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How to write a regex capture group which matches a character 3 or 4 times before a delimiter?

I'm trying to write a regex that splits elements out according to a delimiter. The regex also needs to ensure there are ideally 4, but at least 3 colons : in each match.
Here's an example string:
"Checkers, etc:Blue::C, Backgammon, I say:Green::Pepsi:P, Chess, misc:White:Coke:Florida:A, :::U"
From this, there should be 4 matches:
Checkers, etc:Blue::C
Backgammon, I say:Green::Pepsi:P
Chess, misc:White:Coke:Florida:A
:::U
Here's what I've tried so far:
([^:]*:[^:]*){3,4}(?:, )
Regex 101 at: https://regex101.com/r/O8iacP/8
I tried setting up a non-capturing group for ,
Then I tried matching a group of any character that's not a :, a :, and any character that's not a : 3 or 4 times.
The code I'm using to iterate over these groups is:
String line = "Checkers, etc:Blue::C, Backgammon, I say::Pepsi:P, Chess:White:Coke:Florida:A, :::U";
String pattern = "([^:]*:[^:]*){3,4}(?:, )";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher matcher = r.matcher(line);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Any help is appreciated!
Edit
Using #Casimir's regex, it's working. I had to change the above code to use group(0) like this:
String line = "Checkers, etc:Blue::C, Backgammon, I say::Pepsi:P, Chess:White:Coke:Florida:A, :::U";
String pattern = "(?![\\s,])(?:[^:]*:){3}\\S*(?![^,])";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher matcher = r.matcher(line);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
Now prints:
Checkers, etc:Blue::C
Backgammon, I say::Pepsi:P
Chess:White:Coke:Florida:A
:::U
Thanks again!

I suggest this pattern:
(?![\\s,])(?:[^:]*:){3}\\S*(?![^,])
Negative lookaheads avoid to match leading or trailing delimiters. The second one in particular forces the match to be followed by the delimiter or the end of the string (not followed by a character that isn't a comma).
demo
Note that the pattern doesn't have capture groups, so the result is the whole match (or group 0).

You might use
(?:[^,:]+, )?[^:,]*(?::+[^:,]+)+
(?:[^,:]+, )? Optionally match 1+ any char except a , or : followed by , and space
[^:,]* Match 0+ any char except : or ,
(?: Non Capturing group
:+[^:,]+ Match 1+ : and 1+ times any char except : and ,
)+ Close group and repeat 1+ times
Regex demo

You seem to be making it harder than it needs to be with the lookahead (which won't be satisfied at end-of-line anyway).
([^:]*:){3}[^:,]*:?[^:,]*
Find the first 3 :'s, then start including , in the negative groupings, with an optional 4th :.

What regex should I use to check a string only has numbers and 2 special characters ( - and , ) in Java?

Scenario: I want to check whether string contains only numbers and 2 predefined special characters, a dash and a comma.
My string contains numbers (0 to 9) and 2 special characters: a dash (-) defines a range and a comma (,) defines a sequence.
Tried attempt :
Tried following regex [0-9+-,]+, but not working as expected.
Possible inputs :
1-5
1,5
1-5,6
1,3,5-10
1-5,6-10
1,3,5-7,8,10
The regex should not accept these types of strings:
-----
1--4
,1,5
5,6,
5,4,-
5,6-
-5,6
Please can any one help me to create regex for above scenario?

You may use
^\d+(?:-\d+)?(?:,\d+(?:-\d+)?)*$
See the regex demo
Regex details:
^ - start of string
\d+ - 1 or more digits
(?:-\d+)? - an optional sequence of - and 1+ digits
(?:,\d+(?:-\d+)?)* - zero or more seuqences of:
, - a comma
\d+(?:-\d+)? - same pattern as described above
$ - end of string.

Change your regex [0-9+-,]+ to [0-9,-]+
final String patternStr = "[0-9,-]+";
final Pattern p = Pattern.compile(patternStr);
String data = "1,3,5-7,8,10";
final Matcher m = p.matcher(data);
if (m.matches()) {
System.out.println("SUCCESS");
}else{
System.out.println("ERROR");
}

How to get all integers before hyphen from java String

I want to parse through hyphen, the answer should be 0 0 1 (integer), what could be the best way to parse in java
public static String str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";
Please help me out.

Use the below regex with Pattern and matcher classes.
Pattern.compile("\\d+(?=-)");
\\d+ - Matches one or more digits. + repeats the previous token \\d (which matches a digit character) one or more times.
(?=-) - Only if it's followed by an hyphen. (?=-) Called positive lookahead assertion which asserts that the match must be followed by an - symbol.
String str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";
Matcher m = Pattern.compile("\\d+(?=-)").matcher(str);
while(m.find())
{
System.out.println(m.group());
}

one lazy way: if you already know the pattern of the string, use substring and indexof to locate your word.
String str ="[0-S1|0-S2|1-S3, 1-S1|0-S2|0-S3, 0-S1|1-S2|0-S3]";
integer int1 = Integer.parseInt(str.substring(str.indexOf("["),str.indexOf("-S1")));
and so on.

Java Pattern / Matcher not finding word break

I am having trouble with Java Pattern and Matcher. I've included a very simplified example of what I'm trying to do.
I had expected the pattern ".\b" to find the last character of the first word (or "4" in the example), but as I step through the code, m.find() always returns false. What am I missing here?
Why does the following Java code always print out "Not Found"?
Pattern p = Pattern.compile(".\b");
Matcher m = p.matcher("102939384 is a word");
int ixEndWord = 0;
if (m.find()) {
ixEndWord = m.end();
System.out.println("Found: " + ixEndWord);
} else {
System.out.println("Not Found");
}

You need to escape special characters in the regex: ".\\b"
Basically, in a String the backslash has to be escaped. So "\\" becomes the character '\'.
So the String ".\\b" becomes the litteral String ".\b", which will be used by the Pattern.

To expand upton AntonH's comment, whenever you want the "\" character to appear in a regex expression, you have to escape it so that it first appears in the string you are passing in.
As is, ".\b" is the string of a dot . followed by the special backspace character represented by \b, compared to ".\\b", which is the regex .\b.

How to match just 1 or 2 chars with regex

i want regx to match any word of 2 or 1 characters example ( is , an , or , if, a )
i tried this :-
int scount = 0;
String txt = "hello everyone this is just test aa ";
Pattern p2 = Pattern.compile("\\w{1,2}");
Matcher m2 = p2.matcher(txt);
while (m2.find()) {
scount++;
}
but got wrong matches.

You probably want to use word boundary anchors:
Pattern p2 = Pattern.compile("\\b\\w{1,2}\\b");
These anchors match at the start/end of alphanumeric "words", that is, in positions before a \w character if there is no \w character before that, or after a \w character if there is no \w character after that.

I think that you should be a bit more descriptive. Your current code returns 15 from the variable scount. That's not nothing.
If you want to get a count of the 2 letter words, and that is excluding underscores, digits within this count, I think that you would be better off with negative lookarounds:
Pattern.compile("(?i)(?<![a-z])[a-z]{1,2}(?![a-z])");
With a string input of hello everyone this is just 1 test aa, you get the value of scount as 2 (is and aa) and not 3 (is, 1, aa) as you would have if you were looking for only 1 or 2 consecutive \w.
Also, with hello everyone this is just test aa_, you get a count of 1 with \w (is), but 2 (is, aa)with the lookarounds.