我的访问表如下:
insert into Visit
values(12, to_date('19-JUN-13', 'dd-mon-yy'), to_date('19-JUN-13 12:00 A.M.' , 'dd-mon-yy hh:mi A.M.'));
insert into Visit
values(15, to_date('20-JUN-13', 'dd-mon-yy'), to_date('20-JUN-13 02:00 A.M.' , 'dd-mon-yy hh:mi A.M.'));
insert into Visit
values(18, to_date('21-JUN-13', 'dd-mon-yy'), to_date('21-JUN-13 10:30 A.M.' , 'dd-mon-yy hh:mi A.M.'));
当我尝试查询它时:
select * from Visit
我得到:SQL> select * from visit;
SLOTNUM DATEVISIT ACTUALARRIVALTIME
---------- --------- ------------------------------
12 19-JUN-13 19-JUN-13
15 20-JUN-13 20-JUN-13
18 21-JUN-13 21-JUN-13
SQL> spool off;
怎么没时间?
最佳答案
您还可以设置适用于所有日期的格式,如下所示:
ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-YYYY HH:MI:SS PM';
这样,您的原始查询将以您所追求的格式输出日期,而无需使用
TO_CHAR
.要设置回通常的默认格式,只需执行以下操作:ALTER SESSION SET NLS_DATE_FORMAT = 'DD-MON-RR';
关于甲骨文。如何输出日期和时间?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17223226/