Hi Michael,

please ask for everything you didn't understand. I think you're on a very good way and I don't want to stop you because of too short or bad written comments.

Now I understand, why you're only parsing one xml file. Here is a really complicated, but short solution for this:

opendir THEDIR, "$ipath";
$xml = $ipath.(grep (/$str/, readdir THEDIR))[0];
closedir THEDIR;
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It is your grep (which returns an array of matching items) but as you only expect one element in the array, the grep-result is forced to be an array using the ( ) around grep. The combination of ( ) and &91; &93; allows you to fetch a specific array item without writing it to an array first.

Another comment on your original article already told you that you might not get a array from the XML parser if only one inside object is found. There was a hint for a solution. To check this, you should print the thing you use as a array reference:

print STDERR $config->{'Vehicle'}."\n";
foreach $vehicle ( @{ $config->{'Vehicle'} } ) {
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If this isn't an array reference (shown as ARRAY(0x12345)), you could force it to be one:

$config->{'Vehicle'} = [$config->{'Vehicle'}] if ref($config->{'Vehicl
+e'}) ne 'ARRAY';
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ref() tests if the parameter is a reference and returns the type of reference (in this case, ARRAY). If your value isn't what you expect (an ARRAY - reference), it creates one and fills in the old value into the new array.

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