SQL强化

SQL执行顺序

--举例：
select a.sex,b.city,count(1) as cnt,sum(salary) as sum1
from table1 a
join table2 b on a.id=b.id
where a.name=b.name
group by a.sex,b.city
having cnt>=2
order by a.sex,b.city
limit 10--或者是
select distincta.sex,b.city,a.age
from table1 a
join table2 b on a.id=b.id
where a.name=b.name
order by a.sex,b.city
limit 10

上面的SQL语句的执行顺序是: from (去加载table1 和 table2这2个表 ) -> join -> on -> where -> group by->select 后面的聚合函数count,sum -> having -> distinct -> order by -> limit

--on 和where的先后顺序讨论
--下面用left join 各得到结果，结果不一样。
--下面可知，先执行on，再执行where
select * 
from table1 a
left join table2 b 
on a.id=b.id
where a.name=b.name;
--下面的条数可能会比上面多。
select * 
from table1 a
left join table2 b 
on a.id=b.id
and a.name=b.name;--下面用inner join 各得到结果，结果是一样的
select * 
from table1 a
join table2 b 
on a.id=b.id
where a.name=b.name;select * 
from table1 a
join table2 b 
on a.id=b.id
and a.name=b.name;

hivesql与sparkSQL的区别：

• 子查询hive必须起别名，SparkSQL可以不用起别名

• group by xx，yy，hive不用能用别名，spark可以用别名

• hive不支持临时视图和缓存表，SparkSQL都支持

  • --用SparkSQL的临时视图use interview_db;create or replace temporary view t_view1 asselect *,if(month=1,amount,0) as a1,if(month=2,amount,0) as a2,if(month=3,amount,0) as a3,if(month=4,amount,0) as a4from table2;
    

select year,
sum(a1) as m1,
sum(a2) as m2,
sum(a3) as m3,
sum(a4) as m4
from t_view1
group by year;

–使用SparkSQL的缓存表
cache table cached1 as
select *,
if(month=1,amount,0) as a1,
if(month=2,amount,0) as a2,
if(month=3,amount,0) as a3,
if(month=4,amount,0) as a4
from table2;

select * from cached1;
select year,
sum(a1) as m1,
sum(a2) as m2,
sum(a3) as m3,
sum(a4) as m4
from cached1
group by year;

* 爆炸函数，hive不支持explode与普通字段联合使用，需要用侧视图分开，SparkSQL支持联合使用* ```sql
use interview_db;
select qq,game1 from tableB lateral view explode(split(game,'_')) view1 as game1 ;
--spark还支持这样，但是hive不支持：
select qq,explode(split(game,'_')) game1 from tableB ;

• sparkSQL支持300多种函数，hiveSQL支持200多种函数。sparkSQL函数比hiveSQL要多。

  • 比如SparkSQL有sequence函数，hive就没有

• 先配置环境

• 在pycharm或datagrip或idea中配置hive数据源。也可以配置一个sparkSQL数据源，来加快速度。

• 如果配置hive数据源：

  • 需要提前启动hdfs和yarn，hive的metastore，hive的hiveserver2

  • #启动hdfs和yarn
    start-all.sh  # hive的metastore
    nohup /export/server/hive/bin/hive --service metastore  2>&1 > /tmp/hive-metastore.log &#hive的hiveserver2
    #hiveserver2开启后，等过2分钟后才能生效。
    nohup /export/server/hive/bin/hive --service hiveserver2 2>&1 > /tmp/hive-hiveserver2.log &
    

• 如果遇到下面的问题

• 解决办法

  • hive/conf/hive-env.sh中加入
    export HADOOP_CLIENT_OPTS=" -Xmx512m"
    export HADOOP_HEAPSIZE=1024
    改完重启hiveserver2

• 如果配置SparkSQL数据源

  • 需要提前启动hdfs，hive的metastore，Spark的Thriftserver服务。

  • #启动hdfs和yarn
    start-all.sh  # hive的metastore
    nohup /export/server/hive/bin/hive --service metastore  2>&1 > /tmp/hive-metastore.log &#Spark的Thriftserver服务
    /export/server/spark/sbin/start-thriftserver.sh \--hiveconf hive.server2.thrift.port=10001 \--hiveconf hive.server2.thrift.bind.host=node1 \--master local[*]
    

