Question

我有这个SQL代码：

select count(req_id), SUBMITS from requests where result='REJECT' GROUP BY SUBMITS

我得到的输出是：

1    16-JUN-17
1    19-JUN-17
1    16-JUN-17
1    16-JUN-17

而我想：

3    16-JUN-2017
1    19-JUN-2017

Answer 1

您的SUBMITS列中可能有小时，分钟，秒，您的工具没有向您显示;这样，您认为所有行都具有相同的SUBMITS值，但它们只有相同的日期，但时间不同。

如果是这样，您只需要将信息限制为日期，而不是小时，分钟和秒：

select count(req_id), trunc(SUBMITS)
from requests
where result='REJECT'
GROUP BY trunc(SUBMITS)

例如：

SQL> create table requests (submits date, result VARCHAR2(10), req_id NUMBER);

Table created.

SQL> insert into requests (req_id, submits, result) values (1, sysdate, 'REJECT');

1 row created.

SQL> insert into requests (req_id, submits, result) values (2, sysdate, 'REJECT');

1 row created.

SQL> insert into requests (req_id, submits, result) values (3, sysdate, 'REJECT');

1 row created.    

SQL> select count(req_id), trunc(SUBMITS)
  2  from requests
  3  where result='REJECT' 
  4  GROUP BY trunc(SUBMITS);

COUNT(REQ_ID) TRUNC(SUB
------------- ---------
            3 21-JUN-17

SQL> select count(req_id), SUBMITS
  2  from requests
  3  where result='REJECT'
  4  GROUP BY SUBMITS;

COUNT(REQ_ID) SUBMITS
------------- ---------
            1 21-JUN-17
            1 21-JUN-17
            1 21-JUN-17

SQL：分组显示相同的值？

1 个答案: