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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