Sort one column , based on that sort another column in Oracle

oracle

I have a table :

ID     COUNTER     Date
----   -------  -----------
 1        05    23-07-2019
 2        01    23-06-2017
 3        10    02-06-2019
 4        14    01-12-2016
 5        03    23-11-2018
 6        07    23-11-2015

Now I want to Sort first Date column , based on that sort result sort counter .How to write a query for this. If I have a more than one date for a particular year than , sort according to counter for that particular year records.
Like this :

  COUNTER      Date
  -------   ----------- 
     10       02-06-2019   
     05       23-07-2019     

     03       23-11-2018
     01       23-06-2017
     14       01-12-2016
     07       23-11-2015

Any help would be appreciated!.
Thanks.

Best Answer

order by to_char(date_column, 'YYYY') desc, counter desc

order by trunc(date_column, 'YYYY') desc, counter desc

drop table t1 purge;
create table t1 (counter varchar2(2), date_column date);
insert into t1 values ('05', date'2019-07-23');
insert into t1 values ('01', date'2017-06-23');
insert into t1 values ('10', date'2019-06-02');
insert into t1 values ('14', date'2016-12-01');
insert into t1 values ('03', date'2018-11-23');
insert into t1 values ('07', date'2015-11-23');
commit;

select * from t1 order by trunc(date_column, 'YYYY') desc, counter desc;

CO DATE_COLUM
-- ----------
10 02-06-2019
05 23-07-2019
03 23-11-2018
01 23-06-2017
14 01-12-2016
07 23-11-2015
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