Filtering IP address from ifconfig usefule in scripts
IPADDR=`ifconfig | grep -i inet | awk -F: '{print $2}'| awk '{print $1}'`
echo $IPADDR
193.122.18.101
2 Alternatives + Submit Alt
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0/sbin/ifconfig eth0 | grep 'inet addr:' | cut -d: -f2 | awk '{ print $1}'
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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]}
0print ${$(ifconfig wlan0)[6]}
What do you think?
Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?
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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 :)ifconfig en0 | awk '/inet.[0-9]/ {print $2}'and if you wish to show all except 127.0.0.1ifconfig | awk '/inet.[0-9]/&&!/127.0.0.1/ {print $2}'ifconfig en0 | awk '/inet.[0-9]/ {print $2}'and if you wish to show all except 127.0.0.1ifconfig | awk '/inet.[0-9]/&&!/127.0.0.1/ {print $2}'