java regular expression

You don't need regex at all.

Just use substring: yourString.substring(4,7)

Since you do need to use regex, you can do it like this:

Pattern pattern = Pattern.compile(".{4}(.{3}).*");
Matcher matcher = pattern.matcher("testXXXtest");
matcher.matches();
String whatYouNeed = matcher.group(1);

What does it mean, step by step:

.{4} - any four characters

( - start capturing group, i.e. what you need

.{3} - any three characters

) - end capturing group, you got it now

.* followed by 0 or more arbitrary characters.

matcher.group(1) - get the 1st (only) capturing group.

You should be able to use the substring() method to accomplish this:

string example = "testXXXtest";
string result = example.substring(4,7);

This might help: Groups and capturing in java.util.regex.Pattern.

Here is an example:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class Example {
  public static void main(String[] args) {
    String text = "This is a testWithSomeDataInBetweentest.";
    Pattern p = Pattern.compile("test([A-Za-z0-9]*)test");
    Matcher m = p.matcher(text);
    if (m.find()) {
      System.out.println("Matched: " + m.group(1));
    } else {
      System.out.println("No match.");
    }
  }
}

This prints:

Matched: WithSomeDataInBetween

If you don't want to match the entire pattern rather to the input string (rather than to seek a substring that would match), you can use matches() instead of find(). You can continue searching for more matching substrings with subsequent calls with find().

Also, your question did not specify what are admissible characters and length of the string between two "test" strings. I assumed any length is OK including zero and that we seek a substring composed of small and capital letters as well as digits.

You can use substring for this, you don't need a regex.

yourString.substring(4,7);

I'm sure you could use a regex too, but why if you don't need it. Of course you should protect this code against null and strings that are too short.

Use the String.replaceAll() Class Method

If you don't need to be performance optimized, you can try the String.replaceAll() class method for a cleaner option:

String sDataLine = "testXXXtest";
String sWhatYouNeed = sDataLine.replaceAll( ".{4}(.{3}).*", "$1" );

References

• https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html
• http://www.vogella.com/tutorials/JavaRegularExpressions/article.html#using-regular-expressions-with-string-methods