replacing square brackets of a string with oracle REGEXP_REPLACE function

Question

I want to replace square brackets in string with REGEXP_REPLACE function. Even I escape these chracters it\'s not replacing

select regexp_replace(\'VMI[[DATA         


        

Answer 1

To explain what is happening. Your regular expression [\[\]] is matching:

• A bracket expression [\[\] which matches either a \ character or a [ character or a \ character (since \ is not an escape character in Oracle's regular expression syntax but is treated as a character literal).
• Followed by a ] character.

So your regular expression would match a sub-string that was either \] or [].

---

From the Oracle 12c Documentation:

Bracket expression for specifying a matching list that should match any one of the expressions represented in the list. A non-matching list expression begins with a circumflex (^) and specifies a list that matches any character except for the expressions represented in the list.

To specify a right bracket (]) in the bracket expression, place it first in the list (after the initial circumflex (^), if any).

To specify a hyphen in the bracket expression, place it first in the list (after the initial circumflex (^), if any), last in the list, or as an ending range point in a range expression.

So, if you want to match a closing square bracket in a bracket expression then it needs to be the first character in the list and your regular expression should be [][]. The first [ starts the bracket expression; the ] second character is to match a closing square bracket character; the [ third character matches an opening square bracket character; and the final ] terminates the bracket expression.

Which would give the solution as:

SELECT REGEXP_REPLACE(
         'VMI[[DATA]]INFO',
         '[][]',
         '_'
       )
FROM   DUAL;