  • 下面是spark3集成hive3需要的jar包，如果是spark2集成hive2，则jar包不一样。

  • 

show databases ;
create database if not exists test_sql;
use test_sql;
-- 一些语句会走 MapReduce，所以慢。 可以开启本地化执行的优化。
set hive.exec.mode.local.auto=true;-- (默认为false)
--第1题：访问量统计
CREATE TABLE test_sql.test1 (userId string,visitDate string,visitCount INT )ROW format delimited FIELDS TERMINATED BY "\t";INSERT overwrite TABLE test_sql.test1
VALUES( 'u01', '2017/1/21', 5 ),( 'u02', '2017/1/23', 6 ),( 'u03', '2017/1/22', 8 ),( 'u04', '2017/1/20', 3 ),( 'u01', '2017/1/23', 6 ),( 'u01', '2017/2/21', 8 ),( 'u02', '2017/1/23', 6 ),( 'u01', '2017/2/22', 4 );select *,sum(sum1) over(partition by userid order by month1 /*rows between unbounded preceding and current row*/ ) as `累积`from
(select userid,date_format(replace(visitdate,'/','-'),'yyyy-MM') as month1,sum(visitcount) sum1
from test_sql.test1
group by userid,date_format(replace(visitdate,'/','-'),'yyyy-MM')) as t;-- 第2题：电商场景TopK统计
CREATE TABLE test_sql.test2 (user_id string,shop string )ROW format delimited FIELDS TERMINATED BY '\t';
INSERT INTO TABLE test_sql.test2 VALUES
( 'u1', 'a' ),
( 'u2', 'b' ),
( 'u1', 'b' ),
( 'u1', 'a' ),
( 'u3', 'c' ),
( 'u4', 'b' ),
( 'u1', 'a' ),
( 'u2', 'c' ),
( 'u5', 'b' ),
( 'u4', 'b' ),
( 'u6', 'c' ),
( 'u2', 'c' ),
( 'u1', 'b' ),
( 'u2', 'a' ),
( 'u2', 'a' ),
( 'u3', 'a' ),
( 'u5', 'a' ),
( 'u5', 'a' ),
( 'u5', 'a' );
--(1)每个店铺的UV(访客数)
-- UV和PV
-- PV是访问当前网站所有的次数
-- UV是访问当前网站的客户数(需要去重)
--(2)每个店铺访问次数top3的访客信息。输出店铺名称、访客id、访问次数
select shop,count(distinct user_id) as uv
from test_sql.test2 group by shop ;
--上面的拆解来看，等价于
--distinct后可以接多个字段，表示联合去重
select shop,count(user_id) as uv
from
(select distinct shop,user_id
from test_sql.test2  ) as t
group by shop ;
--也等价于
select shop,count(user_id) as uv
from
(select shop,user_id
from test_sql.test2 group by shop, user_id) as t
group by shop ;select * from
(select *,row_number() over (partition by shop order by cnt desc) as rn
from
(select shop,user_id,count(1) as cnt from test_sql.test2 group by  shop,user_id ) as t) t2
where t2.rn<=3;-- 第3题：订单量统计
CREATE TABLE test_sql.test3 (dt string,order_id string,user_id string,amount DECIMAL ( 10, 2 ) )
ROW format delimited FIELDS TERMINATED BY '\t';INSERT overwrite TABLE test_sql.test3 VALUES('2017-01-01','10029028','1000003251',33.57),('2017-01-01','10029029','1000003251',33.57),('2017-01-01','100290288','1000003252',33.57),('2017-02-02','10029088','1000003251',33.57),('2017-02-02','100290281','1000003251',33.57),('2017-02-02','100290282','1000003253',33.57),('2017-11-02','10290282','100003253',234),('2018-11-02','10290284','100003243',234);-- 	(1)给出 2017年每个月的订单数、用户数、总成交金额。
-- 	(2)给出2017年11月的新客数(指在11月才有第一笔订单)
select date_format(dt,'yyyy-MM') as month1,count(distinct order_id) as cnt1,count(distinct user_id) as cnt2,sum(amount) as amtfrom test_sql.test3where year(dt)=2017
group by date_format(dt,'yyyy-MM');select count(user_id) cnt from
(select user_id,min(date_format(dt,'yyyy-MM')) min_month
from test3 group by user_id) as t where min_month='2017-11';--统计每个月的新客户数
select min_month,count(user_id) cnt
from (select user_id,min(date_format(dt, 'yyyy-MM')) min_monthfrom test3group by user_id) as t
group by min_month;-- 第4题：大数据排序统计
CREATE TABLE test_sql.test4user(user_id string,name string,age int);CREATE TABLE test_sql.test4log(user_id string,url string);INSERT INTO TABLE test_sql.test4user VALUES('001','u1',10),
('002','u2',15),
('003','u3',15),
('004','u4',20),
('005','u5',25),
('006','u6',35),
('007','u7',40),
('008','u8',45),
('009','u9',50),
('0010','u10',65);
INSERT INTO TABLE test_sql.test4log VALUES('001','url1'),
('002','url1'),
('003','url2'),
('004','url3'),
('005','url3'),
('006','url1'),
('007','url5'),
('008','url7'),
('009','url5'),
('0010','url1');select * from test_sql.test4user ;
select * from test_sql.test4log ;--有一个5000万的用户文件(user_id，name，age)，
-- 一个2亿记录的用户看电影的记录文件(user_id，url)，根据年龄段观看电影的次数进行排序？
--取整函数有 round，floor，ceil
select *,round(x,0) as r,--四舍五入floor(x) as f,--向下取整ceil(x) as c--向上取整from
(select 15/10 as x union all
select 18/10 as x union all
select 24/10 as x union all
select 27/10 as x ) as t;select type,sum(cnt) as sum1
from
(select *,concat(floor(age/10)*10,'-',floor(age/10)*10+10) as type
from test_sql.test4user as a
-- join前最好提前减小数据量
join (select user_id,count(url) as cnt from test_sql.test4log group by user_id) as b
on a.user_id=b.user_id) as t
group by type
order by sum(cnt) desc;-- 第5题：活跃用户统计
CREATE TABLE test5(
dt string,
user_id string,
age int)
ROW format delimited fields terminated BY ',';
INSERT overwrite TABLE test_sql.test5 VALUES ('2019-02-11','test_1',23),
('2019-02-11','test_2',19),
('2019-02-11','test_3',39),
('2019-02-11','test_1',23),
('2019-02-11','test_3',39),
('2019-02-11','test_1',23),
('2019-02-12','test_2',19),
('2019-02-13','test_1',23),
('2019-02-15','test_2',19),
('2019-02-16','test_2',19);
select * from test_sql.test5 order by dt,user_id;
--有日志如下，请写出代码求得所有用户和活跃用户的总数及平均年龄。(活跃用户指连续两天都有访问记录的用户)
-- type     总数   平均年龄
-- '所有用户'  3    27
-- '活跃用户'  1    19
with t1 as (select distinct dt, user_id,age from test_sql.test5),t2 as (select *,row_number() over (partition by user_id order by dt) as rn from t1 ),t3 as (select *,date_sub(dt,rn) as dt2 from t2),t4 as (select dt2,user_id,age,count(1) cnt from t3 group by dt2,user_id,age),t5 as (select * from t4 where cnt>=2),t6 as (select distinct user_id,age from t5)
select '所有用户' as type, count(user_id) cnt,avg(age) as avg_age
from (select distinct user_id,age from test_sql.test5) t union all
select '活跃用户' as type, count(user_id) cnt,avg(age) as avg_age from t6;-- 用思路2来分析连续2天登录
with t1 as (select distinct dt, user_id from test_sql.test5),t2 as (select *,date_add(dt,1) as dt2,lead(dt,1)over(partition by user_id order by dt) as dt3from t1)
select count(distinct user_id) from t2 where dt2=dt3;-- 第6题：电商购买金额统计实战
CREATE TABLE test_sql.test6 (userid string,money decimal(10,2),paymenttime string,orderid string);INSERT INTO TABLE test_sql.test6 VALUES('001',100,'2017-10-01','123'),
('001',200,'2017-10-02','124'),
('002',500,'2017-10-01','125'),
('001',100,'2017-11-01','126');select * from test_sql.test6 order by userid,paymenttime;
