java regular expressions: group count fixed

Question

I have a string like that: obj[attr1=val1 attr2=val2 attr3=val3]
i need to extract object name and attributes.

Earlier, i've decided similar task in javascript using next regexp:

/^(\w+)(?:\[(\w+=\w+)(?:\s(\w+=\w+))*\])?$/

Now i have a trouble deciding in java:

Pattern pathPattern = Pattern.compile("^(\\w+)(?:\\[(\\w+=\\w+)(?:\\s+(\\w+=\\w+))*\\])?$");

I'm getting just a object name and first attribute. It seems that Mather class gets group count corresponding to count of "()" without considering symbol "*".

Is exists the possibility to make working java reg exp like js regexp, or i need to make two steps extraction?

thank you

Yes you are right. It does not consider everything matched by *. But just one group count. — Rohit Jain
– Rohit Jain, Commented Oct 23, 2012 at 6:33
I think it is strange way of realisation. I can use reg exp for matching but i can't use data. — Damask
– Damask, Commented Oct 23, 2012 at 6:42
When you use *, the pattern will be continuously replaced by the next match. Hence at last you would have only one matching string. That's why its only one group. — Rohit Jain
– Rohit Jain, Commented Oct 23, 2012 at 6:48

Rohit Jain · Accepted Answer · 2012-10-23 07:43:49Z

4

Matcher.groupCount() only counts the number of opening-brackets and consider them to be a group. So, the number of brackets you open will be the number of group counts (provided you are not using any non-capturing group).

You can use the below pattern to get the value inside the [.*]: -

Pattern pattern = Pattern.compile("(?:\\b)(\\w+?)=(\\w+?)(?:\\b)");
Matcher matcher = pattern.matcher(str);

while (matcher.find()) {
    System.out.println(matcher.group(1) + " : " + matcher.group(2));
}

This will match all the attr=val pair inside the [ and ].

OUTPUT: -

attr1 : val1
attr2 : val2
attr3 : val3

UPDATE: -

Since you don't have to do a boundary check in your above string, the above pattern can even be simplified to: -

Pattern pattern = Pattern.compile("(\\w+?)=(\\w+)");

edited Oct 23, 2012 at 7:43

answered Oct 23, 2012 at 6:44

Rohit Jain

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11 Comments

Damask Over a year ago

This way suppose two step decision. I'm interesting to make that in one step. Seems like in Java RegExp it not possible.

Rohit Jain Over a year ago

@BulatSafin. 2 step decision in the sense? I can't understand.

Damask Over a year ago

First: extract object name and brackets content. Second: parse brackets content.

Rohit Jain Over a year ago

@BulatSafin. That you have to do anyways. How do you want it to work?

Rohit Jain Over a year ago

@BulatSafin. Pattern Matching in Java regex works that way only. First you need to build a matcher object to match to the given string. and then perform the matching operation on that matcher.

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