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How can I retrieve from the command line (or a shell script) only the name of the active network interface, in Linux? If there are several active interfaces, I want just one (selected arbitrarily).

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  • The name change depending of the computer you are running. Commented Mar 15, 2016 at 18:12
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    What if there are more than one? Can you be sure your system will always have only one active? Commented Mar 15, 2016 at 18:12
  • @jasonwryan yes, but it's probably better not to close since ifconfig is getting deprecated in favor of ip and its ilk. Commented Mar 15, 2016 at 18:13
  • no it is not duplicate ... the one you link is for mac and does not fit for me. Commented Mar 15, 2016 at 18:13
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    The answers of the suggested dupe also work for Linux. There's nothing OSX-specific there. However, there are modern alternatives. Please answer my first comment about whether we can be sure that your machine will always only have one active interface. Commented Mar 15, 2016 at 18:14

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The modern way of doing this is using the ip command. For example, on my system with my wireless connection active, I get:

$ ip addr show
1: lo: <LOOPBACK,UP,LOWER_UP> mtu 65536 qdisc noqueue state UNKNOWN group default
    link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
    inet 127.0.0.1/8 scope host lo
       valid_lft forever preferred_lft forever
    inet6 ::1/128 scope host
       valid_lft forever preferred_lft forever
2: eno1: <NO-CARRIER,BROADCAST,MULTICAST,UP> mtu 1500 qdisc fq_codel state DOWN group default qlen 1000
    link/ether 00:26:b9:dd:2c:28 brd ff:ff:ff:ff:ff:ff
3: wlp3s0b1: <BROADCAST,MULTICAST,UP,LOWER_UP> mtu 1500 qdisc mq state UP group default qlen 1000
    link/ether c4:46:19:5f:dc:f5 brd ff:ff:ff:ff:ff:ff
    inet 192.168.1.4/24 brd 192.168.1.255 scope global wlp3s0b1                            ←
       valid_lft forever preferred_lft forever
    inet6 fe80::c646:19ff:fe5f:dcf5/64 scope link
       valid_lft forever preferred_lft forever
16: tun0: <POINTOPOINT,MULTICAST,NOARP,UP,LOWER_UP> mtu 1500 qdisc noqueue state UNKNOWN group default qlen 100
    link/none
    inet 123.167.217.2/24 brd 123.167.217.255 scope global tun0                            ←
       valid_lft forever preferred_lft forever

The active interface(s) have both an inet entry and a broadcast (brd) address. You can show all such interfaces with:

$ ip addr show | awk '/inet.*brd/{print $NF}'
wlp3s0b1
tun0

If you want only one, you can get the first one (only) with:

$ ip addr show | awk '/inet.*brd/{print $NF; exit}'
wlp3s0b1

The exit statement tells awk to stop searching after it finds the first match.

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    @aurelien If this answer solved your issue, please take a moment and accept it by clicking on the check mark to the left. That will mark the question as answered and is the way thanks are expressed on the Stack Exchange sites. Commented Mar 15, 2016 at 18:18
  • I will do ... it says to wait 7 minutes :-/ Commented Mar 15, 2016 at 18:21
  • @aurelien no worries. I just mentioned it because you're new and might not know how the site works. Feel free to not accept and wait for another answer too, if you prefer. Commented Mar 15, 2016 at 18:22
  • Your command just do exactly what I have request for ... So I can just says thanks and accept it :-) Commented Mar 15, 2016 at 18:24
  • not every "active" interface has a broadcast address - e.g. ppp interfaces don't. Commented Mar 15, 2016 at 22:49
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ifconfig | sed 's/[ \t].*//;/^\(lo\|\)$/d'
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    The problem with that approach is that ifconfig is being replaced by ip. Commented Mar 15, 2016 at 18:18
  • ip a | sed 's/[ \t].*//;/^\(lo\|\)$/d' respond an empty thing .. but +1 for this nice try :-) @Maslov-Anton Commented Oct 24, 2018 at 12:39

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