Matcher matches() method in Java with ExamplesThe matches() method of Matcher Class is used to get the result whether this pattern matches with this matcher or not. It returns a boolean value showing the same. Syntax: public boolean matches() Parameters: This method do not takes any parameter.
The matcher() method is used to search for the pattern in a string. It returns a Matcher object which contains information about the search that was performed. The find() method returns true if the pattern was found in the string and false if it was not found.
regex. Matcher ) is used to search through a text for multiple occurrences of a regular expression. You can also use a Matcher to search for the same regular expression in different texts.
In Java, Matcher is a class that is implemented by the MatchResult interface, that performs match operations on a character sequence by interpreting a Pattern. By invoking the pattern's matcher method, a matcher is created from a pattern.
Actually, you misunderstood the documentation. Take a 2nd look at the statement you quoted: -
attempting to query any part of it before a successful match will cause an IllegalStateException to be thrown.
A matcher may throw IllegalStateException
on accessing matcher.group()
if no match was found.
So, you need to use following test, to actually initiate the matching process: -
- matcher.matches() //Or
- matcher.find()
The below code: -
Matcher matcher = pattern.matcher();
Just creates a matcher
instance. This will not actually match a string. Even if there was a successful match.
So, you need to check the following condition, to check for successful matches: -
if (matcher.matches()) {
// Then use `matcher.group()`
}
And if the condition in the if
returns false
, that means nothing was matched. So, if you use matcher.group()
without checking this condition, you will get IllegalStateException
if the match was not found.
Suppose, if Matcher
was designed the way you are saying, then you would have to do a null
check to check whether a match was found or not, to call matcher.group()
, like this: -
The way you think should have been done:-
// Suppose this returned the matched string
Matcher matcher = pattern.matcher(s);
// Need to check whether there was actually a match
if (matcher != null) { // Prints only the first match
System.out.println(matcher.group());
}
But, what if, you want to print any further matches, since a pattern can be matched multiple times in a String, for that, there should be a way to tell the matcher to find the next match. But the null
check would not be able to do that. For that you would have to move your matcher forward to match the next String. So, there are various methods defined in Matcher
class to serve the purpose. The matcher.find()
method matches the String till all the matches is found.
There are other methods also, that match
the string in a different way, that depends on you how you want to match. So its ultimately on Matcher
class to do the matching
against the string. Pattern
class just creates a pattern
to match against. If the Pattern.matcher()
were to match
the pattern, then there has to be some way to define various ways to match
, as matching
can be in different ways. So, there comes the need of Matcher
class.
So, the way it actually is: -
Matcher matcher = pattern.matcher(s);
// Finds all the matches until found by moving the `matcher` forward
while(matcher.find()) {
System.out.println(matcher.group());
}
So, if there are 4 matches found in the string, your first way, would print only the first one, while the 2nd way will print all the matches, by moving the matcher
forward to match the next pattern.
I Hope that makes it clear.
The documentation of Matcher
class describes the use of the three methods it provides, which says: -
A matcher is created from a pattern by invoking the pattern's matcher method. Once created, a matcher can be used to perform three different kinds of match operations:
The matches method attempts to match the entire input sequence against the pattern.
The lookingAt method attempts to match the input sequence, starting at the beginning, against the pattern.
The find method scans the input sequence looking for the next subsequence that matches the pattern.
Unfortunately, I have not been able find any other official sources, saying explicitly Why and How of this issue.
My answer is very similar to Rohit Jain's but includes some reasons why the 'extra' step is necessary.
java.util.regex implementation
The line:
Pattern p = Pattern.compile("foo=(?<foo>[0-9]*),bar=(?<bar>[0-9]*)");
causes a new Pattern object to be allocated, and it internally stores a structure representing the RE - information such as a choice of characters, groups, sequences, greedy vs. non-greedy, repeats and so on.
This pattern is stateless and immutable, so it can be reused, is multi-theadable and optimizes well.
The lines:
String s = "foo=23,bar=42";
Matcher matcher = p.matcher(s);
returns a new Matcher
object for the Pattern
and String
- one that has not yet read the String. Matcher
is really just a state machine's state, where the state machine is the Pattern
.
The matching can be run by stepping the state machine through the matching process using the following API:
lookingAt()
: Attempts to match the input sequence, starting at the beginning, against the patternfind()
: Scans the input sequence looking for the next subsequence that matches the pattern. In both cases, the intermediate state can be read using the start()
, end()
, and group()
methods.
Benefits of this approach
Why would anyone want to do step through the parsing?
Get values from groups that have quantification greater than 1 (i.e. groups that repeat and end up matching more than once). For example in the trivial RE below that parses variable assignments:
Pattern p = new Pattern("([a-z]=([0-9]+);)+");
Matcher m = p.matcher("a=1;b=2;x=3;");
m.matches();
System.out.println(m.group(2)); // Only matches value for x ('3') - not the other values
See the section on "Group name" in "Groups and capturing" the JavaDoc on Pattern
However, on most occasions you do not need to step the state machine through the matching, so there is a convenience method (matches
) which runs the pattern matching to completion.
If a matcher would automatically match the input string, that would be wasted effort in case you wish to find the pattern.
A matcher can be used to check if the pattern matches()
the input string, and it can be used to find()
the pattern in the input string (even repeatedly to find all matching substrings). Until you call one of these two methods, the matcher does not know what test you want to perform, so it cannot give you any matched groups. Even if you do call one of these methods, the call may fail - the pattern is not found - and in that case a call to group
must fail as well.
This is expected and documented.
The reason is that .matches()
returns a boolean indicating if there was a match. If there was a match, then you can call .group(...)
meaningfully. Otherwise, if there's no match, a call to .group(...)
makes no sense. Therefore, you should not be allowed to call .group(...)
before calling matches()
.
The correct way to use a matcher is something like the following:
Matcher m = p.matcher(s);
if (m.matches()) {
...println(matcher.group("foo"));
...
}
My guess is the design decision was based on having queries that had clear, well defined semantics that didn't conflate existence with match properties.
Consider this: what would you expect Matcher queries to return if the matcher has not successfully matched something?
Let's first consider group()
. If we haven't successfully matched something, Matcher shouldn't return the empty string, as it hasn't matched the empty string. We could return null
at this point.
Ok, now let's consider start()
and end()
. Each return int
. What int
value would be valid in this case? Certainly no positive number. What negative number would be appropriate? -1?
Given all this, a user is still going to have to check return values for every query to verify if a match occurred or not. Alternatively, you could check to see if it matches successfully outright, and if successful, the query semantics all have well-defined meaning. If not, the user gets consistent behaviour no matter which angle is queried.
I'll grant that re-using IllegalStateException
may not have resulted in the best description of the error condition. But if we were to rename/subclass IllegalStateException
to NoSuccessfulMatchException
, one should be able to appreciate how the current design enforces query consistency and encourages the user to use queries that have semantics that are known to be defined at the time of asking.
TL;DR: What is value of asking the specific cause of death of a living organism?
You need to check the return value of matcher.matches()
. It will return true
when a match was found, false
otherwise.
if (matcher.matches()) {
System.out.println(matcher.group("foo"));
System.out.println(matcher.group("bar"));
}
If matcher.matches()
does not find a match and you call matcher.group(...)
, you'll still get an IllegalStateException
. That's exactly what the documentation says:
The explicit state of a matcher is initially undefined; attempting to query any part of it before a successful match will cause an IllegalStateException to be thrown.
When matcher.match()
returns false
, no successful match has been found and it doesn't make a lot of sense to get information on the match by calling for example group()
.
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