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Regular expression to determine if the String consists of more than 4 numbers

I want to extract URL strings from a log which looks like below:
<13>Mar 27 11:22:38 144.0.116.31 AgentDevice=WindowsDNS AgentLogFile=DNS.log PluginVersion=X.X.X.X Date=3/27/2019 Time=11:22:34 AM Thread ID=11BC Context=PACKET Message= Internal packet identifier=0000007A4843E100 UDP/TCP indicator=UDP Send/Receive indicator=Snd Remote IP=X.X.X.X Xid (hex)=9b01 Query/Response=R Opcode=Q Flags (hex)=8081 Flags (char codes)=DR ResponseCode=NOERROR Question Type=A Question Name=outlook.office365.com
I am looking to extract Name text which contains more that 5 digits.
A possible way suggested is (\d.*?){5,} but does not seem to work, kindly suggest another way get the field.
Example of string match:
outlook12.office345.com
outlook.office12345.com

You can look for the following expression:
Name=([^ ]*\d{5,}[^ ]*)
Explanation:
Name= look for anything that starts with "Name=", than capture if:
[^ ]* any number of characters which is not a space
\d{5,} then 5 digits in a row
[^ ]* then again, all digits up to a white space

This regular expression:
(?<=Name=).*\d{5,}.*?(?=\s|$)
would extract strings like outlook.office365666.com (with 5 or more consecutive digits) from your example input.
Demo: https://regex101.com/r/YQ5l2w/1

Try this pattern: (?=\b.*(?:\d[^\d\s]*){5,})\S*
Explanation:
(?=...) - positive lookahead, assures that pattern inside it is matched somewhere ahead :)
\b - word boundary
(?:...) - non-capturing group
\d[^\d\s]* - match digit \d, then match zero or more of any characters other than whitespace \s or digit \d
{5,} - match preceeding pattern 5 or more times
\S* - match zero or more of any characters other than space to match the string if assertion is true, but I think you just need assertion :)
Demo
If you want only consecutive numbers use simplified pattern (?=\b.*\d{5,})\S*.
Another demo
Of course, you have to add positive lookbehind: (?<=Name=) to assert that you have Name= string preceeding

Try this regex
([a-z0-9]{5,}.[a-z0-9]{5,})+.com
https://regex101.com/r/OzsChv/3
It Groups,
outlook.office365.com
outlook12.office345.com
also all url strings

Extract exactly n digits in a sentence using REGEX

Example
The no.s 1234 65
Input: n
For n=4, the output should be 1234
For n=2, the output should be : 65 (not 12)
Tried \d{n} which gives 12 and \d{n,} gives 1234 but i want the exact matching one.
Pattern p = Pattern.compile("//\d{n,}");

you need negative lookaround assertion: (?<!..): negative look behind, and (?!..): negative look ahead : regex101
(?<!\d)\d{4}(?!\d)
however not all regex engine supports them, maybe a work around may match also the preceeding character and following character (contrary to look-around which are 0 width matches), (\D matches all excpet a digit)
(?:^|\D)(\d{4})(?:\D|$)

I think what you meant is the \b character.
Hence, the regex you're looking for would be (for n=2):
\b\d{2}\b

From what I understand, you're looking for a regex that will match a number in a string which has n digits, taking into into account the spacing between the numbers. If that's the case, you're looking for something like this:
\b\d{4}\b
The \b will ensure the match is constrained to the start/end of a 'word' where a word is the boundary between anything matched by \w (which includes digits) and anything matched by the opposite, \W (which includes spaces).

I don't code in java but I can try to answer this using regex in general.
If your number is in the format d1d2d3d4 d5d6 and you want to extract digits d5d6, create 3 groups as r'([0-9]+)("/s")([0-9]+)' – each set of parenthesis () represent one group. Now, extract the third group only in another object which is your required output.

Simple Regex phantom matches? [duplicate]

I try to extract the error number from strings like "Wrong parameters - Error 1356":
Pattern p = Pattern.compile("(\\d*)");
Matcher m = p.matcher(myString);
m.find();
System.out.println(m.group(1));
And this does not print anything, that became strange for me as the * means * - Matches the preceding element zero or more times from Wiki
I also went to the www.regexr.com and regex101.com and test it and the result was the same, nothing for this expression \d*
Then I start to test some different things (all tests made on the sites I mentioned):
(\d)* doesn't work
\d{0,} doesn't work
[\d]* doesn't work
[0-9]* doesn't work
\d{4} works
\d+ works
(\d+) works
[0-9]+ works
So, I start to search on the web if I could find an explanation for this. The best I could find was here on the Quantifier section, which states:
\d? Optional digit (one or none).
\d* Eat as many digits as possible (but none if necessary)
\d+ Eat as many digits as possible, but at least one.
\d*? Eat as few digits as necessary (possibly none) to return a match.
\d+? Eat as few digits as necessary (but at least one) to return a match.
The question
As english is not my primary language I'm having trouble to understand the difference (mainly the (but none if necessary) part). So could you Regex expert guys explain this in simple words please?
The closest thing that I find to this question here on SO was this one: Regex: possessive quantifier for the star repetition operator, i.e. \d** but here it is not explained the difference.

The * quantifier matches zero or more occurences.
In practice, this means that
\d*
will match every possible input, including the empty string. So your regex matches at the start of the input string and returns the empty string.

but none if necessary means that it will not break the regex pattern if there is no match. So \d* means it will match zero or more occurrences of digits.
For eg.
\d*[a-z]*
will match
abcdef
but \d+[a-z]*
will not match
abcdef
because \d+ implies that at least one digit is required.

\d* Eat as many digits as possible (but none if necessary)
\d* means it matches a digit zero or more times. In your input, it matches the least possible one (ie, zero times of the digit). So it prints none.
\d+
It matches a digit one or more times. So it should find and match a digit or a digit followed by more digits.

With the pattern /d+ at least one digit will need to be reached, and then the match will return all subsequent characters until a non-digit character is reached.
/d* will match all the empty strings (zero or more), as well at the match. The .Net Regex parser will return all these empty string groups in its set of matches.

Simply:
\d* implies zero or more times
\d+ means one or more times

Java specific regex

I want to write some code that checks whether the input from user matches a specific format. The format is 's' followed by a space and a number, if the input was "s 1 2 3 4", it would call some function. "s 1 2" would also be acceptable. So far I found that this regex works for a specified amount of times:
if (inputLine.matches("s \\d+ \\d+")) { }
works for 2 numbers after the s, but I need to be able to accept any number of numbers after the s.
Any idea on a regex that would suit my needs? Thank you

Change your regex to
if (inputLine.matches("s(?: \\d+)+")) { }
to match s, space and 1+ sequences of a space followed with 1+ digits.
If you allow 0 numbers after s, replace the last + quantifier with * to match zero or more occurrences.
Since the repeated capturing groups overwrite the group contents, it makes no sense using a capturing group here, thus, I suggest using a non-capturing one, (?:...).

Regex in Java: Capture last {n} words

Hi I am trying to do regex in java, I need to capture the last {n} words. (There may be a variable num of whitespaces between words). Requirement is it has to be done in regex.
So e.g. in
The man is very tall.
For n = 2, I need to capture
very tall.
So I tried
(\S*\s*){2}$
But this does not match in java because the initial words have to be consumed first. So I tried
^(.*)(\S*\s*){2}$
But .* consumes everything, and the last 2 words are ignored.
I have also tried
^\S?\s?(\S*\s*){2}$
Anyone know a way around this please?

You had almost got it in your first attempt.
Just change + to *.
The plus sign means at least one character, because there wasn't any space the match had failed.
On the other hand the asterisk means from zero to more, so it will work.
Look it live here: (?:\S*\s*){2}$
Using replaceAll method, you could try this regex: ((?:\\S*\\s*){2}$)|.

Your regex contains - as you already mention - a greedy subpattern that eats up the whole string and sine (\S*\s*){2} can match an empty string, it matches an empty location at the end of the input string.
Lazy dot matching (changing .* to .*?) won't do the whole job since the capturing group is quantified, and the Matcher.group(1) will be set to the last captured non-whitespaces with optional whitespaces. You need to set the capturing group around the quantified group.
Since you most likely are using Matcher#matches, you can use
String str = "The man is very tall.";
Pattern ptrn = Pattern.compile("(.*?)((?:\\S*\\s*){2})"); // no need for `^`/`$` with matches()
Matcher matcher = ptrn.matcher(str);
if (matcher.matches()) { // Group 2 contains the last 2 "words"
System.out.println(matcher.group(2)); // => very tall.
}
See IDEONE demo