perl怎么修改xml

lalalinux

<config logdir="/var/log/foo/" debugfile="/tmp/foo.debug">
    <server name="sahara" osname="solaris" osversion="2.6">
      <address>10.0.0.101</address>
      <address>10.0.1.101</address>
    </server>
    <server name="gobi" osname="irix" osversion="6.5">
      <address>10.0.0.102</address>
    </server>
    <server name="kalahari" osname="linux" osversion="2.0.34">
      <address>10.0.0.103</address>
      <address>10.0.1.103</address>
    </server>
  </config>



#!/usr/bin/perl -w
use IO::File;
use XML::Simple;
use Data:umper;
my $fh = IO::File->new('/home/abc/foo.xml');
my $ref = XMLin($fh);
$ref->{server}->{kalahari}->{osname} = "hello123456";
print "$ref->{server}->{kalahari}->{osname}\n";
print Dumper($ref);
可以输出成：
                        'kalahari' => {
                                        'osversion' => '2.0.34',
                                        'osname' => 'hello123456',



怎么输出成：
  <config logdir="/var/log/foo/" debugfile="/tmp/foo.debug">
    <server name="sahara" osname="solaris" osversion="2.6">
      <address>10.0.0.101</address>
      <address>10.0.1.101</address>
    </server>
    <server name="gobi" osname="irix" osversion="6.5">
      <address>10.0.0.102</address>
    </server>
    <server name="kalahari" osname="hello123456" osversion="2.0.34">
      <address>10.0.0.103</address>
      <address>10.0.1.103</address>
    </server>
  </config>

[ 本帖最后由 lalalinux 于 2010-1-4 16:56 编辑 ]

lalalinux

XMLout()
Takes a data structure (generally a hashref) and returns an XML encoding of that structure. If the resulting XML is parsed using XMLin(), it should return a data structure equivalent to the original (see caveats below).

document说的最多的就是怎么把xml解析出来，怎么就找不到returns an xml 的地方呢? 各位xdjm帮忙指出在哪里啦.
http://search.cpan.org/~grantm/XML-Simple-2.18/lib/XML/Simple.pm

guap514

http://search.cpan.org/~grantm/X ... er%3E_#_out_-_handy
不是写的挺清楚的么？