--请用sql写出所有用户中在今年10月份第一次购买商品的金额，
select userid,paymenttime,money
from
(select *,row_number() over (partition by userid order by paymenttime) as rnfrom test_sql.test6 where date_format(paymenttime,'yyyy-MM')='2017-10' ) as t
where t.rn=1
;-- 第7题：教育领域SQL实战
CREATE TABLE test_sql.book(book_id string,`SORT` string,book_name string,writer string,OUTPUT string,price decimal(10,2));
INSERT INTO TABLE test_sql.book VALUES
('001','TP391','信息处理','author1','机械工业出版社','20'),
('002','TP392','数据库','author12','科学出版社','15'),
('003','TP393','计算机网络','author3','机械工业出版社','29'),
('004','TP399','微机原理','author4','科学出版社','39'),
('005','C931','管理信息系统','author5','机械工业出版社','40'),
('006','C932','运筹学','author6','科学出版社','55');CREATE TABLE test_sql.reader (reader_id string,company string,name string,sex string,grade string,addr string);
INSERT INTO TABLE test_sql.reader VALUES
('0001','阿里巴巴','jack','男','vp','addr1'),
('0002','百度','robin','男','vp','addr2'),
('0003','腾讯','tony','男','vp','addr3'),
('0004','京东','jasper','男','cfo','addr4'),
('0005','网易','zhangsan','女','ceo','addr5'),
('0006','搜狐','lisi','女','ceo','addr6');CREATE TABLE test_sql.borrow_log(reader_id string,book_id string,borrow_date string);INSERT INTO TABLE test_sql.borrow_log VALUES ('0001','002','2019-10-14'),
('0002','001','2019-10-13'),
('0003','005','2019-09-14'),
('0004','006','2019-08-15'),
('0005','003','2019-10-10'),
('0006','004','2019-17-13');select * from test_sql.book;
select * from test_sql.reader;
select * from test_sql.borrow_log;--(8)考虑到数据安全的需要，需定时将“借阅记录”中数据进行备份，请使用一条SQL语句，
-- 在备份用户bak下创建与“借阅记录”表结构完全一致的数据表BORROW_LOG_BAK.
-- 井且将“借阅记录”中现有数据全部复制到BORROW_L0G_ BAK中。
create table test_sql.BORROW_LOG_BAK as select * from test_sql.borrow_log;
select * from test_sql.BORROW_LOG_BAK;--(9)现在需要将原Oracle数据库中数据迁移至Hive仓库，
-- 请写出“图书”在Hive中的建表语句(Hive实现，提示：列分隔符|；
-- 数据表数据需要外部导入：分区分别以month＿part、day＿part 命名)
CREATE TABLE test_sql.book2
(book_id   string,`SORT`    string,book_name string,writer    string,OUTPUT    string,price     decimal(10, 2)
)partitioned by (month_part string,day_part string )row format delimited fields terminated by '|';--(10)Hive中有表A，现在需要将表A的月分区　201505　中　
-- user＿id为20000的user＿dinner字段更新为bonc8920，其他用户user＿dinner字段数据不变，
-- 请列出更新的方法步骤。(Hive实现，提示：Hive中无update语法，请通过其他办法进行数据更新)
--A
-- user_id   user_dinner  part
-- 20000        aaaaa     201505
-- 30000        bbbbb     201505create table A (user_id int,user_dinner string) partitioned by (part string);
insert overwrite table A partition (part = '201505')
values (20000, 'aaaaa'),(30000, 'bbbbb'),(40000, 'ccccc');
select * from A;
--update A set user_dinner='bonc8920' where user_id=20000;insert overwrite table A partition  (part = '201505')
select user_id, 'bonc8920' as user_dinner from A where user_id=20000 and part = '201505' union all
select user_id, user_dinner from A where user_id!=20000 and part = '201505'  ;-- 第8题：服务日志SQL统计
CREATE TABLE test_sql.test8(`date` string,interface string,ip string);INSERT INTO TABLE test_sql.test8 VALUES
('2016-11-09 11:22:05','/api/user/login','110.23.5.23'),
('2016-11-09 11:23:10','/api/user/detail','57.3.2.16'),
('2016-11-09 23:59:40','/api/user/login','200.6.5.166'),
('2016-11-09 11:14:23','/api/user/login','136.79.47.70'),
('2016-11-09 11:15:23','/api/user/detail','94.144.143.141'),
('2016-11-09 11:16:23','/api/user/login','197.161.8.206'),
('2016-11-09 12:14:23','/api/user/detail','240.227.107.145'),
('2016-11-09 13:14:23','/api/user/login','79.130.122.205'),
('2016-11-09 14:14:23','/api/user/detail','65.228.251.189'),
('2016-11-09 14:15:23','/api/user/detail','245.23.122.44'),
('2016-11-09 14:17:23','/api/user/detail','22.74.142.137'),
('2016-11-09 14:19:23','/api/user/detail','54.93.212.87'),
('2016-11-09 14:20:23','/api/user/detail','218.15.167.248'),
('2016-11-09 14:24:23','/api/user/detail','20.117.19.75'),
('2016-11-09 15:14:23','/api/user/login','183.162.66.97'),
('2016-11-09 16:14:23','/api/user/login','108.181.245.147'),
('2016-11-09 14:17:23','/api/user/login','22.74.142.137'),
('2016-11-09 14:19:23','/api/user/login','22.74.142.137');select * from test_sql.test8;
--求11月9号下午14点(14-15点)，访问/api/user/login接口的top10的ip地址
select ip, count(1) cnt
from test_sql.test8
where date_format(`date`, 'yyyy-MM-dd HH') = '2016-11-09 14'and interface = '/api/user/login'
group by ip
order by cnt desc
limit 10
;-- 第9题：充值日志SQL实战
CREATE TABLE test_sql.test9(dist_id string COMMENT '区组id',account string COMMENT '账号',`money` decimal(10,2) COMMENT '充值金额',create_time string COMMENT '订单时间');INSERT INTO TABLE test_sql.test9 VALUES ('1','11',100006,'2019-01-02 13:00:01'),('1','22',110000,'2019-01-02 13:00:02'),('1','33',102000,'2019-01-02 13:00:03'),('1','44',100300,'2019-01-02 13:00:04'),('1','55',100040,'2019-01-02 13:00:05'),('1','66',100005,'2019-01-02 13:00:06'),('1','77',180000,'2019-01-03 13:00:07'),('1','88',106000,'2019-01-02 13:00:08'),('1','99',100400,'2019-01-02 13:00:09'),('1','12',100030,'2019-01-02 13:00:10'),('1','13',100003,'2019-01-02 13:00:20'),('1','14',100020,'2019-01-02 13:00:30'),('1','15',100500,'2019-01-02 13:00:40'),('1','16',106000,'2019-01-02 13:00:50'),('1','17',100800,'2019-01-02 13:00:59'),('2','18',100800,'2019-01-02 13:00:11'),('2','19',100030,'2019-01-02 13:00:12'),('2','10',100000,'2019-01-02 13:00:13'),('2','45',100010,'2019-01-02 13:00:14'),('2','78',100070,'2019-01-02 13:00:15');select * from test_sql.test9 order by dist_id , money desc;
--请写出SQL语句，查询充值日志表2019年01月02号每个区组下充值额最大的账号，要求结果：
--区组id，账号，金额，充值时间
select * from
(select *,row_number() over (partition by dist_id order by money desc) rn
from  test_sql.test9 where to_date(create_time)='2019-01-02') t
where t.rn=1;-- 第10题：电商分组TopK实战
CREATE TABLE test_sql.test10(`dist_id` string COMMENT '区组id',`account` string COMMENT '账号',`gold` int COMMENT '金币');INSERT INTO TABLE test_sql.test10 VALUES ('1','77',18),('1','88',106),('1','99',10),('1','12',13),('1','13',14),('1','14',25),('1','15',36),('1','16',12),('1','17',158),('2','18',12),('2','19',44),('2','10',66),('2','45',80),('2','78',98);select * from test_sql.test10;select * from
(select *,row_number() over (partition by dist_id order by gold desc) rn
from  test_sql.test10  ) t
where t.rn<=10;

