Oracle 12c-SQL查找乱序行 [英] Oracle 12c - sql to find out of order rows

问题描述

我有一个包含以下列的表格:

I have a table with following columns:

FILE_NAME VARCHAR2(30);
STATUS VARCHAR2(2);
DEPT_ID NUMBER;
DEPT_SUB_ID NUMBER;
CREATE_DATE DATE;

样本数据:

FILE_NAME                 STATUS  DEPT_ID   DEPT_SUB_ID  CREATE_DATE
---------                 ------- --------  -----------  ----------
TEST_20180806222127       C       1         10           07-AUG-18 01.04.47.821795000 AM
TEST_20180806221940       C       1         10           07-AUG-18 04.12.20.957400000 AM
TEST_20180806221733       C       1         10           07-AUG-18 03.35.27.809494000 AM
TEST_20180805202020       C       1         20           06-AUG-18 02.24.47.821795000 AM
TEST_20180805201640       C       1         20           06-AUG-18 00.42.20.957400000 AM
TEST_20180805201530       C       1         20           06-AUG-18 03.55.27.809494000 AM

FILE_NAME包含:<TYPE>_<DATETIME> 我想为每个DEPT_ID，DEPT_SUB_ID编写查询，以确定基于FILE_NAME和CREATE_DATE字段中的<DATETIME>乱序创建带有STATUS = 'C'的文件.在此示例中，对于DEPT_SUB_ID = 10，文件TEST_20180806222127是在文件名DATE_TIME之前的其他2个文件之前创建的，因此对于DEPT_SUB_ID = 10，我只需要返回此文件名即可.对于DEPT_SUB_ID = 20，结果应包含TEST_20180805201640和TEST_20180805202020，因为它们都是在TEST_20180805201530之前创建的，这被认为是乱序的.

FILE_NAME consists of: <TYPE>_<DATETIME> I want to write a query for each DEPT_ID, DEPT_SUB_ID to determine which files with STATUS = 'C' were created out of order based on the <DATETIME> on FILE_NAME and CREATE_DATE field. In this example, for DEPT_SUB_ID = 10, file TEST_20180806222127 was created before the other 2 based on the DATE_TIME on the file name so I would need to return only this file name in result for DEPT_SUB_ID = 10. For DEPT_SUB_ID = 20, result should contain TEST_20180805201640 and TEST_20180805202020 since both were created before TEST_20180805201530, which is considered out of order.

查询的预期结果将输出在运行顺序之前创建的所有file_name.

Expected results from query will output all file_name's which were created before it's order of run.

推荐答案

您可以为每行分配两个排名，一个基于文件名中嵌入时间戳的顺序，另一个基于创建日期的顺序:

You can assign two rankings to each row, one based on the order of the timestamp embedded int he file name, or other on the order of the creation date:

select yt.*,
  row_number() over (partition by dept_id, dept_sub_id
    order by to_date(substr(file_name, -14), 'YYYYMMDDHH24MISS')) as rn_file_name,
  row_number() over (partition by dept_id, dept_sub_id
    order by create_date) as rn_create_date
from your_table yt;

FILE_NAME           S    DEPT_ID DEPT_SUB_ID CREATE_DATE                   RN_FILE_NAME RN_CREATE_DATE
------------------- - ---------- ----------- ----------------------------- ------------ --------------
TEST_20180806221733 C          1          10 2018-08-07 03:35:27.809494000            1              2
TEST_20180806221940 C          1          10 2018-08-07 04:12:20.957400000            2              3
TEST_20180806222127 C          1          10 2018-08-07 01:04:47.821795000            3              1
TEST_20180805201530 C          1          20 2018-08-06 03:55:27.809494000            1              3
TEST_20180805201640 C          1          20 2018-08-06 00:42:20.957400000            2              1
TEST_20180805202020 C          1          20 2018-08-06 02:24:47.821795000            3              2

然后过滤以查看不匹配项:

Then filter to see the mismatches:

select file_name, status, dept_id, dept_sub_id, create_date
from (
  select yt.*,
    row_number() over (partition by dept_id, dept_sub_id
      order by to_date(substr(file_name, -14), 'YYYYMMDDHH24MISS')) as rn_file_name,
    row_number() over (partition by dept_id, dept_sub_id
      order by create_date) as rn_create_date
  from your_table yt
)
where rn_file_name > rn_create_date;

FILE_NAME           S    DEPT_ID DEPT_SUB_ID CREATE_DATE                  
------------------- - ---------- ----------- -----------------------------
TEST_20180806222127 C          1          10 2018-08-07 01:04:47.821795000
TEST_20180805201640 C          1          20 2018-08-06 00:42:20.957400000
TEST_20180805202020 C          1          20 2018-08-06 02:24:47.821795000

如果您不想一次看到所有ID，则可以在内部查询或外部查询中为特定ID或子ID添加过滤器.

And you can add a filter for a specific ID or sub-ID, either in the inner or outer query, if you don't want to see them all at once.

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