I have a string
string 1(excluding the quotes) -> "my car number is #8746253 which is actually cool"
conditions - The number 8746253, could be of any length and
- the number can also be immediately followed by an end-of-line.
I want to group-out 8746253 which should not be followed by a dot "."
I have tried,
.*#(\d+)[^.].*
This will get me the number for sure, but this will match even if there is a dot, because [.^] will match the last digit of the number(for example, 3 in the below case)
string 2(excluding the quotes) -> "earth is #8746253.Kms away, which is very far"
I want to match only the string 1 type and not the string 2 types.
To match any number of digits after #
that are not followed with a dot, use
(?<=#)\d++(?!\.)
The ++
is a possessive quantifier that will make the regex engine only check the lookahead (?!\.)
only after the last matched digit, and won't backtrack if there is a dot after that. So, the whole match will get failed if there is a dit after the last digit in a digit chunk.
See the regex demo
To match the whole line and put the digits into capture group #1:
.*#(\d++)(?!\.).*
See this regex demo. Or a version without a lookahead:
^.*#(\d++)(?:[^.\r\n].*)?$
See another demo. In this last version, the digit chunk can only be followed with an optional sequence of a char that is not a .
, CR and LF followed with any 0+ chars other than line break chars ((?:[^.\r\n].*)?
) and then the end of string ($
).
This works like you have described
public class MyRegex{
public static void main(String[] args) {
Pattern patern = Pattern.compile("#(\\d++)[^\\.]");
Matcher matcher1 = patern.matcher("my car number is #8746253 which is actually cool");
if(matcher1.find()){
System.out.println(matcher1.group(1));
}
Matcher matcher2 = patern.matcher("earth is #8746253.Kms away, which is very far");
if(matcher2.find()){
System.out.println(matcher1.group(1));
}else{
System.out.println("No match found");
}
}
}
Outputs:
> 8746253
> No match found
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