行转列(转置)

• 行转列的常规做法是，group by+sum(if())【或count(if())】

华泰证券1

已知

year	month	amount
1991	1	1.1
1991	2	1.2
1991	3	1.3
1991	4	1.4
1992	1	2.1
1992	2	2.2
1992	3	2.3
1992	4	2.4

查成这样一个结果

year	m1	m2	m3	m4
1991	1.1	1.2	1.3	1.4
1992	2.1	2.2	2.3	2.4

解答

• use test_sql;
  set hive.exec.mode.local.auto=true;
  create table table2(year int,month int ,amount double) ;insert overwrite table table2 values(1991,1,1.1),(1991,2,1.2),(1991,3,1.3),(1991,4,1.4),(1992,1,2.1),(1992,2,2.2),(1992,3,2.3),(1992,4,2.4);
  select * from table2;--行转列
  --常规做法是，group by+sum(if())
  --SQLserver中有pivot专门用来行转列
  --原始写法
  select year,sum(a) as m1,sum(b) as m2,sum(c) as m3,sum(d) as m4
  from(select *,if(month=1,amount,0) a,if(month=2,amount,0) b,if(month=3,amount,0) c,if(month=4,amount,0) dfrom table2) t
  group by t.year
  ;
  --简化写法
  select year,sum(if(month=1,amount,0)) m1,sum(if(month=2,amount,0)) m2,sum(if(month=3,amount,0)) m3,sum(if(month=4,amount,0)) m4
  from table2
  group by year;

华泰证券2

• 查询课程编号“2”的成绩比课程编号“1”低的所有同学的学号、姓名。

• 【这是行转列的衍生题】

• create table student(sid int, sname string, gender string, class_id int);
  insert overwrite table student
  values (1, '张三', '女', 1),(2, '李四', '女', 1),(3, '王五', '男', 2);select * from student;create table  course (cid int, cname string, teacher_id int);
  insert overwrite table course
  values (1, '生物', 1),(2, '体育', 1),(3, '物理', 2);
  select * from course;create table score (sid int, student_id int, course_id int, number int);
  insert overwrite table score
  values (1, 1, 1, 58),(4, 1, 2, 50),(2, 1, 2, 68),(3, 2, 2, 89);
  select * from score;with t1 as(select student_id,sum(if(course_id=2,number,0)) as pe, --体育sum(if(course_id=1,number,0)) as bio --生物
  from score
  group by student_id
  having pe<bio)
  select sid, sname
  from t1
  join student
  on t1.student_id = sid
  ;

腾讯游戏

表table如下：

DDate	shengfu
2015-05-09	胜
2015-05-09	胜
2015-05-09	负
2015-05-09	负
2015-05-10	胜
2015-05-10	负
2015-05-10	负

如果要生成下列结果, 该如何写sql语句?

