Mysql 存在空值时，从两个表中减去两个值 选择p.partnerid， 金额（如果c.amount不为空，则c.amount为0结束）作为金额， 求和（c.netamt不为空时，则为c.netamt，否则为0结束）作为总计， 总和（当（c.netamt-d.paidamount）不为空时，则（c.netamt-d.paidamount）否则为0结束）作为剩余的装入， 求和（当d.paidamount不为空时，则d.paidamount否则为0结束）为paidamt 来自customerinfo c 在c.accno=d.accno上左连接每日付款d 在c.partnerid=p.partnerid上右键加入partnerinfo p 其中（d.paiddate为空或（d.paiddate>='2011-3-15'和d.paiddate

使用该函数删除

案例

，并简化计算：

select p.partnerid,
sum(case when c.amount is not null then c.amount else 0 end) as amount,
sum(case when c.netamt is not null then c.netamt else 0 end) as total,
sum(case when (c.netamt - d.paidamount) is not null then (c.netamt - d.paidamount) else 0 end) as remainingamount,
sum(case when d.paidamount is not null then d.paidamount else 0 end) as paidamt
from customerinfo c
left join dailypayments d on c.accno = d.accno
right join partnerinfo p on c.partnerid = p.partnerid
where (d.paiddate is null or (d.paiddate >= '2011-3-15' and d.paiddate <= '2012-6-13')) and  p.manager = 7 group by p.partnerid

sum(c.netamt - ifnull(d.paidamount, 0)) as remainingamount,