MYSQL/HIVESQL笔试题（六）：HIVESQL（六）

本文主要是介绍MYSQL/HIVESQL笔试题（六）：HIVESQL（六），对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

面试题目一

场景:一个日志表中记录了某个商户费率变化状态的所有信息，现在有个需求,要取出按照时间轴顺序，发生了状态变化的数据行；

1.数据如下:

create table datafrog_merchant
(f_merchant_id varchar(20),
f_rate varchar(20),
f_date date
);

insert into datafrog_merchant values
(100,0.1,'2016-03-02'),
(100,0.1,'2016-02-02'),
(100,0.2,'2016-03-05'),
(100,0.2,'2016-03-06'),
(100,0.3,'2016-03-07'),
(100,0.1,'2016-03-09'),
(100,0.1,'2016-03-10'),
(100,0.1,'2016-03-10'),
(200,0.1,'2016-03-10'),
(200,0.1,'2016-02-02'),
(200,0.2,'2016-03-05'),
(200,0.2,'2016-03-06'),
(200,0.3,'2016-03-07'),
(200,0.1,'2016-03-09'),
(200,0.1,'2016-03-10'),
(200,0.1,'2016-03-10');

我们来看看数据长得怎么样:

 2.实现想要的效果
解决问题思路：Lag函数或者Lead函数可以将上一行或者下一行的字段内容获取到本行，这样可以比较字段是否发生变化，进而判断是否状态变化，是否需要提取出该数据行；

select
    t1.f_merchant_id,
    t1.f_rate,
    t1.f_date
from
(
select
    f_merchant_id,
    f_rate,
    f_date,
    lag(f_rate,1,-999) over(partition by f_merchant_id order by f_date) as f_rate2
from 
    datafrog_merchant
) t1
where
    t1.f_rate <> t1.f_rate2

面试题目二

1.题目如下

 2.下面开始建表、插入数据

create table datafrog_test1
(userid varchar(20),
changjing varchar(20),
inttime varchar(20)
);

insert into datafrog_test1 values
(1,1001,1400),
(2,1002,1401),
(1,1002,1402),
(1,1001,1402),
(2,1003,1403),
(2,1004,1404),
(3,1003,1400)
(4,1004,1402),
(4,1003,1403),
(4,1001,1403),
(4,1002,1404)
(5,1002,1402),
(5,1002,1403),
(5,1001,1404),
(5,1003,1405);

3.mysql解答思路:排序及concat连接

select concat(t.userid,'-',group_concat(t.changjing separator'-')) as result
from(
 select userid,changjing,inttime,
     if(@tmp=userid,@rank:=@rank+1,@rank:=1) as new_rank,
   @tmp:=userid as tmp
 from (select userid,changjing, min(inttime) inttime from datafrog_test1 group by userid,changjing)temp
 order by userid,inttime
    )t
where t.new_rank<=2
group by t.userid;

4.输出结果:

5.注意:
有可能大家的代码会有报错现象，主要是ONLY_FULL_GROUP_BY引起的报错，解决办法是运行上面代码时，运行下这个就好set sql_mode='' 。其实mysql 作为查询还是不错的，但是拿来做分析的话，就是有点乏力了，像排序、中位数等都比较麻烦些的，工作中一般用pandas、sqlserver、oracle、hive、spark这些来做分析。这里可不是说mysql没用，反而是特别有用，也容易上手，是其他的基础。

6.大家来看下hive解法

with tmp as  (
select 
            userid,
            changjing,order_num,changjing1 
        from  
            (SELECT  userid ,
                    changjing,
                    row_number() over(partition by userid order by inttime asc) as order_num,
                    lag(changjing,1,'datafrog') OVER(partition by userid order by inttime asc) AS changjing1
                FROM datafrog_test1) as a 
    where changjing!=changjing1)
, tmp2 as (
select userid,changjing,order_num,changjing1,
row_number() over(partition by userid order by order_num ) as changjing_num
from tmp
)
select concat( userid,'-',concat_ws('-', collect_set(changjing)) )
from tmp2 where changjing_num <3
group by userid

这里主要考察了hive sql 中的with as、row_number() over()、lag() 的用法

这篇关于MYSQL/HIVESQL笔试题（六）：HIVESQL（六）的文章就介绍到这儿，希望我们推荐的文章对大家有所帮助，也希望大家多多支持为之网！