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Matcher.start(int group)

Posted on 2007-04-03
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Last Modified: 2013-12-29
Hi Experts ..

java.util.regex.Matcher.start(int group) method throws IllegalStateException if "If no match has been attempted, or if the previous match operation failed."

could someone please give me example for that case??
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Question by:DrAske
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Expert Comment

by:cavey_79
ID: 18846952
When you just create the matcher, without calling methods like matches, or when you call them and they return false.
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by:CEHJ
CEHJ earned 165 total points
ID: 18846956
Pattern p = Pattern.compile("(\\d)");
Matcher m = p.matcher("abc");
m.start(1);
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Expert Comment

by:cavey_79
ID: 18846963
In other words, you will need to successfully do one of these three 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.

before you can call start, group, etc.
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Expert Comment

by:CEHJ
ID: 18846965
There are no digits in "abc" so the match will fail and thus the call to start throws the exception
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by:objects
objects earned 165 total points
ID: 18846989
From the javadoc it "Returns the start index of the subsequence captured by the given group during the previous match operation.".
So if there is no successful previous match operation then it cannot return the start index, in which case it will throw the exception.
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Author Comment

by:DrAske
ID: 18847016
Thanks for all of you :o) so I can avoid it using find() method,
then the following will not throw the exception cause the while condition will be false

Pattern pattern = Pattern.compile("(\\d)");
Matcher matcher = pattern.matcher("abc");
 while(matcher.find()){
            int start = matcher.start(1);
            int end = matcher.end(1);
}

really appreciate your help,
regards,
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cavey_79 earned 170 total points
ID: 18847023
If you don't need to match it repeatedly, you can do this:
if (matcher.matches()){
          int start = matcher.start(1);
          int end = matcher.end(1);
}
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Expert Comment

by:objects
ID: 18847026
that is correct
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Expert Comment

by:CEHJ
ID: 18847031
>>then the following will not throw the exception cause the while condition will be false

Correct
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by:CEHJ
ID: 18847047
:-)
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Author Comment

by:DrAske
ID: 18847285
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