如何根据Oracle中两个列值的差异合并/分组记录

时间：2013-12-01 08:22:03

标签： sql oracle merge group-by

我有以下数据集，

from            to           stayed
01.01.2011     03.04.2011    NY
03.04.2011     25.05.2011    NJ
25.05.2011     04.06.2011    NJ
04.06.2011     20.06.2011    NJ
20.06.2011     30.06.2011    NJ
30.06.2011     05.07.2011    CA
05.07.2011     20.07.2011    CA
20.07.2011     05.08.2011    NY

并且必须转换为，

from            to          stayed
01.01.2011     03.04.2011    NY
03.04.2011     30.06.2011    NJ
30.06.2011     20.07.2011    CA
20.07.2011     05.08.2011    NY

还有另一个数据集，与前一个“to”和当前“from”之间存在差距，应计入新行，如下所示，

from            to           stayed
01.01.2011     03.04.2011    NY
03.04.2011     25.05.2011    NJ
25.05.2011     04.06.2011    NJ
04.06.2011     20.06.2011    NJ
20.06.2011     30.06.2011    NJ
30.06.2011     05.07.2011    CA
05.07.2011     20.07.2011    CA
20.07.2011     05.08.2011    NY
16.09.2011     20.09.2011    NY
20.09.2011     29.09.2011    NY
05.10.2011     20.10.2011    NY

和预期结果如下，    

  from            to          stayed
    01.01.2011     03.04.2011    NY
    03.04.2011     30.06.2011    NJ
    30.06.2011     20.07.2011    CA
    20.07.2011     05.08.2011    NY
    16.09.2011     29.09.2011    NY
    05.10.2011     20.10.2011    NY

进行这样的合并我用分析函数编写了复杂的查询。有一个简单的方法吗？

2 个答案:

答案 0 :(得分：2)

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE tbl ( "from", "to", "stayed" ) AS
          SELECT TO_DATE( '01.01.2011', 'DD.MM.YYYY' ), TO_DATE( '03.04.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL
UNION ALL SELECT TO_DATE( '03.04.2011', 'DD.MM.YYYY' ), TO_DATE( '25.05.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '25.05.2011', 'DD.MM.YYYY' ), TO_DATE( '04.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '04.06.2011', 'DD.MM.YYYY' ), TO_DATE( '20.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '20.06.2011', 'DD.MM.YYYY' ), TO_DATE( '30.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '30.06.2011', 'DD.MM.YYYY' ), TO_DATE( '05.07.2011', 'DD.MM.YYYY' ), 'CA' FROM DUAL
UNION ALL SELECT TO_DATE( '05.07.2011', 'DD.MM.YYYY' ), TO_DATE( '20.07.2011', 'DD.MM.YYYY' ), 'CA' FROM DUAL
UNION ALL SELECT TO_DATE( '20.07.2011', 'DD.MM.YYYY' ), TO_DATE( '05.08.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL;

查询1 ：

WITH groups AS (
  SELECT t.*,
         ROW_NUMBER() OVER ( ORDER BY "from" ) - ROW_NUMBER() OVER ( PARTITION BY "stayed" ORDER BY "from" ) AS grp
  FROM   tbl t
)
SELECT MIN( "from" ) AS "from",
       MAX( "to" ) AS "to",
       "stayed"
FROM   groups
GROUP BY "stayed", grp
ORDER BY "from"

<强> Results ：

|                           FROM |                            TO | STAYED |
|--------------------------------|-------------------------------|--------|
| January, 01 2011 00:00:00+0000 |  April, 03 2011 00:00:00+0000 |     NY |
|   April, 03 2011 00:00:00+0000 |   June, 30 2011 00:00:00+0000 |     NJ |
|    June, 30 2011 00:00:00+0000 |   July, 20 2011 00:00:00+0000 |     CA |
|    July, 20 2011 00:00:00+0000 | August, 05 2011 00:00:00+0000 |     NY |

