How to abstract a sub-string fro ma string by using Regular Expression?

Try the following:

s = s.replaceAll(".*\\d{2]\\.\\d{2}\\.\\d{4}.*", "$1");

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Oops sorry

  	
 
s = s.replaceAll(".*(\\d{2]\\.\\d{2}\\.\\d{4}).*", "$1");

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Yes regex is what you need

String input = "ÖÄÖÜ Stand:¶R **üü 12.02.2009ÿÂ";
Pattern pattern = Pattern.compile(".*([0-9]{2}\\.[0-9]{2}\\.[0-9]{4}).*");
Matcher matcher = pattern.matcher(input);
matcher.find();
System.out.println(matcher.group(1));

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I do as follow:

      private static final String STAND = "ÖÄÖÜ   Stand:¶R **üü 12.02.2009ÿÂ";
      /**
       * @param args
       */
      public static void main(String[] args) {

            String subStr = STAND.replaceAll(".*\\d{2]\\.\\d{2}\\.\\d{4}.*", "$1");
            System.out.println(subStr);
      }


But I get exception as follow:

Exception in thread "main" java.util.regex.PatternSyntaxException: Unclosed counted closure near index 6
.*\d{2]\.\d{2}\.\d{4}.*
      ^

So I correct your regular-expression as follow:
".*\\d{2}\\.\\d{2}\\.\\d{4}.*", "$1"

But I get exception as follow:

Exception in thread "main" java.lang.IndexOutOfBoundsException: No group 1
      at java.util.regex.Matcher.group(Matcher.java:463)
      at java.util.regex.Matcher.appendReplacement(Matcher.java:730)
      at java.util.regex.Matcher.replaceAll(Matcher.java:806)
      at java.lang.String.replaceAll(String.java:2000)
      at com.weichen.app.alltest.FilterSubString.main(FilterSubString.java:11)

See my correction

I think the correct expression should be:
replaceAll(".*(\\d{2}\\.\\d{2}\\.\\d{4}).*", "$1");

The first ']' should be '}'. :-)

Many thanks!

So sorry chenwei - there was still a typo in there, so

  	
 
String date = s.replaceAll(".*(\\d{2}\\.\\d{2}\\.\\d{4}).*", "$1");

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Simply wonderful.

:-)