using java regex how can I replace all when pattern contains left parenthesis

Question

I have a string that looks like this: mynum(1234) and mynum( 123) and mynum ( 12345 ) and lastly mynum(#123)

I want to insert a # in front of the numbers in parenthesis so I have: mynum(#1234) and mynum( #123) and mynum ( #12345 ) and lastly mynum(#123)

How can I do this? Using regex pattern matcher and a replaceAll chokes on the ( in front of the number and I get an

java.util.regex.PatternSyntaxException: Unclosed group near ...

exception.

Bart Kiers · Accepted Answer · 2009-10-09 18:36:48Z

4

Try:

String text = "mynum(1234) and mynum( 123) and foo(123) mynum ( 12345 ) and lastly mynum(#123)";
System.out.println(text.replaceAll("mynum\\s*\\((?!\\s*#)", "$0#"));

A small explanation:

Replace every pattern:

mynum   // match 'mynum'
\s*     // match zero or more white space characters
\(      // match a '('
(?!     // start negative look ahead
  \s*   //   match zero or more white space characters
  #     //   match a '#'
)       // stop negative look ahead

with the sub-string:

$0#

Where $0 holds the text that is matched by the entire regex.

edited Oct 9, 2009 at 18:36

answered Oct 9, 2009 at 17:59

Bart Kiers

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2 Comments

KT. Over a year ago

Awesome!!! That worked. Also, thank you for the explanation. That was very helpful. Now I just need to do some research to understand what is "negative look ahead"

Bart Kiers Over a year ago

@KT: have a look at this tutorial: regular-expressions.info/lookaround.html

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