Assuming you want the part between single quotes, use this regular expression with a Matcher:

"'(.*?)'"

Example:

String mydata = "some string with 'the data i want' inside";
Pattern pattern = Pattern.compile("'(.*?)'");
Matcher matcher = pattern.matcher(mydata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
}

Result:

the data i want

Full example:

private static final Pattern p = Pattern.compile("^([a-zA-Z]+)([0-9]+)(.*)");
public static void main(String[] args) {
    // create matcher for pattern p and given string
    Matcher m = p.matcher("Testing123Testing");

    // if an occurrence if a pattern was found in a given string...
    if (m.find()) {
        // ...then you can use group() methods.
        System.out.println(m.group(0)); // whole matched expression
        System.out.println(m.group(1)); // first expression from round brackets (Testing)
        System.out.println(m.group(2)); // second one (123)
        System.out.println(m.group(3)); // third one (Testing)
    }
}

Since you're looking for the first number, you can use such regexp:

^\D+(\d+).*

and m.group(1) will return you the first number. Note that signed numbers can contain a minus sign:

^\D+(-?\d+).*

The matcher.group() function expects to take a single integer argument: The capturing group index, starting from 1. The index 0 is special, which means "the entire match". A capturing group is created using a pair of parenthesis "(...)". Anything within the parenthesis is captures. Groups are numbered from left to right (again, starting from 1), by opening parenthesis (which means that groups can overlap). Since there are no parenthesis in your regular expression, there can be no group 1.

The javadoc on the Pattern class covers the regular expression syntax.

If you are looking for a pattern that might recur some number of times, you can use Matcher.find() repeatedly until it returns false. Matcher.group(0) once on each iteration will then return what matched that time.

If you want to build one big regular expression that matches everything all at once (which I believe is what you want) then around each of the three sets of things that you want to capture, put a set of capturing parenthesis, use Matcher.match() and then Matcher.group(n) where n is 1, 2 and 3 respectively. Of course Matcher.match() might also return false, in which case the pattern did not match, and you can't retrieve any of the groups.

In your example, what you probably want to do is have it match some preceding text, then start a capturing group, match for digits, end the capturing group, etc...I don't know enough about your exact input format, but here is an example.

Lets say I had strings of the form:

Eat 12 carrots at 12:30
Take 3 pills at 01:15

And I wanted to extract the quantity and times. My regular expression would look something like:

"\w+ (\d+) [\w ]+ (\d{1,2}:\d{2})"

The code would look something like:

Pattern p = Pattern.compile("\\w+ (\\d+) [\\w ]+ (\\d{2}:\\d{2})");
Matcher m = p.matcher(oneline);
if(m.matches()) {
    System.out.println("The quantity is " + m.group(1));
    System.out.println("The time is " + m.group(2));
}

The regular expression means "a string containing a word, a space, one or more digits (which are captured in group 1), a space, a set of words and spaces ending with a space, followed by a time (captured in group 2, and the time assumes that hour is always 0-padded out to 2 digits). I would give a closer example to what you are looking for, but the description of the possible input is a little vague.

The pattern that you tried has parts in it that are not in the example data, and in some parts do not match enough characters.

You could update the pattern to 6 capture groups as:

^([a-zA-Z0-9]+)\s+Confirmed\.on\h+(\d{1,2}/\d{1,2}/\d\d)\h+at\h+(\d{1,2}:\d{1,2})\w*Ksh(\d+(?:\.\d+)?).*?\bfrom\h+(\d+)\h+([^.]+)\.

See a Java demo and a regex demo

final String regex = "^([a-zA-Z0-9]+)\\s+Confirmed\\.on\\h+(\\d{1,2}/\\d{1,2}/\\d\\d)\\h+at\\h+(\\d{1,2}:\\d{1,2})\\w*Ksh(\\d+(?:\\.\\d+)?).*?\\bfrom\\h+(\\d+)\\h+([^.]+)\\.";

final String string2 = "PFEDDTYG0D Confirmed.on 14/6/21 at 12:46PMKsh260.00 received from 254725400049 JOHN DOE. New Account balance is Ksh1,666. Transaction cost, Ksh1";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string2);

if(matcher.find()) {
    String code = matcher.group(1);
    String amountReceived = matcher.group(4);
    String tel = matcher.group(5);
    String from = matcher.group(6);
    String date = matcher.group(2);
    String time = matcher.group(3);

    String format = "code: %s amount received: %s from: %s date: %s time: %s tel: %s";
    System.out.println(String.format(format, code, amountReceived, from, date, time, tel));
}

Output

code: PFEDDTYG0D amount received: 260.00 from: JOHN DOE date: 14/6/21 time: 12:46 tel: 254725400049

It seems to work for me (with this example anyway) if I use this in place of String s = str.replaceAll(regex);

String s = str.replaceAll( ".*§regex=(\\(.*\\)).*", "$1" );

It's just looking for a substring enclosed by parentheses following §regex=.

Use Pattern to compile your regular expression and Matcher to get a particular captured group. The regex I'm using is:

example\((\d+)\)

which captures the digits (\d+) within the parentheses. So:

Pattern p = Pattern.compile("example\\((\\d+)\\)");
Matcher m = p.matcher(text);
if (m.find()) {
  int i = Integer.valueOf(m.group(1));
  ...
}

Here ya go:

^\s*name:\s*([^;]+);

• Start at the beginning of the line.
• Eat as much whitespace as there is.
• Find "name:"
• Eat as much whitespace as there is.
• Capture anything before the next semicolon
• Make sure there's a semicolon