How can I remove gpg key that I added using apt-key add -?

Question

I don't need the key in my server's keyring anymore. Is it possible to remove it? I added the key using this command:

 curl http://repo.varnish-cache.org/debian/GPG-key.txt | apt-key add -

Thanks for helping

Solution 1

First you need to find the key id of the key you added. Do this by the command:

sudo apt-key list

It will list all the keys that you have, with each entry looking like this:

pub   1024R/B455BEF0 2010-07-29
uid                  Launchpad clicompanion-nightlies

Once you have figured out which key to remove, use the command sudo apt-key del <keyid> where <keyid> is replaced with the actual keyid of the key you want to remove from your keyring.

$ sudo apt-key del B455BEF0
$ apt-key list | grep clicompan
$

Solution 2

On 16.10 the short key id is no longer shown when you use the list command, but it is actually the last 8 characters of the long hex.

So for example the key id for the following key

/etc/apt/trusted.gpg.d/ubuntu-keyring-2012-cdimage.gpg
------------------------------------------------------
pub   rsa4096 2012-05-11 [SC]
      8439 38DF 228D 22F7 B374  2BC0 D94A A3F0 EFE2 1092
uid           [ unknown] Ubuntu CD Image Automatic Signing Key (2012) <[email protected]>

The key id will be EFE21092

Solution 3

Update for Ubuntu 20.04

after running

sudo apt-key list

you should get the list of apt keys:

/etc/apt/trusted.gpg
--------------------
pub   rsa4096 2016-04-12 [SC]
      EB4C 1BFD 4F04 2F6D DDCC  EC91 7721 F63B D38B 4796
uid           [ unknown] Google Inc. (Linux Packages Signing Authority) <[email protected]>
sub   rsa4096 2019-07-22 [S] [expires: 2022-07-21]

pub   rsa4096 2017-04-11 [SC] [expired: 2019-09-28]
      D4CC 8597 4C31 396B 18B3  6837 D615 560B A5C7 FF72
uid           [ expired] Opera Software Archive Automatic Signing Key 2017 <[email protected]>

pub   rsa4096 2019-09-12 [SC] [expires: 2021-09-11]
      68E9 B2B0 3661 EE3C 44F7  0750 4B8E C3BA ABDC 4346
uid           [ unknown] Opera Software Archive Automatic Signing Key 2019 <[email protected]>
sub   rsa4096 2019-09-12 [E] [expires: 2021-09-11]

pub   rsa4096 2017-03-13 [SC]
      8CAE 012E BFAC 38B1 7A93  7CD8 C5E2 2450 0C12 89C0
uid           [ unknown] TeamViewer GmbH (TeamViewer Linux 2017) <[email protected]>
sub   rsa4096 2017-03-13 [E]

under uid you have the name of the app, for example:

[ unknown] Opera Software Archive Automatic Signing Key 2019 <[email protected]>

and the key you want to delete is above it:

    D4CC 8597 4C31 396B 18B3  6837 D615 560B A5C7 FF72  <-- THAT'S THE KEY
uid           [ expired] Opera Software Archive Automatic Signing Key 2017 <[email protected]>

and you remove it by putting that key inside double or single quotes like this:

sudo apt-key del "D4CC 8597 4C31 396B 18B3  6837 D615 560B A5C7 FF72"

Solution 4

I made a short script to make things easier and using a string instead of the id.

You can use my script if the key contains a unique string you know.
e.g. in my case for webmin

pub   1024D/11F63C51 2002-02-28
uid                  Jamie Cameron <[email protected]>
sub   1024g/1B24BE83 2002-02-28

I'm sure only the webmin key on my system has jcameron than I use this script to remove the according key.

I saved it as ~/removeAptKey

and run it as

sudo ./removeAptKey jcameron

The ouput should be something like

KEYID: 11F63C51
OK

Here is my script:

#!/bin/bash

function printKeys(){
    echo "Installed keys are"
    echo ""
    sudo apt-key list
}

if [[ $EUID -ne 0 ]]; then
   echo "This script must be run as root" 1>&2
   exit 1
fi

if [[ $# -eq 0 ]]
then
    echo "No key name provided"
    exit 1
fi

UNIQUE=$1

sudo apt-key list | grep "${UNIQUE}" -B 1 > result.temp

LENGTH=$(cat result.temp | wc -l)

if [[ ${LENGTH} -gt 2 ]]
then
    echo "Attention you found more than 1 key. Use a more specific string."
    printKeys
    exit 2
fi

if [[ ${LENGTH} != 2 ]]
then
    echo "Key not found. Doing nothing."
    printKeys
    exit 3
fi

