hivesql与业务解决思路

查看每个城市每天完成订单数，取消订单数，下单订单数，下单用户数

select 
	city_id,
	sum(case when order_status=5 then 1 else 0 end) as cnt_ord_succ_d, 
	sum(case when order_status=3 then 1 else 0 end) as cnt_ord_cacel_d, 
	sum(1) as cnt_ord_d, 
	count(distinct CUST_ID) as cnt_ord_user //用户ID去重 
FROM 
	dw.dw_order 
WHERE 
	dt='${day_01}' 
group by 
	city_id;

获取一天所在的自然周

（但是这样比较费时间，如果有时间维表，直接用时间维表的话更方便，但是用时间维表的时候，要清楚每一个字段的含义和用法，这样在使用的时候，可以直接避免冗余的操作）

select 
	concat(
		date_add(
			statis_day,
			-pmod(
				datediff(    --ds 是星期几
					statis_day,
					'20100104'
				), 
				7
	    	)
		), 
	    '~'
		,date_add(
			statis_day,
            6-pmod(
			    datediff(    --de 是星期几
			        statis_day,
			        '20100104'
		        ), 
		        7
		    )
        )
	) as time_stage,
from 
	t_union_cnt t

近六个月的活跃老用户（近六个月，可以直接根据现在时间，把时间写死）

月份可以用substr函数取，时间数据类型的前六位，即年和月

with target_user as
(
	select t1.six_month as six_month
          ,t1.qimei as qimei
          ,guid
	from 
	(
		select substr(statis_day,0,6) as six_month
	                 ,qimei
	                 ,guid
	    from u_wsd::t_od_qidian_cl_daumap_hx_c1
	    where statis_day>=20200701
    ) t1
    join
	(
		select substr(statis_day,0,6) as six_month
			   ,qimei
	    from yw_dh_app::t_ed_ywdh_hx_ai_wide_c1
		where is_new = 0 and is_active > 0 and statis_day>=20200701
	) t2
	on t1.six_month = t2.six_month and t1.qimei = t2.qimei
	group by t1.six_month, t1.qimei, guid
)

每日新访客

insert into table etl_user_new_day partition(day="20170101")
select 
   a.uid,a.commit_time,a.city,a.release_channel,a.app_ver_name
from etl_user_active_day a
left join
etl_history_user b
on a.uid=b.uid
where a.day="20170101" and b.uid is null;

年度销售商品的前十

在select查找的数据中，使用row_number函数，并且使用partition对时间分区，和使用order by对于商品进行排序，使得row_number的值就成为了这个时间区的商品销售名次

select

       ref_host,

       ref_host_cnts,

       concat(month,hour,day),

       row_number() over (partition by concat(month,hour,day) order by ref_host_cnts desc) as od

from

       dw_ref_host_visit_cnts_h

商品复购率

需求列出的商品的7日,15日,30复购率，目的了解这几款商品的周期. 计算口径:当日购买部分商品的用户数/7日重复购买此商品的用户数。每天查看每个城市每个商品当日购买用户数，7日15日30日复购率。

SELECT 
	t3.atdate AS cdate,
	t3.city_id,
	t3.goods_id, 
	COUNT(DISTINCT CASE WHEN days=0 THEN t3.cust_id END) AS cnt_buy_cust_d, 
	COUNT(DISTINCT CASE WHEN days>0 AND days<=7 THEN t3.cust_id END) AS cnt_buy_cust_7_d, 
	COUNT(DISTINCT CASE WHEN days>0 AND days<=15 THEN t3.cust_id END) AS cnt_buy_cust_15_d, 
	COUNT(DISTINCT CASE WHEN days>0 AND days<=30 THEN t3.cust_id END) AS cnt_buy_cust_30_d 
FROM 
( 
	SELECT 
		t1.atdate,
		t1.city_id,
		t1.cust_id,
		t1.goods_id, 
		DATEDIFF(t2.atdate, t1.atdate) days 
	FROM (
	//查出成功订单的。 
		SELECT 
			o.order_date AS atdate,
			o.city_id, 
			o.cust_id,
			og.goods_id 
		FROM 
			dw.dw_order o 
		INNER JOIN 
			dw.dw_order_goods og 
		ON 
			o.order_id=og.order_id AND 
			o.ORDER_STATUS = 5 AND 
			og.source_id=1 AND 
			o.dt = '20151010' 
	) t1 
	INNER JOIN (
	//查看n天后成功订单的 
		SELECT 
			o.order_date AS atdate,
			o.city_id, 
			o.cust_id,
			og.goods_id, 
			og.goods_name 
		FROM 
			dw.dw_order o 
		INNER JOIN 
			dw.dw_order_goods og 
		ON 
			o.order_id=og.order_id AND 
			o.ORDER_STATUS = 5 AND 
			og.source_id=1 
	) t2 
	ON 
t1.cust_id=t2.cust_id AND 
t1.goods_id=t2.goods_id 
) t3 
GROUP BY 
	t3.atdate,
	t3.city_id,
	t3.goods_id;

Tips：

join的时候，可能会有对应的维度表，尽量join维度表这种小标

全量表

	select cbid,site,authorid from  t
	where site in (2,3,22,23) --and (regexp_replace(substr(trim(updatetime),1,10),'-','') between 20200101 and 20200102) 
	and statis_day = 20201231 and status != 50 --(全量表，一天的数据就含有了以前的历史信息，用当天的数据即可，不要用<=)
	and ext1=1 and auditstatus=19

回溯数据任务完成步骤

1先看表中有没有当天数据

2再看这个表是不是可回溯（每天增量还是全量）

3看是否有前置数据，（venus前面的组件）

4venus或者ide insert，加入数据

hivesql与业务解决思路

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