Have done a lot of search before asking this question, like How to convert a string into date format, How to convert a string into date, but still can't figure it out.
So, here's the question, how to convert these string dates into date in Oracle:
"2016-08-15 10:45:30" (String type) -> 20160815 (Date type)
"20160815104530" (String type) -> 20160815 (Date type)
Any idea will be appreciated.
You are mixing two things here:
The first is the conversion of a String data type, in Oracle VARCHAR2 into a DATE data type.
The DATE data type has a precision of seconds, you can't change that. A DATE data type will always give you the date and time component, i.e year, month, day, hours, minutes and seconds: Oracle SQL Data Type Documentation
However, the second part of what you are asking is about how to format the date when retrieved. This is helpful when running reports, or other kinds of visual display of dates. For example, in the US you would most likely want your date columns appear in the format MM/DD/YYYY while everywhere else in the world you most likely want to stick with DD/MM/YYYY. Oracle lets you do that by telling it what NLS_DATE_FORMAT you want to use. You can set that parameter for each individual session as well as on database level, it is up to you (and your DBA) to decide where and when you want to set that. In your case you can apply this via the ALTER SESSION command:
SQL> ALTER SESSION SET nls_date_format='YYYY-MM-DD';
Session altered.
SQL> SELECT TO_DATE('2016-08-15 10:45:30', 'YYYY-MM-DD HH24:MI:SS') FROM DUAL;
TO_DATE(
----------
2016-08-15
SQL> SELECT TO_DATE('20160815104530', 'YYYYMMDDHH24MISS') FROM DUAL;
TO_DATE(
----------
2016-08-15
You use to_date():
select to_date(substr(str1, 1, 10), 'YYYY-MM-DD')
select to_date(substr(str2, 1, 8), 'YYYYMMDD')
2016-08-15 00:00:00 when execute select to_date(substr(str2, 1, 8), 'YYYYMMDD'), how can I just get the date according to the format of YYYYMMDD?
to_char(). In Oracle, the date type includes the time as well.