KEYID=$(cat result.temp | grep 'pub' | cut -d " " -f 4 | cut -d "/" -f 2)
echo "KEYID: "$KEYID

apt-key del ${KEYID}

rm result.temp

First I get the upper two lines of my key's block:

• sudo apt-key list: lists the apt keys as usual
• grep '${UNIQUE}' -B 1: take only the line containing the unique key string jcameron and -B 1 the line before
• > result.temp: Save it in a file (which is later removed)

If this returns exactly 2 lines (-> got exactly 1 key) I move on:

• grep 'pub': Now take only the line with the pup key id
• cut -d " " -f 4: take the 4th word of that line (the first is pub than come two spaces, than the string we are after ``)
• cut -d "/" -f 2: take only the part after /

And finally delete this key and cleanup

• apt-key del ${KEYID} (in my case 11F63C51)
• rm result.temp: don't need this file anymore

Solution 5

I know I might be late, but just wanted to share this one-line command to achieve this.

NOTE: This will only work if the output is an unique key.

---

Ubuntu versions up to 16.04 (UPDATED 2018-12-22):

apt-key del $(apt-key list | awk 'NR=='$(apt-key list | grep --line-number --regexp "FOOBAR" | cut --fields 1 --delimiter ":")'{print;exit}' | awk '{print $2}' | cut --fields 2 --delimiter "/")

where FOOBAR is the UID name.

---

Ubuntu versions from 16.10:

apt-key del $(apt-key list | awk 'NR=='`expr $(apt-key list | grep --line-number --regexp "FOOBAR" | cut --fields 1 --delimiter ":") - 1`'{print;exit}')

where FOOBAR is the UID name.

View more solutions

Comments

[Nitin Venkatesh]

@Raymond No problemo :)

[ctrl-alt-delor]

Ah I see the unique Id is on the line labelled pub, not the line labelled uid.

[mxdsp]

on ubuntu 16.10 results seems a little different : pub rsa4096 2012-05-11 [SC] 8439 .... uid ....

[SColvin]

Very helpful, thank you. This is extremely unhelpful UX.

[baptx]

@SColvin you can just do sudo apt-key del "8439 38DF 228D 22F7 B374 2BC0 D94A A3F0 EFE2 1092" and I think it is safer to use the whole fingerprint, the keyid could have duplicates (at least when you use PGP for emails, I read you should share your whole fingerprint and not just the keyid).

[Hartmut Pfarr]

Very helpful, very true also for 17.10!

[Russ Bateman]

And for 18.04.1.

[Gabriel Fair]

I'm not sure why, but I got an error when I tried to run the 16.04 command you provided. awk: line 1: syntax error at or near { But the angle braces match, so I'm not sure why this doesn't work

[Gabriel Fair]

I ran this without a parameter and it just wiped out all my keys. ;(

[David Tabernero M.]

@GabrielFair Thanks for noticing, back in june it has been working (I copy-pasted it from my console) but now seems that has been updated and the apt-key list format has changed. Now it seems to be working again. (EDIT: Remember to run this as superuser)

[Gabriel Fair]

If anyone else has their keys wipped, I was able to fix it by following these instructions: askubuntu.com/a/145933/13693

[derHugo]

Hu? How did this happen? It should exit with an "No key name provided" if there was no parameter..

[kennyB]

Looks like this needs to be updated for 18.04

[defuzed]

@derHugo not sure but could it be that the # in that if clause is the culprit?

[derHugo]

@defuzed if you mean $# then no. It returns the amount of given parameters.

[Corey]

After deleting the key, I encountered a new problem The following signatures couldn't be verified because the public key is not available: NO_PUBKEY, what should I do next ?

[Corey]

I found the answer, adding a # before the repo in /etc/apt/sources.list, then re-run sudo apt update, works for me on Ubuntu 18.04.

[Purefan]

it also worked for me to take the last 8 characters in the key and remove the space in between: sudo apt-key del A5C7FF72

[lewis4u]

yes... I think it would even work if you take any part of that key because it searches the pattern inside that whole key without spaces

[Amos Folarin]

has to be one of the stupidest --help listing (totally unclear what the id is)

[sorrow poetry]

On 16.10 the short key id is no longer shown when you use the list command, but it is actually the last 8 characters of the long hex. from @wesam answer.