1. 定義
城市等級(city_rank)小於3且GMV大於6000或者城市等級大與3且GMV大於5000定義爲高消費(gq)
城市等級(city_rank)小於3且廣告收入大於360或者城市等級大與3且廣告收入大於300定義爲高收入(pq)
flow_rank: 0低流量 1中流量 2高流量
合作商跨多個城市，選擇city_rank最小值爲其city_rank，毛收入多城市取和，廣告毛收入按合作商收取
2. 要求查詢
GMV城市最高小於10000、總和小於30000、非首次合作、如GMV有量毛利大於0.02
高消費額、低收入、非高流量；低消費額、高收入、非高流量；低消費額、低收入、中流量；低消費額、低收入、低流量，且合作商gmv超過2000
3. 實現
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. 關鍵
4.1 t2
a) tc子查詢計算流量、是否首次合作
b) t子查詢計算毛利、毛收入
c) 計算合作商消費額、流量、收入類型，查詢滿足要求1的合作商
d) having子查詢過濾，case子句限制毛利率
4.2 要求2
where過濾
5. 總結
大量使用了聚合計算、過濾，業務功能使用一條SQL實現
如果用代碼實現類似的功能，複雜程度可以想象，每個聚合都會對應一大坨代碼