CODEWITHSUNDEEP
javaregex
Damask
Damask

Reputation: 1254

java regular expressions: group count fixed

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

Upvotes: 2

Views: 3816

Answers (1)

Rohit Jain
Rohit Jain

Reputation: 213263

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+)");

Upvotes: 4

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