-- 收费业务分析
-- 收费数据分析
-- 收费金额
select
		 t.car_park_name,
		 case when sum(amount) is null then 0 else round(sum(amount)/100,2) end +30 as reality_amount, -- 应收金额
		 case when sum(amount) is null then 0 else round(sum(amount)/100,2) end as amount -- 实收金额
FROM
	luo_parking_information t
	left join 
(select a.park_id,a.pay_time,b.amount
from luo_parking_chargerecord a
inner join luo_parking_chargerecord_subject b
on a.order_no=b.order_no) c
ON t.id = c.park_id 
group by 1
HAVING sum(amount)>0