您可以通过动态sql来实现这一点，例如，首先查找不同的名称值，然后围绕它们构建其余的代码
鉴于

MariaDB [sandbox]> select * from t;
+------------+-----------+
| day        | name      |
+------------+-----------+
| 2018-01-01 | test      |
| 2018-01-01 | test      |
| 2018-01-01 | example   |
| 2018-01-01 | somevalue |
| 2018-01-02 | test      |
| 2018-01-03 | test      |
+------------+-----------+
6 rows in set (0.00 sec)

set @sql = concat('select day, ',
(select group_concat(maxstr)
from
(select concat('max(case when name = ', char(39),name,char(39),' then  name else null end) as ', concat('name',@rn:=@rn+1)) maxstr
from  
(select distinct name from t) t,(select @rn:=0) rn
) s
)
,
' from t group by day')
;

生成此代码

select day, max(case when name = 'test' then  name else null end) as name1,
        max(case when name = 'example' then  name else null end) as name2,
        max(case when name = 'somevalue' then  name else null end) as name3 
from t group by day;

当运行

+------------+-------+---------+-----------+
| day        | name1 | name2   | name3     |
+------------+-------+---------+-----------+
| 2018-01-01 | test  | example | somevalue |
| 2018-01-02 | test  | NULL    | NULL      |
| 2018-01-03 | test  | NULL    | NULL      |
+------------+-------+---------+-----------+
3 rows in set (0.00 sec)

使用动态sql的优点是，它非常容易出错，并且忘记了代码将捕获的任何新值。不过，要注意团体的限制。
像这样执行动态sql-

prepare sqlstmt from @sql;
execute sqlstmt;
deallocate prepare sqlstmt;

在这段代码中，您按日期和名称分组，并按升序排序

SELECT * FROM table1
group by day, name
ORDER BY day ASC;

检查图像

您可以使用groupby和groupconcat来执行此操作，如下所示

select t.day,left(t.data,length(t.data)-1)
from
(
SELECT day,replace(group_concat(concat(name,'-')),',','')as data
FROM table1
group by day
  )t