Question

i am having a XML file which i need to transfer to a list with Perl (without using XSLT).

This is my (simplyfied, removed like 10 more attributes to make it easier to read!) XML:

...
<XMLTAG ID="1" name="NAME1" status="0" date1="24.05.2012 13:37:00" date2="25.05.2012 13:37:00" />
<XMLTAG ID="2" name="NAME2" status="1" date1="24.05.2012 13:37:00" date2="25.05.2012 13:37:00" />
<XMLTAG ID="3" name="NAME3" status="0" date1="24.05.2012 13:37:00" date2="25.05.2012 13:37:00" />
...

What i got so far:

my $input = in.xml;
my $output = out.txt;

# open input
open( INPUT, $input )
  || die "Can't find $input: $_";

# open output
open( OUTPUT, ">$output" )
  || die "Can't find $output: $_";

    # run until perl returns undef (at the end of the file)
    while (<INPUT>) {
        if ($_ == /date1=\"[0-3]?[0-9].[0-3]?[0-9].(?:[0-9]{2})?[0-9]{2} [0-5][0-9]:[0-5][0-9]:[0-5][0-9]\"/) {
        print OUTPUT $_;};
    }
    close(INPUT);
    close(OUTPUT);

The output file should look like this:

date1="24.05.2012 13:37:00"
date1="24.05.2012 13:37:01"
date1="24.05.2012 13:37:02"
...

Thanks in advance,
Marley

Answer 1

use XML::LibXML qw();
my $dom = XML::LibXML->load_xml(location => 'in.xml');
printf qq(date1="%s"\n), $_->getAttribute('date1')
    for $dom->findnodes('//XMLTAG');