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!
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 linksboolean matches(): Attempts to match the entire region against the pattern. .* meansAs 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).
.*? and .* for regexNote 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".
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?String API quick cheat sheetHere's a quick cheat sheet that lists which methods are regex-based and which aren't:
String replace(char oldChar, char newChar)String replace(CharSequence target, CharSequence replacement)boolean startsWith(String prefix)boolean endsWith(String suffix)boolean contains(CharSequence s)String replaceAll(String regex, String replacement)String replaceFirst(String regex, String replacement)String[] split(String regex)boolean matches(String regex)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.
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