DDate	胜	负
2015-05-09	2	2
2015-05-10	1	2

--建表
create table table1(DDate string, shengfu string) ;
insert overwrite table table1 values ('2015-05-09', "胜"),('2015-05-09', "胜"),('2015-05-09', "负"),('2015-05-09', "负"),('2015-05-10', "胜"),('2015-05-10', "负"),('2015-05-10', "负");select DDate,SUM(case when shengfu = '胜' then 1 else 0 end) `胜`,SUM(case when shengfu = '负' then 1 else 0 end) `负`
from table1
group by DDate;

腾讯QQ

假设tableA如表5, tableB如表6,

表5

qq号(字段名：qq)	游戏(字段名：game)
10000	a
10000	b
10000	c
20000	c
20000	d

表6

qq号(字段名：qq)	游戏(字段名：game)
10000	a_b_c
20000	c_d

请写出以下sql逻辑：

a, 将tableA输出为tableB的格式； 【行转列】

b, 将tableB输出为tableA的格式; 【列转行】

create table tableA(qq string, game string) 
insert overwrite table tableA values (10000, 'a'),(10000, 'b'),(10000, 'c'),(20000, 'c'),(20000, 'd');create table tableB(qq string, game string) ;
insert overwrite table tableB values 
(10000, 'a_b_c'),
(20000, 'c_d');--将tableA输出为tableB的格式；  
select qq,concat_ws('_', collect_list(game)) game
from tableA
group by qq;   --将tableB输出为tableA的格式;    
select qq,tmp.game
from tableB lateral view explode(split(game, '_')) tmp as game;

连续N天登陆

• 思路分析过程

  • --核心代码
    ->distinct
    -> row_number 
    -> date_sub(dt,rn) as dt2
    -> group by dt2,name 
    -> where count(1)>=N天 
    -> distinct name 
    -> count(name)

  • 思路2

  • [外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-NKpl2VCc-1666088291563)(images/1659759179547.png)]

  • --核心代码
    ->distinct
    ->date_add(dt,N-1) as date2
    ->lead(dt,N-1) over(partition by userid order by dt) as date3
    ->where date2=date3
    ->distinct

OPPO

3、以下为用户登陆游戏的日期，用一条sQL语句查询出连续三天登录的人员姓名

name	date
张三	2021-01-01
张三	2021-01-02
张三	2021-01-03
张三	2021-01-02
李四	2021-01-01
李四	2021-01-02
王五	2021-01-03
王五	2021-01-02
王五	2021-01-02

create table game(name string,  `date` string);
insert overwrite table game values
('张三','2021-01-01'),
('张三','2021-01-02'),
('张三','2021-01-03'),
('张三','2021-01-02'),('张三','2021-01-07'),
('张三','2021-01-08'),
('张三','2021-01-09'),('李四','2021-01-01'),
('李四','2021-01-02'),
('王五','2021-01-03'),
('王五','2021-01-02'),
('王五','2021-01-02');
with t1 as ( select distinct  name,date from game),t2 as ( select *,row_number() over (partition by name order by date) rnfrom t1),t3 as ( select *,date_sub(date,rn) date2 from t2 )select distinct name from t3 group by name,date2 having count(1)>=3;--方案二
select * from game;
with t1 as (select distinct name,`date` from game
),t2 as (select *,date_add(`date`,3-1) as date2,lead(`date`,3-1) over(partition by name order by `date`) as date3from t1)
select distinct name from t2 where date2=date3;
--方案二的写法2
with t1 as (select distinct name,`date` from game
),t2 as (select *,lead(`date`,3-1) over(partition by name order by `date`) as date3from t1)
select distinct name from t2 where datediff(date3,`date`)=2 ;

脉脉

用户每日登陆脉脉会访问app不同的模块，现有两个表

表1记录了每日脉脉活跃用户的uid和不同模块的活跃时长

表2记录了脉脉所有注册用户的一些属性

表1：maimai.dau

d	uid	module	active_duration	列说明
2020-01-01	1	jobs	324	d：活跃的日期uid：用户的唯一编码module：用户活跃模块actre.duration：该模块下对应的活跃时长(单位：s)
2020-01-01	2	feeds	445
2020-01-01	3	im	345
2020-01-02	2	network	765
2020-01-02	3	jobs	342
…	…	…	…

在过去一个月内,曾连续两天活跃的用户

-- 建表
-- 表1 dau   记录了每日脉脉活跃用户的uid和不同模块的活跃时长
create table dau(d string, uid int, module string, active_duration int);
insert overwrite table dau
values ('2020-01-01', 1, 'jobs', 324),('2020-01-01', 2, 'feeds', 445),('2020-01-01', 3, 'im', 345),('2020-01-02', 2, 'network', 765),('2020-01-02', 3, 'jobs', 342);
select *from dau;with t1 as (select DISTINCT d, uid from dau),t2 as (select *,date_sub(d, (row_number() over (partition by uid order by d))) disfrom t1where d <= `current_date`()and d >= date_sub((`current_date`()), 30)),
t3 as (select uid,min(d)   `开始日期`,max(d)   `结束日期`,count(1) `连续登入天数`from t2group by uid,dishaving count(*) >= 2 
)
select DISTINCT uid from t3 ;

