select sum(money) s,name,date_format(time,'%Y-%m-%d') d from table group by name,d;
然后根据这里转换

非要用sql做的话，得写存储过程，然后创建个临时表了。
没必要，把数据取出来后用服务端去做吧。效率高很多。

create table tt1(id int ,name VARCHAR(100),money FLOAT,time DATETIME) insert into tt1 select 1,'mike',6,'2016-09-01 12:58:00' UNION select 2,'mike',10,'2016-09-01 13:52:56' UNION select 3,'leo',10,'2016-09-02 00:05:05' UNION select 4,'mike',6,'2016-09-03 08:06:05' select name,sum(case when datediff(time,'2016-09-01')=0 then money else 0 end) as '2016-09-01' ,sum(case when datediff(time,'2016-09-02')=0 then money else 0 end) as '2016-09-02' ,sum(case when datediff(time,'2016-09-03')=0 then money else 0 end) as '2016-09-03' from tt1 GROUP BY name

@ch21st 给的就是典型的行转列SQL写法
因为只需要表扫描一次，因此一般情况下不存在性能问题，除非是特别大的表。

这种写法本身只是一个解决思路，如果列不固定的话，可以在应用端通过php或java动态生成sql语句。

写个生成脚本的sql就完了嘛

select "select name," union select concat("sum(case when datediff(time,'",date(time),"')=0 then money else 0 end) as '",date(time),"',") a from tt1 group by a union select " from tt1 group by name;"; PS：union中间段最后一个逗号记得去掉

SELECT name, sum( if(date_format(time,'%Y-%m-%d') = '2016-09-01', money, 0 ) ) AS '2016-09-01', sum( if(date_format(time,'%Y-%m-%d') = '2016-09-02', money, 0 ) ) AS '2016-09-02', sum( if(date_format(time,'%Y-%m-%d') = '2016-09-03', money, 0 ) ) AS '2016-09-03' FROM money_table GROUP BY name;