Re: [Boost-users] [serialization] removing excess XML tags

On Thu, Feb 5, 2009 at 2:47 PM, Robert Ramey wrote:

Make your own archive derived from xml_?archive.

Robert Ramey

"Paul Heil" wrote in message news:2fc4ff370902051118v6b811275rdad4dd46c04d40c9@mail.gmail.com... I'd like to use the boost::serialization library to generate XML that must be read by a third party application. Unfortunately, this application does not like the extra stuff that the serialization library generates.

The code below generates XML that looks like this: <?xml version="1.0" encoding="UTF-8" standalone="yes" ?> <!DOCTYPE boost_serialization> <MYTAG class_id="0" tracking_level="0" version="0"></MYTAG>

What I need is XML that looks more like this: <?xml version="1.0" encoding="UTF-8" standalone="yes" ?> <MYTAG></MYTAG>

What can I do to eliminate the extra stuff?

Thanks, PaulH

class MyClass { private: friend class boost::serialization::access;

template<class Archive> void serialize(Archive &ar, const unsigned int version){};

public: MyClass(){}; friend std::ostream & operator<<(std::ostream &os, const MyClass &gp); virtual ~MyClass(){}; };

int main( int argc, _TCHAR* argv[] ) { MyClass mc;

std::ofstream ofs("MyFile.xml"); boost::archive::xml_oarchive oa(ofs); oa << boost::serialization::make_nvp( "MYTAG", mc );

return 0; }

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That looks like the right way to go if I'm going to use Boost.Serialize to do this. I guess I was hoping I was just missing a flag or something easy. I'm not really looking to re-implement that much stuff. Thanks