select sum(money) s,name,date_format(time,'%Y-%m-%d') d from table group by name,d;
Then convert according to here

If you have to use SQL, you have to write a stored procedure and then create a temporary table.
No need, just take out the data and use the server to do it. Much more efficient.

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 gave you a typical row-to-column SQL writing method
Because the table only needs to be scanned once, there is generally no performance problem unless it is a particularly large table.

This writing method itself is just a solution idea. If the columns are not fixed, the sql statement can be dynamically generated through php or java on the application side.

Just write the sql that generates the script and you’re done

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;