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Useful in scripts while you just need an IP address in a variable.
There are 7 alternatives - vote for the best!
A method for aquiring the ip address using zsh. If you prefer the use of iproute2 (which, frankly, you should) then the following should provide the same (ip outputs CIDR addresses):
print ${$(ip -o -4 a s eth0)[4]}
we could also pass a qualifier to take only the IP and not the (CIDR) mask
print ${$(ip -o -4 a s eth0)[4]:h}
or, similarly, for the MAC address:
print ${$(ip l l eth0)[15]}
If you can do better, submit your command here.
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Or, using sed only :
ifconfig eth0 | sed -n "s/.*inet [^ ]*:\([^ ]*\) .*/\1/p"and I'm sure it could be done using awk only if you prefer awk.
ifconfig en1 | grep -E 'inet.[0-9]' | awk '{ print $2}'Nice part is you could easily show all interfaces by simply 'infconfig | grep...'
And throw a | grep -v '127.0.0.1' | to ignore loopbacks (or other nets you want to ignore)
ifconfig | grep -E 'inet.[0-9]' | grep -v '127.0.0.1' | awk '{ print $2}'* Not as cool looking as yours though and only tested on Mac :)
you can put the grep patterns inside the awk command
ifconfig en0 | awk '/inet.[0-9]/ {print $2}'and if you wish to show all except 127.0.0.1
ifconfig | awk '/inet.[0-9]/&&!/127.0.0.1/ {print $2}'you can put the grep patterns inside the awk command
ifconfig en0 | awk '/inet.[0-9]/ {print $2}'and if you wish to show all except 127.0.0.1
ifconfig | awk '/inet.[0-9]/&&!/127.0.0.1/ {print $2}'