HiveSQL——连续增长问题

注：参考文章：

SQL连续增长问题--HQL面试题35_sql判断一个列是否连续增长-CSDN博客文章浏览阅读2.6k次，点赞6次，收藏30次。目录0 需求分析1 数据准备3 小结0 需求分析假设我们有一张订单表shop_order shop_id,order_id,order_time,order_amt 我们需要计算过去至少3天销售金额连续增长的商户shop_id。数据如下：shop_idorder_amtorder_time11002021-05-10 10:03:5411012021-05-10 10:04:5413002021-0_sql判断一个列是否连续增长https://blog.csdn.net/godlovedaniel/article/details/119080882

0 需求分析

  现有一张订单表shop_order ,含有字段shop_id,order_id,order_time,order_amt， 需要统计过去至少连续3天销售金额连续增长的商户shop_id。

1 数据准备

create table shop_order(shop_id int,order_amt int,order_time string
)
row format delimited fields terminated by '\t';
load data local inpath "/opt/module/hive_data/shop_order.txt" into table shop_order;

2 数据分析

   完整的代码如下：

with tmp as (selectshop_id,to_date(order_time) as dt,sum(order_amt)      as amtfrom shop_ordergroup by shop_id, to_date(order_time)
)
selectshop_id
from (select *,-- 判断日期是否连续date_sub(dt, row_number() over (partition by shop_id order by dt )) as order_date_difffrom (selectshop_id,dt,amt,--判断销售额是否增长-- 当前行的销售金额与上一行的销售金额之间的差值 order_amt_diffamt - lag(amt, 1, 0) over (partition by shop_id order by dt) as order_amt_diff from tmp) t1-- 差值大于0的代表销售额增长where order_amt_diff > 0) t2
group by shop_id, order_date_diff
having count(1) >=3;

输出结果为 shop_id 为2

上述代码分析：

 step1: 求出每家商户销售金额连续增长的记录

with tmp as (selectshop_id,to_date(order_time) as dt,sum(order_amt)      as amtfrom shop_ordergroup by shop_id, to_date(order_time)
)select *
from (selectshop_id,dt,amt,--判断销售额是否增长-- 当前行的销售金额与上一行的销售金额之间的差值 order_amt_diffamt - lag(amt, 1, 0) over (partition by shop_id order by dt) as order_amt_difffrom tmp) t1-- 差值大于0的代表销售额增长
where order_amt_diff > 0

 step2: 求出每家商户至少连续3天销售金额连续增长，在step1的基础上，还要求dt是连续的

with tmp as (selectshop_id,to_date(order_time) as dt,sum(order_amt)      as amtfrom shop_ordergroup by shop_id, to_date(order_time)
)select *,-- 判断日期是否连续date_sub(dt, row_number() over (partition by shop_id order by dt )) as order_date_diff
from (selectshop_id,dt,amt,--判断销售额是否增长-- 当前行的销售金额与上一行的销售金额之间的差值 order_amt_diffamt - lag(amt, 1, 0) over (partition by shop_id order by dt) as order_amt_difffrom tmp) t1-- 差值大于0的代表销售额增长
where order_amt_diff > 0

step3: 对商户shop_id以及日期差值order_date_diff这两个字段分组，求出最终结果

with tmp as (selectshop_id,to_date(order_time) as dt,sum(order_amt)      as amtfrom shop_ordergroup by shop_id, to_date(order_time)
)
selectshop_id
from (select *,-- 判断日期是否连续date_sub(dt, row_number() over (partition by shop_id order by dt )) as order_date_difffrom (selectshop_id,dt,amt,--判断销售额是否增长-- 当前行的销售金额与上一行的销售金额之间的差值 order_amt_diffamt - lag(amt, 1, 0) over (partition by shop_id order by dt) as order_amt_diff --判断是否增长from tmp) t1-- 差值大于0的代表销售额增长where order_amt_diff > 0) t2
group by shop_id, order_date_diff
having count(1) >=3;

3 小结

   date_sub(日期减少函数)

• 语法：date_sub(string startdate,int days)
• 返回值：string
• 说明：返回   开始日期startdate 减去days天后的日期
• 举例：select  date_sub('2024-02-01',3) --->2024-01-29

lag

• 语法：lag(column,n,default) over(partition by ....order by....)
• 说明：取得column列前边的第n行数据，如果存在则返回，如果不存在，返回默认值default

     针对【日期连续】等类型的题型，一般处理思路：先计算date_sub(dt, row_number() over (partition by shop_id order by dt )) as dt_diff ，再对dt_diff 分组，求count()值

    针对【xx连续增长】等类型的题型，一般处理思路：利用前后函数lag或者lead往前/往后取一行，计算两者的差值diff，再利用 if( diff >0,1,0) as flag 等条件判断函数 进行打标签，基于标签再进行后续的分组计算.......