Question

我一直在搜索没有得体的答案。

我想改变这个表：


Client_id    Date
-----------  ------------  
1            2013-02-03    
1            2013-02-10
1            2013-05-12
2            2013-02-03
2            2013-07-15

要：


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

Answer 1

为了获得此结果，您需要 pivot 数据。 MySQL没有pivot函数，但你可以使用带有CASE表达式的聚合函数。

如果已知日期数，则可以对查询进行硬编码：

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 Fiddle with Demo

我实现了用户变量，为client_id组中的每条记录分配行号。

如果您的日期数量未知，那么您需要使用预准备语句动态创建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 Fiddle with Demo。

他们都给出结果：

| 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) |

mysql数据透视表日期（垂直于水平数据）

1 个答案: