Regex.Match Whole Words

Regex match entire words only

Use word boundaries:

/\b($word)\b/i

Or if you're searching for "S.P.E.C.T.R.E." like in Sinan Ünür's example:

/(?:\W|^)(\Q$word\E)(?:\W|$)/i

regex match whole word and punctuation with it using re.search()

There are two issues here.

1. In regex . is special. It means "match one of any character". However, you are trying to use it to match a regular period. (It will indeed match that, but it will also match everything else.) Instead, to match a period, you need to use the pattern \.. And to change that to match either a period or a hyphen, you can use a class, like [-.].
2. You are using \b at the end of your pattern to match the word boundary, but \b is defined as being the boundary between a word character and a non-word character, and periods and spaces are both non-word characters. This means that Python won't find a match. Instead, you could use a lookahead assertion, which will match whatever character you want, but won't consume the string.

Now, to match a whole word - any word - you can do something like \w+, which matches one or more word characters.

Also, it is quite possible that there won't be a match anyway, so you should check whether a match occurred using an if statement or a try statement. Putting it all together:

txt = "The indian in. Spain."
pattern = r"\w+[-.]"
x = re.search(r"\b" + pattern + r"(?=\W)", txt)
if x:
    print(x.start(), x.end())

Edit

There is one problem with the lookahead assertion above - it won't match the end of the string. This means that if your text is The rain in Spain. then it won't match Spain., as there is no non-word character following the final period.

To fix this, you can use a negative lookahead assertion, which matches when the following text does not include the pattern, and also does not consume the string.

x = re.search(r"\b" + pattern + r"(?!\w)", txt)

This will match when the character after the word is anything other than a word character, including the end of the string.

How to make regex match only whole words and not break the words down?

You could use {2,6} and make sure to use word boundaries\b so that there are not 2 matches, one for ABSTRA and the other for CT

\b[A-Z]{2,6}(?:-[0-9]+)?\b

Regex demo

In python:

regex = r"\b[A-Z]{2,6}(?:-[0-9]+)?\b"

If in thispart -*[0-9]* the hyphen is not optional you could turn it into an optional group (?:-[0-9]+)?

If there should not be anything on the left or right, you could use:

(?<!\S)[A-Z]{2,6}-?[0-9]*(?!\S)

Note that -* will match 0 or more hyphens and -? matches an optional one.

Regex demo

Regex.Match whole words

You should add the word delimiter to your regex:

\b(shoes|shirt|pants)\b

In code:

Regex.Match(content, @"\b(shoes|shirt|pants)\b");

Java Regex : match whole word with word boundary

It appears you only want to match "words" enclosed with whitespace (or at the start/end of strings).

Use

String pattern = "(?<!\\S)" + Pattern.quote(word) + "(?!\\S)";

The (?<!\S) negative lookbehind will fail all matches that are immediately preceded with a char other than a whitespace and (?!\s) is a negative lookahead that will fail all matches that are immediately followed with a char other than whitespace. Pattern.quote() is necessary to escape special chars that need to be treated as literal chars in the regex pattern.

How to match a whole word or sentence after a specific character with regexp

Here are 2 options depending on whether you want to include the colon in the pattern that you are capturing.

• with the colon
  ^:\w*
• with a lookback for the colon
  (?<=^:)\w*
  
  This will match a word after the colon.
  
  You may want any number of any character .* or any combination of word characters and spaces `[\w\s]*

Matching whole words that start or end with special characters

The \b word boundary construct is ambiguous. You need to use unambiguous constructs that will make sure there are non-word chars or start/end of string to the left/right of the word matched.

You may use

/(?:^|\W)\?FOO\?(?!\w)/g

Here, (?:^|\W) is a non-capturing group that matches either the start of a string or any non-word char, a char other than an ASCII letter, digit and _. (?!\w) is a negative lookahead that fails the match if, immediately to the right of the current location, there is a word char.

Or, with ECMAScript 2018 compatible JS environments,

/(?<!\w)\?FOO\?(?!\w)/g

See this regex demo.

The (?<!\w) is a negative lookbehind that fails the match if there is a word char immediately to the left of the current location.

In code, you may use it directly with String#match to extract all occurrences, like s.match(/(?<!\w)\?FOO\?(?!\w)/g).

The first expression needs a capturing group around the word you need to extract:

var strs = ["?FOO is cool", "I love ?FOO", "FOO is cool", "FOO?is cool", "aaFOO?is cool"];
var rx = /(?:^|\W)(\?FOO)(?!\w)/g;
for (var s of strs) {
  var res = [], m;
  while (m=rx.exec(s)) {
    res.push(m[1]);
  }
  console.log(s, "=>", res);
}

const text = 'This is your sample message that is really messy.';
const pattern = /(\w*mes\w*)/gi;
const results = [...text.matchAll(pattern)];
console.log(results)

rlike '(?i)\\bFECHADO\\b|\\bCIERRE\\b|\\bCLOSED\\b'