regex program alteration excluding whitespace

why must it be a an overly complicated regex?

String string = "XPA  DREP EDS";
String[] s = string.split("\\s+");
for( String str: s){
  if ( str.contains("P") ){
     System.out.println( str );
  }
}

you can try and use the \s pattern (match whitespace). Look at this regexp page for java.

Thr reex being ^[^P]P[^P]$

Such a regex finds only string containing exactly one P, which may or may not be what you want. I suppose you want .*P.* instead.

For finding all words containing at least one P you can use \\S+P\\S+, where \S stands for non-blank character. You may consider \w instead.

For finding all words containing exactly one P you can use [^\\sP]+P[^\\sP]+(?=\\s) which is more complicated. Here, \s stands for blank, [^abc] matches everything expect for abc, (?=...) is lookahead. Without the lookahead, you'd find in "APBPC" two "words": "APB" and "PC".

Try adding whitespace characters (\s) in your negated character classes, and you'll also want to remove the ^ and $ anchors:

[^P\s]*P[^P\s]*

or as a Java String literal:

"[^P\\s]*P[^P\\s]*"

Note that the above does not work on Unicode, only ASCII (as tchrist mentioned in the comments).

\b[^P\s]*P[^P\s]*\b

will match all words that contain exactly one P. Don't forget to double the backslashes when constructing your regex from a Java string.

Explanation:

\b      # Assert position at start/end of a word
[^P\s]* # Match any number of characters except P and whitespace
P       # Match a P
[^P\s]* # Match any number of characters except P and whitespace
\b      # Assert position at start/end of a word

Please note that \b doesn't match all word boundaries correctly when dealing with Unicode string (thanks tchrist for reminding me). If that is the case for you, you might want to replace the \bs with (don't look):

(?:(?<=[\pL\pM\p{Nd}\p{Nl}\p{Pc}[\p{InEnclosedAlphanumerics}&&\p{So}]])(?![\pL\pM\p{Nd}\p{Nl}\p{Pc}[\p{InEnclosedAlphanumerics}&&\p{So}]])|(?<![\pL\pM\p{Nd}\p{Nl}\p{Pc}[\p{InEnclosedAlphanumerics}&&\p{So}]])(?=[\pL\pM\p{Nd}\p{Nl}\p{Pc}[\p{InEnclosedAlphanumerics}&&\p{So}]]))

(taken from this question's winning answer)