy-mm-dd'));
insert into t_user_login(user_id, login_date) values(4,to_date('20200808','yyyy-mm-dd') - 4);
insert into t_user_login(user_id, login_date) values(4,to_date('20200808','yyyy-mm-dd') - 2);
insert into t_user_login(user_id, login_date) values(4,to_date('20200808','yyyy-mm-dd') - 1);
insert into t_user_login(user_id, login_date) values(4,to_date('20200808','yyyy-mm-dd'));
insert into t_user_login(user_id, login_date) values(5,to_date('20200808','yyyy-mm-dd'));
commit;
select * from t_user_login;
--查询用户最长连续登陆的天数
/*
例如：
user_id series_day start_date end_date
1 4 20200805 20200808
2 3 20200806 20200808
3 3 20200803 20200805
4 3 20200806 20200808
*/
select user_id,
min(login_date) as start_date,
max(login_date) as end_date,
count(1) as series_day
from (select user_id,
login_date,
row_number() over(partition by user_id order by login_date) as sort
from t_user_login) a
group by a.user_id, a.login_date - a.sort
having count(a.login_date - a.sort) > 1
order by a.user_id;
--连续纳税题
create table test(
year number,
ent varchar2(10),
amt number
);
insert into test (year, ent, amt) values(2010,'AA',2000);
insert into test (year, ent, amt) values(2011,'AA',2500);
insert into test (year, ent, amt) values(2012,'AA',2300);
insert into test (year, ent, amt) values(2013,'AA',1900);
insert into test (year, ent, amt) values(2014,'AA',2200);
insert into test (year, ent, amt) values(2015,'AA',2300);
insert into test (year, ent, amt) values(2016,'AA',2006);
insert into test (year, ent, amt) values(2010,'BB',2400);
insert into test (year, ent, amt) values(2011,'BB',2401);
insert into test (year, ent, amt) values(2012,'BB',1800);
insert into test (year, ent, amt) values(2013,'BB',2100);
insert into test (year, ent, amt) values(2014,'BB',2101);
insert into test (year, ent, amt) values(2015,'BB',2102);
insert into test (year, ent, amt) values(2016,'BB',2020);
commit;
/*
查询出在此期间连续3年纳税金额大于等于2000万的企业有哪些，
其连续的年份的起止时间是多少，平均纳税金额是多少，结果如下：
企业名称 开始时间 结束时间 平均纳税
AA 2010 2012 2266.7
AA 2014 2016 2168.7
BB 2013 2016 2080.8
*/
select * from test;
with tmp_a as(
select t.year,
t.ent,
t.amt,
row_number() over(partition by t.ent order by t.year) rn
from test t
where t.amt >= 2000)
select t.ent 企业名称,
count(t.year - rn) 连续时间,
min(t.year) 开始时间,
max(t.year) 结束时间,
round(avg(t.amt),1) 平均纳税
from tmp_a t
group by t.ent,t.year - rn
having count(t.year - rn) > 2
order by 企业名称,开始时间;