bash command fails when used with variables

Question

I use a Linux (Ubuntu) bash. When I ping an address dirrectly - all work fine. But when I use it with a variable that gets it's value from calculation - it fails.

what am i doing wrong?

network=`ip r | grep default | awk '{print $3;}'`
echo $network   # prints: 10.0.0.138
ping $network   # prints: ping: 10.0.0.138: Name or service not known (ERROR???)
ping 10.0.0.138 # prints: PING 10.0.0.138 (10.0.0.138) 56(84) bytes of data. (OK)

There's probably some sort of nonprinting character in $network causing trouble. Rather than echo $network, try printf %s "$network" | xxd and see what that shows. — Gordon Davisson
– Gordon Davisson, Commented Aug 1, 2022 at 8:57
00000000: 1b5b 3335 6d31 302e 302e 302e 3133 38 .[35m10.0.0.138 — DuduArbel
– DuduArbel, Commented Aug 1, 2022 at 9:00

DuduArbel · Accepted Answer · 2022-08-01 09:20:13Z

2

it's becaues the result is colored... ipis aliased to ip -c

so - this had solved the issue:

 network=`/usr/bin/ip r | grep default | awk '{print $3;}'`

answered Aug 1, 2022 at 9:20

DuduArbel

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user3811862 · Accepted Answer · 2022-08-01 08:29:25Z

-1

That sequence should work OK at the command line (it'd be interesting to see a screenshot). However you can lose the grep and use a more rigorous awk script:

defaultgw=$(ip r | awk '{if ($1=="default") print $3}')

Then just send a single echo-request without any name resolution:

ping -n -c 1 $defaultgw

answered Aug 1, 2022 at 8:29

user3811862

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2 Comments

DuduArbel Over a year ago

dudu@dudu-pc:~$ defaultgw=$(ip r | awk '{if ($1=="default") print $3}') dudu@dudu-pc:~$ ping -n -c 1 $defaultgw ping: 10.0.0.138: Name or service not known

DuduArbel Over a year ago

it should, but it doesn't

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