bash - Completely removing Node.js: automatic vs manual -
recently used following list of files associated node removing on linux mint 18.1:
sudo rm -rf /usr/local/lib/node* /usr/local/include/node* ~/.local/node* ~/.lib/node* ~/.include/node* /usr/local/bin/node*
on other hand there solution uses apt-get:
sudo apt-get purge --auto-remove nodejs
is there difference in them in context of full , total removing nodejs local machine? 1 more correct/safe/complete?
thanks in advance!
it's safer , more reliable use apt-get purge
that. otherwise have know of package files located. includes not limited to: configuration files, man pages, libraries, etc. in addition, package structure may change during time (like new files added or directories changed). apt-get
utility execute set of post-removal scripts in /var/lib/dpkg/info
. if keep removing files manually, system cluttered , may end in inconsistent state.
here visual example (for ubuntu):
root@ubuntu-xenial:~# rm -rf /usr/lib/node* /usr/include/node* /usr/bin/node* root@ubuntu-xenial:~# updatedb root@ubuntu-xenial:~# locate nodejs | head -n 3 /etc/profile.d/nodejs.sh /usr/share/nodejs /usr/share/doc/nodejs # there still more 200 files left related nodejs package. root@ubuntu-xenial:~# locate nodejs | wc -l 212
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