通过本文的5个需求分析，可以看出SQL窗口函数的功能十分强大，不仅能够使我们编写的SQL逻辑更加清晰，而且在某种程度上可以简化需求开发。

数据准备

本文主要分析只涉及一张订单表orders，操作过程在Hive中完成，具体数据如下：

-- 建表CREATE TABLE orders(    order_id int,    customer_id string,    city string,    add_time string,    amount decimal(10,2));-- 准备数据                              INSERT INTO orders VALUES(1,"A","上海","2020-01-01 00:00:00.000000",200),(2,"B","上海","2020-01-05 00:00:00.000000",250),(3,"C","北京","2020-01-12 00:00:00.000000",200),(4,"A","上海","2020-02-04 00:00:00.000000",400),(5,"D","上海","2020-02-05 00:00:00.000000",250),(5,"D","上海","2020-02-05 12:00:00.000000",300),(6,"C","北京","2020-02-19 00:00:00.000000",300),(7,"A","上海","2020-03-01 00:00:00.000000",150),(8,"E","北京","2020-03-05 00:00:00.000000",500),(9,"F","上海","2020-03-09 00:00:00.000000",250),(10,"B","上海","2020-03-21 00:00:00.000000",600);

在业务方面，第m1个月的收入增长计算如下：100 *（m1-m0）/ m0

其中，m1是给定月份的收入，m0是上个月的收入。因此，从技术上讲，我们需要找到每个月的收入，然后以某种方式将每个月的收入与上一个收入相关联，以便进行上述计算。计算当时如下：

WITHmonthly_revenue as (    SELECT    trunc(add_time,'MM') as month,    sum(amount) as revenue    FROM orders    GROUP BY 1),prev_month_revenue as (    SELECT     month,    revenue,    lag(revenue) over (order by month) as prev_month_revenue -- 上一月收入    FROM monthly_revenue)SELECT   month,  revenue,  prev_month_revenue,  round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growthFROM prev_month_revenueORDER BY 1

结果输出

我们还可以按照按城市分组进行统计，查看某个城市某个月份的收入增长情况

WITHmonthly_revenue as (    SELECT     trunc(add_time,'MM') as month,    city,    sum(amount) as revenue    FROM orders    GROUP BY 1,2),prev_month_revenue as (    SELECT     month,    city,    revenue,    lag(revenue) over (partition by city order by month) as prev_month_revenue    FROM monthly_revenue)SELECT month,city,revenue,round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growthFROM prev_month_revenueORDER BY 2,1

结果输出

需求2：累计求和

累计汇总，即当前元素和所有先前元素的总和，如下面的SQL：

WITHmonthly_revenue as (    SELECT    trunc(add_time,'MM') as month,    sum(amount) as revenue    FROM orders    GROUP BY 1)SELECT month,revenue,sum(revenue) over (order by month rows between unbounded preceding and current row) as running_totalFROM monthly_revenueORDER BY 1

结果输出

我们还可以使用下面的组合方式进行分析，SQL如下：

SELECT   order_id,   customer_id,   city,   add_time,   amount,   sum(amount) over () as amount_total, -- 所有数据求和   sum(amount) over (order by order_id rows between unbounded preceding and current row) as running_sum, -- 累计求和   sum(amount) over (partition by customer_id order by add_time rows between unbounded    preceding and current row) as running_sum_by_customer,    avg(amount) over (order by add_time rows between 5 preceding and current row) as  trailing_avg -- 滚动求平均FROM ordersORDER BY 1

结果输出：

需求3：处理重复数据

从上面的数据可以看出，存在两条重复的数据**(5,"D","上海","2020-02-05 00:00:00.000000",250), (5,"D","上海","2020-02-05 12:00:00.000000",300),**显然需要对其进行清洗去重，保留最新的一条数据，SQL如下：

我们先进行分组排名，然后保留最新的那条数据即可：

SELECT *FROM (    SELECT *,    row_number() over (partition by order_id order by add_time desc) as rank    FROM orders) tWHERE rank=1

结果输出：

经过上面的清洗过程，对数据进行了去重。重新计算上面的需求1，正确SQL脚本为：

WITHorders_cleaned as (    SELECT *    FROM (        SELECT *,        row_number() over (partition by order_id order by add_time desc) as rank        FROM orders    )t    WHERE rank=1),monthly_revenue as (    SELECT    trunc(add_time,'MM') as month,    sum(amount) as revenue    FROM orders_cleaned    GROUP BY 1),prev_month_revenue as (    SELECT     month,    revenue,    lag(revenue) over (order by month) as prev_month_revenue    FROM monthly_revenue)SELECT month,revenue,round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growthFROM prev_month_revenueORDER BY 1

结果输出：

将清洗后的数据创建成视图，方便以后使用

CREATE VIEW orders_cleaned ASSELECT    order_id,     customer_id,     city,     add_time,     amountFROM (    SELECT *,    row_number() over (partition by order_id order by add_time desc) as rank    FROM orders)tWHERE rank=1

分组取topN是最常见的SQL窗口函数使用场景，下面的SQL是计算每个月份的top2订单金额，如下：

WITH orders_ranked as (    SELECT    trunc(add_time,'MM') as month,    *,    row_number() over (partition by trunc(add_time,'MM') order by amount desc, add_time) as rank    FROM orders_cleaned)SELECT     month,    order_id,    customer_id,    city,    add_time,    amountFROM orders_rankedWHERE rank <=2ORDER BY 1

下面的SQL计算重复购买率：重复购买的人数/总人数*100%以及第一笔订单金额与第二笔订单金额之间的典型差额:avg(第二笔订单金额/第一笔订单金额)

WITH customer_orders as (    SELECT *,    row_number() over (partition by customer_id order by add_time) as customer_order_n,    lag(amount) over (partition by customer_id order by add_time) as prev_order_amount    FROM orders_cleaned)SELECTround(100.0*sum(case when customer_order_n=2 then 1 end)/count(distinct customer_id),1) as repeat_purchases,-- 重复购买率avg(case when customer_order_n=2 then 1.0*amount/prev_order_amount end) as revenue_expansion -- 重复购买较上次购买差异，第一笔订单金额与第二笔订单金额之间的典型差额FROM customer_orders

结果输出：

WITH结果输出：

最终结果输出：