java regex to accept any word other than none

Question

I need a regular expression to match any string other than none. I tried using regular exp ="^[^none]$", But it does not work.

Answer 1

You can use the regular expression (?!^none$).*. See this question for details: Regex inverse matching on specific string?

The reason "^[^none]$" doesn't work is that you are actually matching all strings except the strings "n", "o", or "e".

Of course, it would be easier to just use String.equals like so: !"none".equals(testString).

Answer 2

If you are matching a String against a specific word in Java you should use equals(). In this case you want to invert the match so your logic becomes:

if(!theString.equals("none")) {
  // do stuff here
}

Much less resource hungry, and much more intuitive.

If you need to match a String which contains the word "none", you are probably looking for something like:

if(theString.matches("\\bnone\\b")) {
  /* matches theString if the substring "none" is enclosed between 
   * “word boundaries”, so it will not match for example: "nonetheless"
   */
}

Or if you can be fairly certain that “word boundaries” mean a specific delimiter you can still evade regular expressions by using the indexOf() method:

int i = theString.indexOf("none");
if(i > -1) {
    if(i > 0) {
       // check theString.charAt(i - 1) to see if it is a word boundary 
       // e.g.: whitespace
    }
    // the 4 is because of the fact that "none" is 4 characters long.
    if((theString.length() - i - 4) > 0) {
       // check theString.charAt(i + 4) to see if it is a word boundary 
       // e.g.: whitespace
    }
}
else {
    // not found.
}

Answer 3

Actually this is the regex to match all words except "word":

Pattern regex = Pattern.compile("\\b(?!word\\b)\\w+\\b");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
    // matched text: regexMatcher.group()
    // match start: regexMatcher.start()
    // match end: regexMatcher.end()
}

You must use word boundaries so that "word" is not contained in other words.

Explanation:

"
\b          # Assert position at a word boundary
(?!         # Assert that it is impossible to match the regex below starting at this position (negative lookahead)
   Lorem       # Match the characters “Lorem” literally
   \b          # Assert position at a word boundary
)
\w          # Match a single character that is a “word character” (letters, digits, etc.)
   +           # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\b          # Assert position at a word boundary
"

Answer 4

This is the regex you are looking for:

Pattern p = Pattern.compile("^(?!none$).*$");
Matcher m = p.matcher("your string");
System.out.println(s + ": " + (m.matches() ? "Match" : "NO Match"));

Having that said, if you are not forced to use a regex that matches everything but "none", the more simple, fast, clear, and easy to write and understand is this:

Pattern p = Pattern.compile("^none$");

Then, you just exclude the matches.

Matcher m = p.matcher("your string");
System.out.println(s + ": " + (m.matches() ? "NO Match" : "Match"));