Linux/bash: How to get interface's IPv6 address?

There are lots of ways you could do this.

Here is one:

ip addr show dev eth0 | sed -e's/^.*inet6 \([^ ]*\)\/.*$/\1/;t;d'

It is similar to Robert's answer, except strips out the address only.

You could use:

ip -6 addr

It will return all the IPv6 adresses you have configured.

If you're under Linux you also can parse: /proc/net/if_inet6

First column is the full IPv6 without colons. Sixth column is the interface name.

Here a rather lengthy not optimized version (I only now the basics of awk) Maybe python/perl is a better choice.

for i in "$(grep enp0s25 /proc/net/if_inet6)"; do
    echo "$i" | awk '{
        split($1, _, "[0-9a-f]{,4}", seps)
        joined = sep = ""
        for (i=1; i in seps; i++) {
            joined = joined sep seps[i]
            sep = ":"
        }
        print joined
    }'
done

If you have GNU awk (gawk) this can be shortened to:

for i in "$(grep enp0s25 /proc/net/if_inet6)"; do
    echo "$i" | gawk '@include "join"
    {
        split($1, _, "[0-9a-f]{,4}", seps)
        print join(seps, 1, length(seps), ":")
    }'
done

You may put it in a {ba,z,}sh function to use it later.