How to have regexp_replace start with a string replacement at column 15 or beyond

This Oracle query: select distinct regexp_replace(My_column,'(-.*$)','')
looks for a hypen and removes the hypen and any text after the hypen.

What I need is to modify this for the search for the hypen to start at character 15.
Any text in characters 1 to 14 needs to be returned as-is.

Example input: abcdef-abcdefg   x-gorgumplatz
should return:  abcdef-abcdefg   x

Any help on this is greatly appreciated.
toddvorosAsked:
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sdstuberCommented:
select regexp_replace('abcdef-abcdefg   x-gorgumplatz','-.*$',null,15) from dual;

I recommend not using () in your expressions unless you must.  Unlike most code where they simply act as grouping, in a regexp, they cause sub-expression evaluation which requires more resources (cpu and memory.)
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sdstuberCommented:
another option,

  REGEXP_SUBSTR(my_column, '^.{1,14}[^-]*')


compare with first option...


SELECT REGEXP_REPLACE(my_column,'-.*$',NULL,15),
       REGEXP_SUBSTR(my_column, '^.{1,14}[^-]*')
  FROM (SELECT 'abcdef-abcdefg   x-gorgumplatz' my_column FROM DUAL);
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awking00Commented:
Without regular expressions, this should work for your sample data -
select replace(col1,substr(col1,instr(col1,'-',-1)))
Note = this removes everything starting with the last hyphen. If there are more than one hyphen after the 15th position, I think you will have to revert to regular expressions.
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sdstuberCommented:
the replace won't work if the text does not have a "-"  character in it


SELECT REPLACE(col1, SUBSTR(col1, INSTR(col1, '-', -1)))
  FROM (SELECT 'abc' col1 FROM DUAL)
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