Java regex: match expression delimited by whitespaces -

phone number like:

+38097-12-34-123 +380971234-123 380971234-123 

i should match phone numbers in string assuming separated whitespaces. not matched strings:

45+38097-12-34-123 +38097-12-34-123=test number:38097-12-34-123 

working regex phone number without whitespace delimiter requirements is:

\\+?380\\d\\d(?:-?\\d){7} 

i try tests \b / \b boundary matcher don't work if leading + sign used.

update fixed examples because type memory , forget put 8 sign.

update2 @washington guedes regex (i strip \ \-):

\b(\+?\d{4}(?:-\d{2}-|\d{2})\d{2}-\d{3})\b 

doesn't work because when + in matching string:

bash# groovysh  groovy:000> "45 +8097-12-34-123".matches(".*\\b(\\+?\\d{4}(?:-\\d{2}-|\\d{2})\\d{2}-\\d{3})\\b.*") ===> true groovy:000> "45+8097-12-34-123".matches(".*\\b(\\+?\\d{4}(?:-\\d{2}-|\\d{2})\\d{2}-\\d{3})\\b.*") ===> true 

i expect expession without whitepsace before + sign should fail.

@stribizhev same suggestion:

groovy:000> "45+38097-12-34-123".matches(".*(?<!\\w)\\+?380\\d\\d(?:-?\\d){7}(?!\\w).*") ===> true groovy:000> "45 +38097-12-34-123".matches(".*(?<!\\w)\\+?380\\d\\d(?:-?\\d){7}(?!\\w).*") ===> true 

the main difficulty have making word boundary work before optional subpattern \+?. trailing word boundary, can use either \b or (?!\w) (which preferable when context unknown, here, have obligatory digit, so, \b enough).

due backtracking, \+? "gobbled" , lookbehind (?<!\w) checks if first digit not preceded non-word character. since in 45+38097-12-34-123, 3 preceded non-word +, matched.

a solution can using alternative in lookahead:

string re = ".*\\+?(?<!\\w|\\w\\+)380\\d\\d(?:-?\\d){7}\\b.*";  string s = "45+38097-12-34-123";  system.out.println(s.matches(re)); 

see ideone demo

see regex demo input examples.