Question

代码：

$.ajax({
    url : url,
    success : function(data){
        var responseHtml = $.parseHTML(data);
        songTitle = $.trim(
                $(responseHtml).filter('div.head:first').text());
        alert(songTitle);
    }
});

我想将结果分成如下：

CREATE TABLE table1 (
  day DATE,
  name VARCHAR(40)
  );
INSERT INTO table1 (day, name)
VALUES 
('2018-01-01', 'test1'),
('2018-01-01', 'test2'),
('2018-01-01', 'example'),
('2018-01-01', 'somevalue'),
('2018-01-02', 'test3'),
('2018-01-03', 'test4');

而不是在选择数据时复制日期。

预期结果：

day - name1 - name2 - namex  
DATE - value - value - value

SQL FIDDLE

Answer 1

您可以通过动态sql执行此操作，首先查找不同的名称值然后围绕它们构建其余代码，例如

给定的

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')
;

构建此代码

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;

运行时产生此结果

+------------+-------+---------+-----------+
| 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的优点是它非常火，并且忘记代码会捕获任何新值。但请注意group_concat限制。

像这样执行动态sql -

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

Answer 2

在此代码中，您按日分组和名称，并使用升序

对其进行排序

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

检查image

Answer 3

你可以使用group by和group concat按照下面的方式执行此操作

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

将值拆分为其他几列

3 个答案: