Oracle Rank()在（子）组内的作用

Question

Oracle Rank()在（子）组内的作用

6

我正在尝试获取子组内正确的排名，并且在结果方面遇到了问题。我需要根据时间戳获取每次组更改的排名。

例如，使用此表：

create table syntrans (
    transid     number,
    launchtime  timestamp,
    status      varchar2(10)
);

insert into syntrans values ( 1, '19-APR-17 07.34.05.824875 PM','FAIL');
insert into syntrans values (  1, '19-APR-17 07.34.06.828753 PM','FAIL');
insert into syntrans values (  1, '19-APR-17 07.34.08.567579 PM','SUCCESS');
insert into syntrans values (  1, '19-APR-17 08.07.31.731745 PM','SUCCESS');
insert into syntrans values (  1, '19-APR-17 08.07.32.735582 PM','SUCCESS');
insert into syntrans values (  2, '19-APR-17 08.17.51.332804 PM','FAIL');
insert into syntrans values (  2, '19-APR-17 08.17.52.336530 PM','FAIL');
insert into syntrans values (  2, '19-APR-17 08.19.27.993327 PM','SUCCESS');
insert into syntrans values (  2, '19-APR-17 08.25.54.860077 PM','FAIL');
insert into syntrans values (  2, '19-APR-17 08.25.55.862830 PM','SUCCESS');

我现在得到的是：

SELECT transid,
       launchtime,
       status,
       rank() over (partition by status order by launchtime) rnk
  FROM syntrans 
 order by transid, launchtime, status;

   TRANSID LAUNCHTIME                     STATUS            RNK
---------- ------------------------------ ---------- ----------
         1 19-APR-17 07.34.05.824875 PM   FAIL                1
         1 19-APR-17 07.34.06.828753 PM   FAIL                2
         1 19-APR-17 07.34.08.567579 PM   SUCCESS             1
         1 19-APR-17 08.07.31.731745 PM   SUCCESS             2
         1 19-APR-17 08.07.32.735582 PM   SUCCESS             3
         2 19-APR-17 08.17.51.332804 PM   FAIL                3
         2 19-APR-17 08.17.52.336530 PM   FAIL                4
         2 19-APR-17 08.19.27.993327 PM   SUCCESS             4
         2 19-APR-17 08.25.54.860077 PM   FAIL                5
         2 19-APR-17 08.25.55.862830 PM   SUCCESS             5

需要的输出结果应该像这样：

   TRANSID LAUNCHTIME                     STATUS            RNK
---------- ------------------------------ ---------- ----------
         1 19-APR-17 07.34.05.824875 PM   FAIL                1
         1 19-APR-17 07.34.06.828753 PM   FAIL                2
         1 19-APR-17 07.34.08.567579 PM   SUCCESS             1
         1 19-APR-17 08.07.31.731745 PM   SUCCESS             2
         1 19-APR-17 08.07.32.735582 PM   SUCCESS             3
         2 19-APR-17 08.17.51.332804 PM   FAIL                1
         2 19-APR-17 08.17.52.336530 PM   FAIL                2
         2 19-APR-17 08.19.27.993327 PM   SUCCESS             1
         2 19-APR-17 08.25.54.860077 PM   FAIL                1
         2 19-APR-17 08.25.55.862830 PM   SUCCESS             1

每次状态值改变时（按日期排列），"rank"就会重新开始计数。我理解我得到的输出是按整体状态组排名，但我没有找到任何函数组合可以得到所需的输出。

这个方法几乎能做到，但还不够：

SELECT transid, launchtime, status, rnk
  FROM (SELECT transid,
              status,
              launchtime,
              RANK() OVER (PARTITION BY transid, status ORDER BY launchtime) rnk
         FROM syntrans )
ORDER BY transid, launchtime;

- dlivings

1个回答

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- Vamsi Prabhala · Accepted Answer

你可以使用行号差法将连续相同状态的行分类为一组。（运行内部查询以查看如何分配组。）然后使用 row_number 与这些组。

SELECT transid, launchtime, status
,ROW_NUMBER() over(PARTITION BY transid,grp ORDER BY launchtime) as rnk
FROM (SELECT  transid,
              status,
              launchtime,
              ROW_NUMBER() OVER (PARTITION BY transid ORDER BY launchtime)
              -ROW_NUMBER() OVER (PARTITION BY transid, status ORDER BY launchtime) as grp
         FROM syntrans) t