Hi, Holger Levsen wrote (13 Feb 2014 15:48:59 GMT) : > right. I'd be happy to do this, but I dont know how to remove those hundred > keys from my keyring...
OK, so here's a small gift for you! The attached script should list the keys that belong to both of the two keyrings passed on the command-line. It depends on an internal parcimonie library, and on the libset-object-perl Debian package. *Caution*: please backup the original ~/.gnupg/pubring.gpg before deleting all these keys from your keyring, as you will lose some information in this process, such as local-only key certifications, your own public keys, and probably more. Once you're sure you want to do it, I assume that piping the output into "xargs -n 1 gpg --batch --delete-key" should do the job. Cheers, -- intrigeri | GnuPG key @ https://gaffer.ptitcanardnoir.org/intrigeri/intrigeri.asc | OTR fingerprint @ https://gaffer.ptitcanardnoir.org/intrigeri/otr.asc
#!/usr/bin/perl
=head1 NAME
keyrings_intersection - list keys that belong to the intersection of two
keyrings
=cut
use strict;
use warnings FATAL => 'all';
use 5.10.1;
use App::Parcimonie qw{gpgPublicKeys};
use Set::Object qw{set};
my $usage = 'Usage: keyrings_intersection KEYRING_1 KEYRING_2';
sub keys_in {
my $keyring = shift;
gpgPublicKeys({ extra_args => [
'--no-default-keyring', "--keyring=$keyring",
]});
}
if (@ARGV == 1 and $ARGV[0] eq '--help') {
say STDERR $usage;
exit 0;
}
elsif (@ARGV != 2) {
die "$usage";
}
my $keyring_1 = shift;
my $keyring_2 = shift;
-e $keyring_1 || die "'$keyring_1' does not exist.";
-e $keyring_2 || die "'$keyring_2' does not exist.";
say join("\n",
(set(keys_in $keyring_1) * set(keys_in $keyring_2))->members
);
pgpoTDWzrx5yf.pgp
Description: PGP signature