广州银行

有一张表C_T(列举了部分数据)表示持卡人消费记录，表结构如下：

CARD NER	VARCHAR2	卡号，
C_MONTH	NUMBER	消费月份，
C_DATE	DATE	消费日期，
C_TYPEVAR	CHAR2	消费类型
C_ATM	NUMBER	消费金额

每个月每张卡连续消费的最大天数(如卡在当月只有一次消费则为1)。

连续消费天数：指一楼时间内连续每天都有消费，同一天有多笔消费算一天消费，不能跨月份统计。

create table c_t
(card_nbr string,c_month  string,c_date   string,c_type   string,c_atm    decimal
);
insert overwrite table c_t values(1,'2022-01','2022-01-01','网购',100),(1,'2022-01','2022-01-02','网购',200),(1,'2022-01','2022-01-03','网购',300),(1,'2022-01','2022-01-15','网购',100),(1,'2022-01','2022-01-16','网购',200),(2,'2022-01','2022-01-06','网购',500),(2,'2022-01','2022-01-07','网购',800),(1,'2022-02','2022-02-01','网购',100),(1,'2022-02','2022-02-02','网购',200),(1,'2022-02','2022-02-03','网购',300),(2,'2022-02','2022-02-06','网购',500),(2,'2022-02','2022-02-07','网购',800);
with t1 as (select distinct card_nbr,c_month,c_date from c_t),t2 as (select *,row_number() over (partition by card_nbr,c_month order by c_date) rn from t1  ),t3 as (select *,date_sub(c_date,rn) dt2 from t2  ),t4 as (select  dt2,card_nbr,c_month,count(1) as cnt from t3 group by dt2,card_nbr,c_month),t5 as ( select *,row_number() over (partition by card_nbr,c_month order by cnt desc) as rn from t4)
select card_nbr,c_month,cnt from t5 where rn=1

N日留存率

• 核心代码

• -> where 日期 in (首日,1天后,7天后)
  -> group by 用户
  ->count(if(日期=首日,1,null))  as cntcount(if(日期=1天后,1,null)) as cnt2count(if(日期=7天后,1,null)) as cnt8
  ->having cnt>0
  ->count(user_id) as 首日总数count(if(cnt2>0,1,null)) as 次日留存数count(if(cnt8>0,1,null)) as 7日留存数
  ->次日留存数/首日总数 as 次日留存率7日留存数/首日总数 as 7日留存率
  

• 先按用户分组，得到每个用户的各相关日期的登录情况。

  • select cuid,count(if(event_day='2020-04-01',1,null)) as cnt,count(if(event_day='2020-04-02',1,null)) as cnt2,count(if(event_day='2020-04-08',1,null)) as cnt8from tb_cuid_1d--提前过滤数据where event_day in ('2020-04-01','2020-04-02','2020-04-08')
    group by cuid
    -- 2020-04-01必须登录，剔除掉2020-04-01没登录的
    having cnt>0
    

    效果如下

    

• 再对上面的用户汇总

  • select count(cnt) as uv,count(if(cnt2!=0,1,null)) as uv2,count(if(cnt8!=0,1,null)) as uv8
    

• 最后再用 【后续日期的留存数】除以【首日总数】，就是【留存率】

• 方案二，性能慢，但是步骤比较简单

  select count(a.cuid) uv,count(b.cuid) uv2,count(c.cuid) uv7
  from (select distinct event_day, cuid from tb_cuid_1d where event_day='首日') as a
  left join (select distinct event_day, cuid from tb_cuid_1d where event_day='次日') as b on a.cuid=b.cuid
  left join (select distinct event_day, cuid from tb_cuid_1d where event_day='7日后') as c on a.cuid=c.cuid;
  

腾讯视频号游戏直播

表：tableA

ds(日期)	device	user_id	is_active
2020-03-01	ios	0001	0
2020-03-01	ios	0002	1
2020-03-01	android	0003	1
2020-03-02	ios	0001	0
2020-03-02	ios	0002	0
2020-03-02	android	0003	1

20200301的ios设备用户活跃的次日留存率是多少？

use test_sql;
set hive.exec.mode.local.auto=true;
--腾讯视频号游戏直播
drop table if exists tableA;
create table tableA
(ds string comment '(日期)'  ,device string,user_id string,is_active int) ;
insert overwrite table  tableA values
('2020-03-01','ios','0001',0),
('2020-03-01','ios','0002',1),
('2020-03-01','ios','0004',1),
('2020-03-01','android','0003',1),
('2020-03-02','ios','0001',0),
('2020-03-02','ios','0002',0),
('2020-03-02','android','0003',1),
('2020-03-02','ios','0005',1) ,
('2020-03-02','ios','0004',1) ;--方案1，过程见下面的顺序编号
with t1 as (select user_id,--3-一个用户如果在'2020-03-01'活跃，则cnt1>0count(if(ds = '2020-03-01', 1, null)) cnt1,--4-一个用户如果在'2020-03-02'活跃，则cnt2>0count(if(ds = '2020-03-02' and is_active = 1, 1, null)) cnt2from tableA--1-预先全局过滤where device = 'ios'and ( (ds='2020-03-01' and is_active = 1) or ds='2020-03-02')--2-按用户分组group by user_id--6-只筛选'2020-03-01'活跃的用户，他在'2020-03-02'是否活跃，看cnt2=0则不活跃，>0则活跃having cnt1 > 0
)
select count(cnt1)                               sum1,--'2020-03-01'的活跃数count(if(cnt2 > 0, user_id, null))        sum2,----并且在次日依然活跃的用户数count(if(cnt2 > 0, user_id, null)) / count(cnt1) rate--次日留存率
from t1;