修改

要回答您的后续编辑 - 这是解决问题的一种方法：

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE tbl ( "from", "to", "stayed" ) AS
          SELECT TO_DATE( '01.01.2011', 'DD.MM.YYYY' ), TO_DATE( '03.04.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL
UNION ALL SELECT TO_DATE( '03.04.2011', 'DD.MM.YYYY' ), TO_DATE( '25.05.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '25.05.2011', 'DD.MM.YYYY' ), TO_DATE( '04.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '04.06.2011', 'DD.MM.YYYY' ), TO_DATE( '20.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '20.06.2011', 'DD.MM.YYYY' ), TO_DATE( '30.06.2011', 'DD.MM.YYYY' ), 'NJ' FROM DUAL
UNION ALL SELECT TO_DATE( '30.06.2011', 'DD.MM.YYYY' ), TO_DATE( '05.07.2011', 'DD.MM.YYYY' ), 'CA' FROM DUAL
UNION ALL SELECT TO_DATE( '05.07.2011', 'DD.MM.YYYY' ), TO_DATE( '20.07.2011', 'DD.MM.YYYY' ), 'CA' FROM DUAL
UNION ALL SELECT TO_DATE( '20.07.2011', 'DD.MM.YYYY' ), TO_DATE( '05.08.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL
UNION ALL SELECT TO_DATE( '16.09.2011', 'DD.MM.YYYY' ), TO_DATE( '20.09.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL
UNION ALL SELECT TO_DATE( '20.09.2011', 'DD.MM.YYYY' ), TO_DATE( '29.09.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL
UNION ALL SELECT TO_DATE( '05.10.2011', 'DD.MM.YYYY' ), TO_DATE( '20.10.2011', 'DD.MM.YYYY' ), 'NY' FROM DUAL;

查询2 ：

WITH change_in_groups AS (
  SELECT t.*,
         CASE WHEN t."from"   = LAG( t."to"     ) OVER ( ORDER BY "from" )
              AND  t."stayed" = LAG( t."stayed" ) OVER ( ORDER BY "from" )
              THEN 0
              ELSE 1
              END AS has_changed_group
  FROM   tbl t
),
groups AS (
  SELECT "from",
         "to",
         "stayed"
         ,SUM( has_changed_group ) OVER ( ORDER BY "from" ) AS grp
  FROM   change_in_groups
)
SELECT MIN( "from" ) AS "from",
       MAX( "to" ) AS "to",
       MAX( "stayed" ) AS "stayed"
FROM   groups
GROUP BY grp
ORDER BY "from"

<强> Results ：

|                             FROM |                               TO | STAYED |
|----------------------------------|----------------------------------|--------|
|   January, 01 2011 00:00:00+0000 |     April, 03 2011 00:00:00+0000 |     NY |
|     April, 03 2011 00:00:00+0000 |      June, 30 2011 00:00:00+0000 |     NJ |
|      June, 30 2011 00:00:00+0000 |      July, 20 2011 00:00:00+0000 |     CA |
|      July, 20 2011 00:00:00+0000 |    August, 05 2011 00:00:00+0000 |     NY |
| September, 16 2011 00:00:00+0000 | September, 29 2011 00:00:00+0000 |     NY |
|   October, 05 2011 00:00:00+0000 |   October, 20 2011 00:00:00+0000 |     NY |

使用分层查询的替代解决方案：

查询3 ：

WITH leaves AS (
  SELECT CONNECT_BY_ROOT "from" AS "from",
         "to",
         "stayed",
         CONNECT_BY_ISLEAF AS leaf
  FROM   tbl t
  CONNECT BY PRIOR "stayed" = "stayed"
         AND PRIOR "to"     = "from"
)
SELECT MIN( "from" ) AS "from",
       "to",
       MIN( "stayed" ) AS "stayed"
FROM   leaves
WHERE  leaf = 1
GROUP BY "to"

<强> Results ：

|                             FROM |                               TO | STAYED |
|----------------------------------|----------------------------------|--------|
|     April, 03 2011 00:00:00+0000 |      June, 30 2011 00:00:00+0000 |     NJ |
|   January, 01 2011 00:00:00+0000 |     April, 03 2011 00:00:00+0000 |     NY |
|      June, 30 2011 00:00:00+0000 |      July, 20 2011 00:00:00+0000 |     CA |
|      July, 20 2011 00:00:00+0000 |    August, 05 2011 00:00:00+0000 |     NY |
| September, 16 2011 00:00:00+0000 | September, 29 2011 00:00:00+0000 |     NY |
|   October, 05 2011 00:00:00+0000 |   October, 20 2011 00:00:00+0000 |     NY |

您可能希望对查询进行概要分析，但是从解释计划中，分层查询看起来更有效。

答案 1 :(得分：0)

Select distinct 
   min(from) over (parttion by stayed order by from) as from_stayed,
   max(to) over (parttion by stayed order by to) as to_stayed,
   stayed
from table

好的，试试这个更正的

with t2 as
(select t1.*,
  case lag(stayed,1,stayed) over (order by dt)
  when stayed then 0 else 1 end as stayed_mod
from table),
t3 as 
  (select t2.*,
    sum(stayed_mod) over (order by from) as group_id
  from t2)
select distinct stayed,
  min(from) over (partition by GROUP_ID) as from_min,
  max(to) over (partition by GROUP_ID) as to_max
from t3
order by 2;
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