Question on regular expression.

Thread view:

Paul - 15 Mar 2007 16:22 GMT

Here is a pattern and the code snippet I have:

String pat = "#foo#([+-][0-9]+)*";
String input = "#foo#+1-2";
Pattern pattern = Pattern.compile(pat);
Matcher matcher = pattern.matcher(input);

while (matcher.find())
{
String matched = matcher.group(1);
...
}

The result I'm hoping for is to get "+1" and "-2" returned for
"matched" string. However, I'm only getting "-2" and then matcher is
done.

Can any shed some light on this?

Thanks,
Paul.

Reply to this Message

Patricia Shanahan - 15 Mar 2007 16:49 GMT

> Here is a pattern and the code snippet I have:
>
[quoted text clipped - 17 lines]
> Thanks,
> Paul.

There is only one "#foo#" to match the start of the expression, so only
the first find() call can succeed.

What does matcher.groupCount() return after the successful find()? Just
looking at it, I would have expected 2, with group(1) equal to "+1" and
group(2) equal to "-2", but you only show a call for group(1).

Patricia

Reply to this Message

Lars Enderin - 15 Mar 2007 16:55 GMT

Paul skrev:

> Here is a pattern and the code snippet I have:
>
[quoted text clipped - 14 lines]
>
> Can any shed some light on this?

Since #foo# is part of pat, you cannot get more than one match. Remove
#foo# from pat and also remove the *.

Reply to this Message

Paul - 15 Mar 2007 17:23 GMT

> Paul skrev:
>
[quoted text clipped - 21 lines]
>
> - Show quoted text -

The input can be "#foo#", "#foo#+1", "#foo#+1-2", etc. The current
approach I took is to modify the input and reset the matcher each time
I got a match. I don't like this approach. It's kind of like what you
indicated here. Basically, you have to use 2 patterns during the
process. Right?

Patricia,

It only returns 1 group. I was hoping it behaves like what you
described.

Thanks,
Paul.

Reply to this Message

Steve Wampler - 15 Mar 2007 18:02 GMT

> Here is a pattern and the code snippet I have:
>
[quoted text clipped - 14 lines]
>
> Can any shed some light on this?

I believe that only one group is defined in the above pattern, and since
'*' is greedy, that group gets assigned the last successful match of
the contained subpattern. The following varient shows this better:
---------------------------------------------------
import java.util.regex.*;

public class Foo {

public static void main(String[] args) {
String pat = "#foo#([+-][0-9]+)([+-][0-9]+)*";
String input = "#foo#+1-2+3-4";
Pattern pattern = Pattern.compile(pat);
Matcher matcher = pattern.matcher(input);

if (matcher.matches()) {
System.out.println("Found "+matcher.groupCount()+" groups");
for (int i = 1; i <= matcher.groupCount(); ++i) {
System.out.println(""+i+": "+matcher.group(i));
}
}

System.exit(0);
}

}
--------------------------------------------------
When run, the output is:
--------------------------------------------------
->java Foo
Found 2 groups
1: +1
2: -4
->
--------------------------------------------------

Unfortunately, I can't see how to specify an arbitrary number of groups...

Signature

Steve Wampler -- swampler@noao.edu
The gods that smiled on your birth are now laughing out loud.

Reply to this Message

Patricia Shanahan - 15 Mar 2007 18:35 GMT

>> Here is a pattern and the code snippet I have:
>>
[quoted text clipped - 14 lines]
>>
>> Can any shed some light on this?

...

> Unfortunately, I can't see how to specify an arbitrary number of groups...

I think it is going to have to be done with two patterns.
As I understand the situation, the current pattern does a good job of
representing the substring that should be matched, so how about keeping
something like it, but adding another pattern for splitting up the
repeated groups?

Pattern outer = Pattern.compile("#foo#(([+-][0-9]+)*)");
Pattern inner = Pattern.compile("[+-][0-9]+");

void test(String input) {
Matcher outerMatcher = outer.matcher(input);
if (outerMatcher.find()) {
System.out.printf("Matched %s in %s%n",outerMatcher.group(0),input);
Matcher innerMatcher = inner.matcher(outerMatcher.group(1));
while(innerMatcher.find()){
System.out.printf("Group %s%n",innerMatcher.group(0));
}
} else {
System.out.println("No match: " + input);
}
}

Reply to this Message

Steve Wampler - 15 Mar 2007 19:08 GMT

>> Unfortunately, I can't see how to specify an arbitrary number of
>> groups...
>
> I think it is going to have to be done with two patterns.

I agree. I *should* have said "...arbitrary number of groups
in a single pattern".

Signature

Steve Wampler -- swampler@noao.edu
The gods that smiled on your birth are now laughing out loud.

Reply to this Message

Question on regular expression.

Free Magazines