百度

有1张表

create table if not exists tb_cuid_1d
(cuid         string comment '用户的唯一标识',os           string comment '平台',soft_version string comment '版本',event_day    string comment '日期',visit_time    int comment '用户访问时间戳',duration     decimal comment '用户访问时长',ext          array<string> comment '扩展字段'
);
insert overwrite table tb_cuid_1d values(1,'android',1,'2020-04-01',1234567,100,`array`('')),(1,'android',1,'2020-04-02',1234567,100,`array`('')),(1,'android',1,'2020-04-08',1234567,100,`array`('')),(2,'android',1,'2020-04-01',1234567,100,`array`('')),(3,'android',1,'2020-04-02',1234567,100,`array`(''));

写出用户表 tb_cuid_1d的 20200401 的次日、次7日留存的具体HQL ：

一条sql统计出以下指标 (4.1号uv，4.1号在4.2号的留存uv，4.1号在4.8号的留存uv);

--一个理解简单，但是性能不快的做法
select count(a.cuid) uv,count(b.cuid) uv2,count(c.cuid) uv7
from (select distinct event_day, cuid from tb_cuid_1d where event_day='2020-04-01') as a
left join (select distinct event_day, cuid from tb_cuid_1d where event_day='2020-04-02') as b on a.cuid=b.cuid
left join (select distinct event_day, cuid from tb_cuid_1d where event_day='2020-04-08') as c on a.cuid=c.cuid;
--另一个理解稍微复杂，但是性能快的做法
with t1 as (select cuid,count(if(event_day='2020-04-01',1,null)) as cnt1,count(if(event_day='2020-04-02',1,null)) as cnt2,count(if(event_day='2020-04-08',1,null)) as cnt8from tb_cuid_1dwhere event_day in ('2020-04-01','2020-04-02','2020-04-08')group by cuidhaving cnt1 >0
),t2 as (select count(cuid)                  as uv1,count(if(cnt2 > 0, 1, null)) as uv2,count(if(cnt8 > 0, 1, null)) as uv7from t1)
select *,uv2 / uv1 as `次日留存率`,uv7 / uv1 as `7日留存率`
from t2

分组内top前几

• 需求常见词：【每组xxx的第一个yyy的zzz】【每组xxx的最后一个】

  【每组xxx的前n个】【每组最xx的前n个】

• 公式：row_number() over(partition by 组名) as rn，再筛选rn<=N名

跨越物流

员工表结构

员工表数据

题目描述

求出每个部门工资最高的前三名员工，并计算这些员工的工资占所属部门总工资的百分比。

结果

create table emp(empno string ,ename string,hiredate string,sal int ,deptno string);
insert overwrite table emp values
('7521', 'WARD', '1981-2-22', 1250, 30),
('7566', 'JONES', '1981-4-2', 2975, 20),
('7876', 'ADAMS', '1987-7-13', 1100, 20),
('7369', 'SMITH', '1980-12-17', 800, 20),
('7934', 'MILLER', '1982-1-23', 1300, 10),
('7844', 'TURNER', '1981-9-8', 1500, 30),
('7782', 'CLARK', '1981-6-9', 2450, 10),
('7839', 'KING', '1981-11-17', 5000, 10),
('7902', 'FORD', '1981-12-3', 3000, 20),
('7499', 'ALLEN', '1981-2-20', 1600, 30),
('7654', 'MARTIN', '1981-9-28', 1250, 30),
('7900', 'JAMES', '1981-12-3', 950, 30),
('7788', 'SCOTT', '1987-7-13', 3000, 20),
('7698', 'BLAKE', '1981-5-1', 2850, 30);select * from emp;--求出每个部门工资最高的前三名员工，并计算这些员工的工资占所属部门总工资的百分比。
select a.empno,a.sal,a.deptno,a.rn,a.sum_sal,round(a.sal/a.sum_sal,2) as rate
from
(select *,
--每个部门工资排名row_number() over (partition by deptno order by sal desc) as rn, 
--每个部门的总工资sum(sal) over(partition by deptno ) as sum_sal 
from emp) a
where rn<=3;

小米电商

订单表，torder. 字段，user_id, order_id, ctime(10位时间戳)，city id，sale_num，sku_id(商品)

问题：20201201至今每日订单量top10的城市及其订单量(订单量对order id去重)(在线写)

create table t_order (user_id string, order_id string, ctime string, city_id string, sale_num int , sku_id string) ;
with t1 as (select to_date(ctime) cdate, city_id, count(distinct order_id) cntfrom t_orderwhere to_date(ctime) >= '2020-12-01'and to_date(ctime) <= `current_date`()group by to_date(ctime), city_id),t2 as (select *, row_number() over (partition by cdate order by cnt desc) rn from t1)
select cdate, city_id, cnt
from t2
where rn <= 10;

窗口函数

• 窗口函数最重要的特点是有over关键字，代表定义窗口

  • 函数名(字段名) over(partition by xxx,yyy order by zzz)

• 聚合类的窗口函数

  • sum() over()
  • count/avg/max/min

• 排序类的窗口函数

  • row_number，rank，dense_rank

• 偏移类的，跨行的

  • lag / lead

• 【了解】first_value和last_value

• 【了解】ntile

交通银行

Emp表的表数据如下：

NAME	MONTH	AMT
张三	01	100
李四	02	120
王五	03	150
赵六	04	500
张三	05	400
李四	06	350
王五	07	180
赵六	08	400

