I would like to know the regex to match words such that the words have a maximum length. for eg, if a word is of maximum 10 characters in length, I would like the regex to match, but if the length exceeds 10, then the regex should not match.
I tried
^(\w{10})$
but that brings me matches only if the minimum length of the word is 10 characters. If the word is more than 10 characters, it still matches, but matches only first 10 characters.
I think you want \b\w{1,10}\b. The \b matches a word boundary.
Of course, you could also replace the \b and do ^\w{1,10}$. This will match a word of at most 10 characters as long as its the only contents of the string. I think this is what you were doing before.
Since it's Java, you'll actually have to escape the backslashes: "\\b\\w{1,10}\\b". You probably knew this already, but it's gotten me before.
^\w{0,10}$ # allows words of up to 10 characters.
^\w{5,}$ # allows words of more than 4 characters.
^\w{5,10}$ # allows words of between 5 and 10 characters.
Length of characters to be matched.
{n,m} n <= length <= m
{n} length == n
{n,} length >= n
And by default, the engine is greedy to match this pattern. For example, if the input is 123456789, \d{2,5} will match 12345 which is with length 5.
If you want the engine returns when length of 2 matched, use \d{2,5}?
Word boundaries would work perfectly here, such as with:
\b\w{3,8}\b
\b\w{2,}
\b\w{,10}\b
\b\w{5}\b
Some languages such as Java and C++ are double-escape required:
\\b\\w{3,8}\\b
\\b\\w{2,}
\\b\\w{,10}\\b
\\b\\w{5}\\b
PS: \\b\\w{,10}\\b may not work for all languages or flavors.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegularExpression{
public static void main(String[] args){
final String regex = "\\b\\w{3,8}\\b";
final String string = "words with length three to eight";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
}
}
}
Full match: words
Full match: with
Full match: length
Full match: three
Full match: eight
Another good-to-know method is to use negative lookarounds:
(?<!\w)\w{3,8}(?!\w)
(?<!\w)\w{2,}
(?<!\w)\w{,10}(?!\w)
(?<!\w)\w{5}(?!\w)
(?<!\\w)\\w{3,8}(?!\\w)
(?<!\\w)\\w{2,}
(?<!\\w)\\w{,10}(?!\\w)
(?<!\\w)\\w{5}(?!\\w)
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegularExpression{
public static void main(String[] args){
final String regex = "(?<!\\w)\\w{1,10}(?!\\w)";
final String string = "words with length three to eight";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
}
}
}
Full match: words
Full match: with
Full match: length
Full match: three
Full match: to
Full match: eight
jex.im visualizes regular expressions:
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
Even, I was looking for the same regex but I wanted to include the all special character and blank spaces too. So here is the regex for that:
^[A-Za-z0-9\s$&+,:;=?@#|'<>.^*()%!-]{0,10}$
Simple, complete and tested java code, for finding words of certain length n:
int n = 10;
String regex = "\\b\\w{" + n + "}\\b";
String str = "Hello, this is a test 1234567890";
ArrayList<String> words = new ArrayList<>();
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
words.add(matcher.group(0));
}
System.out.println(words);
For more explanations and different options - see other answers.
Liked Pardeep's answer but I needed whole word bounds in a string/title that can be any messed up string an advertising dept. can think up .
**\b\w(**[A-Za-z0-9\s$&+,:;=?@#|'<>.^*()%!-]{1,22}**)\b**
should iterate through a string ( tested notepad++ ) and get the largest group of words in the range i.e. 1,22 chars here without splitting mid word.
Here was the final command for me in python to add some LF's
name = re.sub(r"\b(\w[A-Za-z0-9\s$&+,:;=?@#|'<>.^*()%!-]{1,22})\b","\\\1\\\n",name)
