Parsing zipped xml data


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Parsing zipped xml data

by space_monk (Hermit)

on Jan 08, 2013 at 10:13 UTC ( #1012212=perlquestion: print w/ replies, xml )

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space_monk has asked for the wisdom of the Perl Monks concerning the following question:

So I wrote this little application at work using XML::Parser, and then my line manager said "that works great, but the files are huge and I'd be much happier if it took .gz files as input. Now as far as I know, XML::Parser is based on expat, and if I had chosen to write my app using XML::LibXML::Parser instead, which uses the C libxml2 library, it would have taken .gz files as input automagically.

What's the easiest way to get my XML::Parser based app to process .xml.gz files with the minimum of changes?

A Monk aims to give answers to those who have none, and to learn from those who know more.

Comment on Parsing zipped xml data
Re: Parsing zipped xml data by Anonymous Monk on Jan 08, 2013 at 10:54 UTC
gzcat, but you knew that already :)	[reply]
Re^2: Parsing zipped xml data by space_monk (Hermit) on Jan 08, 2013 at 11:09 UTC
I probably did, but the grey cells that hold that information have been killed off by the influx of alcohol between XMas and New Year :-) Plus, I was interested to see various TMTOWTDI alternatives if there are any.... A Monk aims to give answers to those who have none, and to learn from those who know more.	[reply]
Re: Parsing zipped xml data by Jenda (Monsignor) on Jan 09, 2013 at 00:07 UTC
The parse() method accepts an opened filehandle so something like `use PerlIO::gzip; open my $IN, "<:gzip", "file.gz" or die $!; $parser->parse($IN); ...` [download] should work. (untested) Jenda Enoch was right! Enjoy the last years of Rome.	[reply] [d/l]

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