我一直在寻找小时,没有一个像样的答案。
我想要改造这个表:
Client_id Date
----------- ------------
1 2013-02-03
1 2013-02-10
1 2013-05-12
2 2013-02-03
2 2013-07-15
至:
Client_id Date1 Date2 Date3 Date4, Date5, Date6...
----------- ------------ ------------ ------------ ------------
1 2013-02-03 2013-02-10 2013-05-12
2 2013-02-03 2013-07-15
为了得到这样的结果,你会想透视数据。 MySQL不具有旋转功能,但你可以使用聚合函数与CASE
表达式。
如果日期的数量是已知的,那么你可以硬编码查询:
select client_id,
max(case when rownum = 1 then date end) Date1,
max(case when rownum = 2 then date end) Date2,
max(case when rownum = 3 then date end) Date3
from
(
select client_id,
date,
@row:=if(@prev=client_id, @row,0) + 1 as rownum,
@prev:=client_id
from yourtable, (SELECT @row:=0, @prev:=null) r
order by client_id, date
) s
group by client_id
order by client_id, date
请参阅SQL拨弄演示
我实现了用户变量行号分配到内每个记录client_id
组。
如果你有数目不详的日期,那么你将需要使用一份声明中动态创建的SQL:
SET @sql = NULL;
SELECT
GROUP_CONCAT(DISTINCT
CONCAT(
'MAX(CASE WHEN rownum = ',
rownum,
' THEN date END) AS Date_',
rownum
)
) INTO @sql
from
(
select client_id,
date,
@row:=if(@prev=client_id, @row,0) + 1 as rownum,
@prev:=client_id
from yourtable, (SELECT @row:=0) r
order by client_id, date
) s
order by client_id, date;
SET @sql
= CONCAT('SELECT client_id, ', @sql, '
from
(
select client_id,
date,
@row:=if(@prev=client_id, @row,0) + 1 as rownum,
@prev:=client_id
from yourtable, (SELECT @row:=0) r
order by client_id, date
) s
group by client_id
order by client_id, date');
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;
请参阅SQL拨弄演示 。
他们都给出了结果:
| CLIENT_ID | DATE_1 | DATE_2 | DATE_3 |
--------------------------------------------------------------------------------------------------------------
| 1 | February, 03 2013 00:00:00+0000 | February, 10 2013 00:00:00+0000 | May, 12 2013 00:00:00+0000 |
| 2 | February, 03 2013 00:00:00+0000 | July, 15 2013 00:00:00+0000 | (null) |