There's probably a more concise regex, but this will certainly work:

string.replaceAll("[^a-zA-Z0-9]", "");

Try this code:

String str = "a12.334tyz.78x";
str = str.replaceAll("[^\\d.]", "");

Now str will contain "12.334.78".

You can use the String.replaceAll method with a regular expression like this:

input.replaceAll("-?[^\\d]", "");

Adding a . will allow decimal:

input.replaceAll("-?[^\\d.]", "");

Edited to support negative numbers.

Sure, add the additional characters (ie. "-") to keep to the character class of things to keep, which is already created and used.

At the end of the character class the "-" means itself (although it could also be escaped). Thus the matching pattern becomes:

"[^a-zA-Z0-9-]"

(This says, match - to remove - everything that is not an English letter, a decimal digit, or a dash.)

You would use regex something like:

str.replaceAll("[^ABCD]", "");

Should do it

Narrow your regex to only include non-alphabetic characters (and the space, so you can split) instead.

String[] words = str.replaceAll("[^A-za-z ]", "").split("\\s+");

I would say forget using Double (release numbers are not floating point numbers at all) and create your own comparable class called ReleaseNumber:

//details left out for brevity
public class ReleaseNumber implements Comparable<ReleaseNumber> {
    private final int majorVersion;
    private final int minorVersion;
}

You could then use a regular expression to get the fields of release number from each String:

//...
List<String> fileNames /* = ... */;

String latestVersionFileName = Collections.max(fileNames, new Comparator<String>() {
    public int compare(String file1, String file2) {
        return getReleaseNumberFromFileName(file1)
                 .compareTo(getReleaseNumberFromFileName(file2));
    }
}

//...

Pattern releaseNumberPattern = Pattern.compile("(\\d+)\\.(\\d+)");

public ReleaseNumber getReleaseNumberFromFileName(String fileName) {
    Matcher releaseNumberMatcher = releaseNumberPattern.matcher(fileName);
    if (releaseNumberMatcher.find()) {
        int majorVersion = Integer.parseInt(releaseNumberMatcher.group(1));
        int minorVersion = Integer.parseInt(releaseNumberMatcher.group(2));
        ReleaseNumber version = new ReleaseNumber(majorVersion, minorVersion);
        return version;
    }
    throw new RuntimeException();
}

In this case, we just use a simple regular expression \d+\.\d+ which means "one or more decimal digits, followed by a dot, followed by one or more decimal digits" and we search the file contents for the first occurrence of that pattern.

Could you try this one?

System.out.println("ja.v_,a".replaceAll("[^a-zA-Z]", "")) //java

You can make use of the ^. It considers everything apart from what you have infront of it.

So if you have [^y] its going to filter everything apart from y. In your case you would do something like

String value = string.replaceAll("[^0-9]","");

where string is a variable holding the actual text!