method matches not work well

Question

I don't understand why with this regex the method returns false;

Pattern.matches("\\bi", "an is");

the character i is at a word boundary!

Answer 1

In Java, matches attempts to match a pattern against the entire string.

This is true for String.matches, Pattern.matches and Matcher.matches.

If you want to check if there's a match somewhere in a string, you can use .*\bi.*. In this case, as a Java string literal, it's ".*\\bi.*".

java.util.regex.Matcher API links

• boolean matches(): Attempts to match the entire region against the pattern.

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What .* means

As used here, the dot . is a regex metacharacter that means (almost) any character. * is a regex metacharacter that means "zero-or-more repetition of". So for example something like A.*B matches A, followed by zero-or-more of "any" character, followed by B (see on rubular.com).

References

• regular-expressions.info/Repetition with Star and Plus and The Dot Matches (Almost) Any Character

Related questions

• Difference between .*? and .* for regex

Note that both the . and * (as well as other metacharacters) may lose their special meaning depending on where they appear. [.*] is a character class that matches either a literal period . or a literal asterisk *. Preceded by a backslash also escapes metacharacters, so a\.b matches "a.b".

• regular-expressions.info/Character Class and Literal Characters and Metacharacters

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Related problems

Java does not have regex-based endsWith, startsWith, and contains. You can still use matches to accomplish the same things as follows:

• matches(".*pattern.*") - does it contain a match of the pattern anywhere?
• matches("pattern.*") - does it start with a match of the pattern?
• matches(".*pattern") - does it end with a match of the pattern?

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String API quick cheat sheet

Here's a quick cheat sheet that lists which methods are regex-based and which aren't:

• Non-regex methods:
  • String replace(char oldChar, char newChar)
  • String replace(CharSequence target, CharSequence replacement)
  • boolean startsWith(String prefix)
  • boolean endsWith(String suffix)
  • boolean contains(CharSequence s)
• Regex methods:
  • String replaceAll(String regex, String replacement)
  • String replaceFirst(String regex, String replacement)
  • String[] split(String regex)
  • boolean matches(String regex)

Answer 2

The whole string has to match if you use matches:

Pattern.matches(".*\\bi.*", "an is")

This allows 0 or more characters before and after. Or:

boolean anywhere = Pattern.compile("\\bi").matcher("an is").find();

will tell you if any substring matches (true in this case). As a note, compiling regexes then keeping them around can improve performance.

Answer 3

I don't understand why Java decided to go in the opposite direction from languages like Perl that has supported regex natively for years. I threw the standard Java regex away and started using my own perl-style regex lib for Java called MentaRegex. See below how regex can make sense in Java.

The method matches returns a boolean saying whether we have a regex match or not.

matches("Sergio Oliveira Jr.", "/oliveira/i" ) => true

The method match returns an array with the groups matched. So it not only tells you whether you have a match or not but it also returns the groups matched in case you have a match.

match("aa11bb22", "/(\\d+)/g" ) => ["11", "22"]

The method sub allows you perform substitutions with regex.

sub("aa11bb22", "s/\\d+/00/g" ) => "aa00bb00"

Support global and case-insensitive regex.

match("aa11bb22", "/(\\d+)/" ) => ["11"]
match("aa11bb22", "/(\\d+)/g" ) => ["11", "22"]
matches("Sergio Oliveira Jr.", "/oliveira/" ) => false
matches("Sergio Oliveira Jr.", "/oliveira/i" ) => true

Allows you to change the escape character in case you don't like to see so many '\'.

match("aa11bb22", "/(\\d+)/g" ) => ["11", "22"]
match("aa11bb22", "/(#d+)/g", '#' ) => ["11", "22"]