Oracle 按分钟分组

Question

我编写了一个查询来获取一天中每小时收到的数据。

SELECT To_CHAR( A.req_start_time , 'DD/MM/YYYY HH24') as input , count(A.REQUEST_ID)
FROM ILBULK.SAS_RE_TASK_MESSAGE A,ILBULK.SAS_RE_REQUEST_MESSAGE
WHERE A.NE_TYPE = 'HLR'
LIKE '%Synchronous%'
AND A.REQUEST_ID = ILBULK.SAS_RE_REQUEST_MESSAGE.REQUEST_ID and A.REQ_START_TIME > to_DATE ('12/26/2014 00', 'MM/DD/YYYY HH24') and A.REQ_START_TIME < to_DATE('12/27/2014 00', 'MM/DD/YYYY HH24')
GROUP BY To_CHAR(A.REQ_START_TIME, 'DD/MM/YYYY HH24');

我收到以下回复

26/12/2014 02   13823
26/12/2014 14   4681
26/12/2014 12   2939
26/12/2014 18   457
26/12/2014 03   34327
26/12/2014 04   15673
26/12/2014 19   28885
26/12/2014 06   70699
26/12/2014 10   10743

现在我想获取按升序排列的每分钟的数据，我尝试分割时间但没有任何效果。我怎么做？

Answer 1

使用to_char(A.req_start_time , 'DD/MM/YYYY HH24:MI')，其中MI是分钟部分。将相同的内容添加到 asSELECT子句中GROUP BY。

SELECT To_char(A.req_start_time, 'DD/MM/YYYY HH24:MI') AS input, 
       Count(A.request_id) 
FROM   ilbulk.sas_re_task_message A, 
       ilbulk.sas_re_request_message 
WHERE  A.ne_type = 'HLR' LIKE '%Synchronous%' 
       AND A.request_id = ilbulk.sas_re_request_message.request_id 
       AND A.req_start_time > To_date ('12/26/2014 00', 'MM/DD/YYYY HH24') 
       AND A.req_start_time < To_date('12/27/2014 00', 'MM/DD/YYYY HH24') 
GROUP  BY To_char(A.req_start_time, 'DD/MM/YYYY HH24:MI'); 

例如，

SQL> SELECT to_char(SYSDATE, 'DD/MM/YYYY HH24:MI') GRP_MIN
  2  FROM DUAL
  3  GROUP BY to_char(SYSDATE, 'DD/MM/YYYY HH24:MI')
  4  /

GRP_MIN
----------------
29/12/2014 11:50

SQL>