CODEWITHSUNDEEP
javaregexsubstring
d-man
d-man

Reputation: 58083

Java Regex to find substring

I have following string "Class (102) (401)" and "Class (401)" i want to find regex to find substring which always return me last bracket value in my case it is '(401) '

Following is my code

Pattern MY_PATTERN = Pattern.compile(".*(\\(\\d+\\))");
    Matcher mat = MY_PATTERN.matcher("Class (102) (401)");
    while (mat.find()){
        System.out.println(mat.group());
    }

It is returning

--( --) --( --)

Upvotes: 3

Views: 774

Answers (3)

Hachi
Hachi

Reputation: 3289

how about the expression: .*\\(([^\\(\\)]+)\\)[^\\(\\)]*$

it finds a ( followed by non-brackets [^\\(\\)] (your desired string) followed by a ) and after that are only non-brackets allowed, so it must be the last

Upvotes: 1

Cylian
Cylian

Reputation: 11182

Try this:

(?<=\()[^\)(]+(?=\)[^\)\(]+$)

Explanation:

<!--
(?<=\()[^\)(]+(?=\)[^\)\(]+$)

Options: ^ and $ match at line breaks; free-spacing

Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=\()»
   Match the character “(” literally «\(»
Match a single character NOT present in the list below «[^\)(]+»
   Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
   A ) character «\)»
   The character “(” «(»
Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=\)[^\)\(]+$)»
   Match the character “)” literally «\)»
   Match a single character NOT present in the list below «[^\)\(]+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
      A ) character «\)»
      A ( character «\(»
   Assert position at the end of a line (at the end of the string or before a line break character) «$»
-->

Upvotes: 1

codaddict
codaddict

Reputation: 455312

You can use:

Pattern MY_PATTERN = Pattern.compile(".*(\\(\\d+\\))");

See it

Upvotes: 2

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