Use format pattern from regexp_replace in to_char - sql

Question

I have a SQL stored in a column where the format for date is in square brackets:
sql_column
------------------------------------------------------------------------------
select col1 from table1 where col2 = to_date('[DD-MON-YYYY]', 'DD-MON-YYYY');
select col1 from table2 where col2 between [YYYYMMDD]00000 and [YYYYMMDD]99999 and col3 = to_date('[DD-MON-YYYY]', 'DD-MON-YYYY');
....
I don't know which format can be there beforehand, but it is always a date format.
Is there a way to use regexp_replace (or regexp_substr or even with regexp_* functions) to find and replace the pattern with the result of to_char of my given date and pattern taken from the db column.
Perhaps something like this (which doesn't work obviously):
select sql_column,
regexp_replace(sql_column, '\[(.+?)\]', to_char(some_date, '\1'))
from my_table;
Could you please help?

Answer 1

There is no way to parse ALL the possible date formats. For example, if you see '01/11/2017' in a varchar field you cannot say if it refers to November 1st or January 11th.
Said that, you can use a common table expression to choose the best pattern and then use it to convert the string value to date. For example:
select * from
(select
case
when REGEXP_LIKE(col2, '^[0-9]{2}/[0-9]{2}/[0-9]{4}$') then 'DD/MM/YYYY'
when REGEXP_LIKE(col2, '^[0-9]{14}$') then 'YYYYMMDDHH24MISS'
end pattern, table1.*
from table1) x
where to_date(x.col2, x.pattern) = to_date('01/11/2017','DD/MM/YYYY')
This approach would probably lead to a full table scan so it is far from being efficient. If the original table is large, I advice creating a materialized view with the converted date to improve performance.

Answer 2

You'll need a bit of dynamic SQL to solve this problem.
I do not present a complete solution, but this should give you a hint how to approach it.
Let's approach it bottom up.
What you actually need is a REPLACE statement, that transforms you SQL text in the required form with the paramater date.
This could be for your example this statement for teh first example (parameter date is 30.11.2017)
select replace(sql_column, '[DD-MON-YYYY]','30-NOV-2017') from my_table;
If you run the first statement on your first row you get the expected result:
select col1 from table1 where col2 = to_date('30-NOV-2017', 'DD-MON-YYYY');
So how to get those REPLACE statements. One possibility is to write a PL/SQL function.
The function has two parameters - the original SQL string and teh parameter date.
Using regexp you scrap the date format mask.
With dynamic SQL (EXECUTE IMMEDIATE) you format you parameter DATE as string with propper format.
Finally the REPLACE statement is returned.
create or replace function format_date(i_txt IN VARCHAR2, i_date DATE) return VARCHAR2 is
v_date_format VARCHAR2(4000);
v_form_date VARCHAR2(4000);
v_param VARCHAR2(4000);
v_form VARCHAR2(4000);
v_sql VARCHAR2(4000);
BEGIN
v_param := regexp_substr(i_txt, '\[(.+?)\]');
v_date_format := replace(replace(v_param,'[',null),']',null);
v_sql := 'select to_char(:d,'''||v_date_format||''') as my_dt from dual';
execute immediate v_sql into v_form_date using i_date;
v_form := 'select replace(sql_column, '''||v_param||''','''||v_form_date||''') from my_table';
return (v_form);
END;
/
NOTE that I handle only the first date mask in the string, you'll need to loop on all occcureces to get correct the second example!