如何修复MySQL查询？

Question

创建表test。tableone（

idINT NOT NULL AUTO_INCREMENT,

placeVARCHAR(45)为空,

主键(ID))；

创建表test。TableTwo（

回执十进制（6,2）为空，

付款十进制（6,2）为空，

statusVARCHAR(45)为空,

插入test.tableone(empno,name,balance,place)值(101、'joselin','1000','dexcity')；

插入test.tableone(empno,name,balance,place)值(102、“rusfal”,“0”,“donxes”）；在test.tableone(empno,name,balance,place)中插入值(103、“raser”、“100”、“versity”）；

插入test.tableone(empno,name,balance,place)值(104、'rse','2500','sew')；

插入test.tabletwo(date,empno,receive,payment,status)值('2016-08-15',100,'1000,'0','OK')；插入test.tabletwo(date,empno,receive,payment,status)值('2016-08-15',100,'0','1000','OK')；

插入test.tabletwo(date,empno,receive,payment,status)值('2016-08-17',101,'0','2000','OK')；

插入test.tabletwo(date,empno,receive,payment,status)值（“2016-08-18”,103,“100”,“0”,“Not OK”）；插入test.tabletwo(date,empno,receive,payment,status)值('2016-08-19',100,'1500,'0','OK')；插入test.tabletwo(date,empno,receive,payment,status)值('2016-08-20',100,'0','1000','OK')；

然后我需要输出这个

开始日期“2016-08-18”结束日期“2016-08-20”

Empno Name Place OB cb
100 John Wasinton DC 2400 2000 101 Joselin Dexcity 1000 0 102 Rusfal Donxes 0 0

etc 103 104

也

Answer 1

使用此：

SELECT a.id,a.name,sum(amount1)as amount1,sum(Receipt) as Receipt,sum(payment) as payment,
sum(amount1)+sum(Receipt)-sum(payment) as Total,nameid,b.name 
FROM test.mas as a left join test.trans as b on a.id=b.nameid GROUP BY a.id;