通过下面的极其复杂的sql语句，如下
SELECT d.id AS uid,(SELECT paystudate_id FROM erp_demand_meter as b WHERE demand_id = d.id ORDER BY addtime DESC LIMIT 1) as pp,
(SELECT uerdates FROM erp_demand_meter as b WHERE demand_id = d.id and b.paystudate_id = 14 ORDER BY addtime DESC LIMIT 1) as uerdat,
(SELECT plan_time FROM erp_demand_meter AS b where Demand_id = d.id ORDER BY addtime DESC limit 1) AS plan_time,
(SELECT addtime FROM erp_demand_meter WHERE Demand_id = d.id ORDER BY addtime DESC LIMIT 1) AS time ,
d.exec_name,d.paytime,d.paystudate,d.facttime,d.Demand_type_id,d. NAME,d.pid,d.addtime,d.company_type,d.demand_desc,d.pername,d.Demand_file_id, DATE_FORMAT(d.endtime, '%Y-%m-%d') AS edtime,d.state,d.perid,d.demand_level,d.demand_file,t.type_name,t.company_id,t.id
FROM erp_demand d LEFT JOIN erp_demand_type t ON d.demand_type_id = t.id
where d.paystudate = 0 and demand_type_id = 1
ORDER BY d.update_time desc,uid desc limit 0,20
我们得到了一个多集合的结果集，我们这次的任务是pp为NULL和0的筛选出来
用到：
（1）IFNULL(expr1,expr2)的用法：
假如expr1 不为 NULL，则 IFNULL() 的返回值为 expr1;
（2）用select * from （结果集）as result where IFNULL(result.pp,0) = 0 【可以找出null，0】
如下：
select * from
(SELECT d.id AS uid,(SELECT paystudate_id FROM erp_demand_meter as b WHERE demand_id = d.id ORDER BY addtime DESC LIMIT 1) as pp,
(SELECT uerdates FROM erp_demand_meter as b WHERE demand_id = d.id and b.paystudate_id = 14 ORDER BY addtime DESC LIMIT 1) as uerdat,
(SELECT plan_time FROM erp_demand_meter AS b where Demand_id = d.id ORDER BY addtime DESC limit 1) AS plan_time,
(SELECT addtime FROM erp_demand_meter WHERE Demand_id = d.id ORDER BY addtime DESC LIMIT 1) AS time ,
d.exec_name,d.paytime,d.paystudate,d.facttime,d.Demand_type_id,d. NAME,d.pid,d.addtime,d.company_type,d.demand_desc,d.pername,d.Demand_file_id, DATE_FORMAT(d.endtime, '%Y-%m-%d') AS edtime,d.state,d.perid,d.demand_level,d.demand_file,t.type_name,t.company_id,t.id
FROM erp_demand d LEFT JOIN erp_demand_type t ON d.demand_type_id = t.id
where d.paystudate = 0
ORDER BY d.update_time desc,uid desc)a where IFNULL(a.pp,0) = 0
现在筛选出来了