问题：请写出可以得到以下的结果SQL

NAME	总金额	排名	占比
赵六	900	1	40.91%
张三	500	2	22.73%
李四	470	3	21.36%
王五	330	4	15.00%

create table emp(name string , month string, amt int);
insert overwrite table emp values ('张三', '01', 100),('李四', '02', 120),('王五', '03', 150),('赵六', '04', 500),('张三', '05', 400),('李四', '06', 350),('王五', '07', 180),('赵六', '08', 400);
--rank 1224
--dense_rank 1223
with t1 as (select name,sum(amt) as sum_amtfrom empgroup by name),t2 as (select name,sum_amt,row_number() over (order by sum_amt desc) rn,sum_amt/sum(sum_amt) over () as ratefrom t1)
select name, sum_amt, rn, concat(round(rate*100,2),'%') rate from t2

跨越物流

题目描述

在第一题员工表的基础上，统计每年入职总数以及截至本年累计入职总人数。

截至本年累计入职总人数=本年总入职人数 + 本年之前所有年的总入职人数之和

结果

select *,sum(cnt) over (order by year1) cnt2
from
(select year(hiredate) as year1,count(1) as cnt
from emp
group by year(hiredate)) a;

带条件的聚合统计

• 一般的做法是group by xx,yy 再多次的sum(if(…))
• 好处是避免多次加载表，得到多个指标，可以只加载一次表就得到多个指标。

腾讯数据提取

用户行为表：t_user_video_action_d分区：ds(格式 yyyyMMdd)

主键：user_id、video_id

含义：一个 user 对一个视频的所有行为聚合，每天增量字段：

字段名	字段含义	类型
user_id	用户 id	string
video_id	视频 id	string
expose_cnt	曝光次数	int
like_cnt	点赞次数	int

视频表：t_video_d

分区：ds(格式 yyyyMMdd)主键：video_id

含义：当天全量视频数据字段：

字段名	字段含义	类型	枚举
video_id	视频 id	string
video_type	视频类型	string	娱乐、新闻、搞笑
video_user_id	视频创建者 user_id	string
video_create_time	视频创建时间	bigint

​

作者表：t_video_user_d

分区：ds(格式 yyyyMMdd)主键：video_user_id

含义：当天全量视频创建者数据

字段名	字段含义	类型	枚举
video_user_id	视频创建者 user_id	string
video_user_name	名称	string
video_user_type	视频创建者类型	string	娱乐、新闻、搞笑

需求方需要视频号搞笑类型视频的曝光点赞时长等数据，请提供一张 ads 表。搞笑类型视频定义：视频类型为搞笑或者视频创建者类型为搞笑

需要产出字段：视频 id，视频创建者 user_id，视频创建者名称、当天曝光次数、当天点赞次数、近 30 天曝光次数、近 30 天点赞次数

create table if not exists t_user_video_action_d
(user_id    string comment "用户id",video_id   string comment "视频id",expose_cnt int comment "曝光次数",like_cnt   int comment "点赞次数"
) partitioned by (ds string);drop table t_video_d;
create table if not exists t_video_d
(video_id          string comment '视频id',video_type        string comment '视频类型',video_user_id     string comment '视频创建者user_id',video_create_time bigint comment '视频创建时间'
) partitioned by (ds string);create table if not exists t_video_user_d
(video_user_id   string comment '视频创建者user_id',video_user_name string comment '名称',video_user_type string comment '视频创建者类型'
) partitioned by (ds string);--假设当天是2022-07-31
select t1.*,t2.video_user_id,t2.video_user_name
from (select video_id,sum(case when ds = '2022-07-30' then expose_cnt else 0 end),--当天曝光次数、sum(case when ds = '2022-07-30' then like_cnt else 0 end),-- 当天点赞次数、sum(expose_cnt) as sum_expose,-- 近 30 天曝光次数、sum(like_cnt)-- 近 30 天点赞次数from t_user_video_action_dwhere ds between '2022-07-01' and '2022-07-30'group by video_id) as t1
join (select d.video_id, d.video_user_id, u.video_user_namefrom t_video_d djoin t_video_user_d u on d.video_user_id = u.video_user_idwhere (d.video_type like '%搞笑%' or u.video_user_type like '%搞笑%')and d.ds = '2022-07-30'and u.ds = '2022-07-30') as t2 on t1.video_id = t2.video_id

小米电商

要求：编写SQL能运行，数据正确且符合规范，如遇到自定义函数或不记得的函数可以用XX代替

1.已知有如下两个表表sale：字段如下

Create table sale_order(Order_id bigint comment '订单ID',User_id bigint comment '用户ID',Order_status int,Create_time string,Last_update_time string,Product_id bigint,Product_num bigint 
);

用户注册表：

Create table user_info(user_id bigint comment'用户ID，唯一主键',sex string.age int
);

问题：用一条SQL生成完整的用户画像表，包含如下字段：

user_id, sex, age, d7order_num, d14_order_num，后面两个字段分别为近7天订单数量，近14天订单数量。

create table sale_order(order_id bigint comment '订单ID',user_id bigint comment '用户ID',order_status int ,create_time string,last_update_time string,product_id bigint,product_num bigint
);
create table user_info(user_id bigint comment '用户ID,唯一主键',sex string,age int
);select u.user_id,s.d7order_num,s.d14order_num
from user_info u
left join (select user_id,count(if(create_time >= '7天前'  and create_time <= '今天', order_id,null)) as d7order_num,count(if(create_time >= '14天前' and create_time <= '今天', order_id,null)) as d14order_numfrom sale_orderwhere create_time >= '14天前'group by user_id) s on u.user_id = s.user_id;