1. 定义
城市等级(city_rank)小于3且GMV大于6000或者城市等级大与3且GMV大于5000定义为高消费(gq)
城市等级(city_rank)小于3且广告收入大于360或者城市等级大与3且广告收入大于300定义为高收入(pq)
flow_rank: 0低流量 1中流量 2高流量
合作商跨多个城市，选择city_rank最小值为其city_rank，毛收入多城市取和，广告毛收入按合作商收取
2. 要求查询
- GMV城市最高小于10000、总和小于30000、非首次合作、如GMV有量毛利大于0.02
- 高消费额、低收入、非高流量；低消费额、高收入、非高流量；低消费额、低收入、中流量；低消费额、低收入、低流量，且合作商gmv超过2000
3. 实现
select distinct
tc.partner_id as partnerId,
tc.contract_id as contractId,
tc.contract_num as contractNum,
tc.bd_id as bdId,
tc.org_id as orgId,
if(tc.org_scale=‘NULL‘,‘0‘,tc.org_scale) as orgScale
from table tc
join (
select
tc.partner_id,
case
when min(city_rank)<=3 and avg(t.gmv)>=6000 then 1
when min(city_rank)>3 and avg(t.gmv)>=5000 then 1
else 0
end as gq,
case
when min(city_rank)<=3 and avg(t.gross_profit+t.advertisement_gross_profit)>=360 then 1
when min(city_rank)>3 and avg(t.gross_profit+t.advertisement_gross_profit)>=300 then 1
else 0
end as pq,
max(tc.flow_rank) as fq,
sum(t.gmv) as gmv
from (
select partner_id,poi_id,min(city_rank) as city_rank,
max(tc.flow_rank) as flow_rank,
sum(is_old) as is_old
from table tc
where tc.partner_id>=#{start} and tc.partner_id<=#{end}
group by partner_id,poi_id
) tc
join (
select poi_id,
sum(gmv) as gmv,
sum(gross_profit) as gross_profit,
avg(advertisement_gross_profit) as advertisement_gross_profit
from table tc
where tc.partner_id>=#{start} and tc.partner_id<=#{end}
group by poi_id
) t on tc.poi_id=t.poi_id
group by tc.partner_id
having max(t.gmv)<10000 and sum(t.gmv)<30000 and sum(tc.is_old)>0 and
case
when sum(gmv)>0 then sum(gross_profit)/sum(gmv)>0.02
else 1=1
end
) t2 on t.partner_id=tc.partner_id
where
]]>
<if test="type ==0">
((t2.gq=1 and t2.pq=0 and fq!=2) or (t2.gq=0 and t2.pq=1 and fq!=2) or (t2.gq=0 and t2.pq=0 and fq=1) or (t2.gq=0 and t2.pq=0 and fq=0 and t2.gmv>2000))
</if>
4. 关键
4.1 t2
a) tc子查询计算流量、是否首次合作
b) t子查询计算毛利、毛收入
c) 计算合作商消费额、流量、收入类型，查询满足要求1的合作商
d) having子查询过滤，case子句限制毛利率
4.2 要求2
where过滤
5. 总结
大量使用了聚合计算、过滤，业务功能使用一条SQL实现
如果用代码实现类似的功能，复杂程度可以想象，每个聚合都会对应一大坨代码
时间： 2024-10-06 06:06:41