http://qs1969.pair.com?node_id=597452

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

Searching from a top tree, how can I tell if the entry under the top tree is a non-leaf node(node which contain other entry udner it)? Thanks.

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Re: Net LDAP
by shmem (Chancellor) on Jan 31, 2007 at 01:14 UTC
    You take that entry's dn as search base and search from there with filter '(objectClass=*)', restricting your search scope to one-level (see ldapsearch of openldap; dunno right now how's that in Net::LDAP). If you get results, it's a non-leaf-node.

    <update>

    After RTFMing a bit, here's an example: searching 'dc=example,dc=com'

    my $mesg = $ldap->search( base => "dc=example,dc=com", filter => '(objectClass=*)', scope => 'one', ); foreach my $entry($mesg->all_entries) { my $res = 0; my $indent = 0; while($res = is_leaf($ldap,$entry)) { print " " x $indent, 'non-leaf: ', $entry->dn, "\n"; $indent++; $entry = $res; } print " " x $indent, 'leaf: ', $entry->dn, "\n" unless $res; # $r +es should always be 0; } sub is_leaf { my ($ldap, $entry) = @_; my $res = $ldap->search( base => $entry->dn, filter => '(objectClass=*)', sizelimit => 1, scope => 'one', ); if(my @l = $res->all_entries) { return $l[0]; } else { return 0; } }
    this will print out all entries and their "leafness" under the base dn. Non-leaf entries are followed, searching the first child, until a leaf node is found.

    </update>

    But I might be parling heck.
    --shmem

    _($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                                  /\_¯/(q    /
    ----------------------------  \__(m.====·.(_("always off the crowd"))."·
    ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}