Re: find function of String

Thank you for all your responses. I guess I am not very clear in my
question. Actually it is nothing complicated. All text except [ or ]
needs to be extracted. there is no complex stuff of subsets etc...

Anywayz I found a simple way using boost::regex to match only the
characters followed by numbers

int main(int argc, char* argv[])
{
    try
    {
        cout << "[START REGEXP CHECKING]" << endl;

        // String to be validated
        string myString = "[SBD 65 [INI 25 [PUP 0 ][RPN 25 ]][ANA 1 ][BUIPNR
0 ][BUIPAX 1 ]";

        // Regular Expression to be used.
        string myRegexp = "[a-zA-Z]+\\s\\d+";
        boost::regex aMyRegExp(myRegexp);

        // Needed to extract some part from myString, considering the sub-
expressions grouped by "()"
        boost::cmatch matches;

        cout << endl << "String: " << myString;
        cout << endl << "Regular Expression: " << myRegexp << endl;

        // Checking if myString matches the Regular Expression.
        // Use boost::regex_match(myString.c_str(), aMyRegExp) if the
extraction is not needed
        std::string::const_iterator start, end;
        start = myString.begin();
        end = myString.end();
        boost::match_results<std::string::const_iterator> what;
        boost::match_flag_type flags = boost::match_default;
        while (regex_search(start, end, what, aMyRegExp, flags)){
            cout << "Size: " << what.size() << endl;
            cout << "Result: " << what << endl;

            // update start position
            start = what[0].second;
        }

        cout << endl << endl << "[END REGEXP CHECKING]" << endl;
        return 0;
    }
    catch (...)
    {
        cout << "EXCEPTION!!!!" << endl;
        return 1;
    }

}

This prints out only text in the format of XYZ 34 etc

Regards,
Prashant

On Oct 20, 1:38 am, "Jim Langston" <tazmas...@rocketmail.com> wrote:


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