Bash if statement not catching

Question

Good afternoon,

im playing around with bash and wanted to write a script that would perform a traceroute and check if the second hop was a specific IP address. below is my script:

Failover_check= traceroute 8.8.8.8 -n | grep 192.168.0.2 | awk '{print  $2;}'

if [ $Failover_check = "192.168.0.2" ]
then
    echo "ip found"
else
    echo "ip not ound"
fi

every time I run the script it always hits the else statement

Thanks

See also http://stackoverflow.com/questions/2268104/bash-script-variable-declaration-command-not-found — tripleee, Jun 03 '16 at 13:59
And [useless use of `grep`](http://www.iki.fi/era/unix/award.html#grep) — tripleee, Jun 03 '16 at 14:02

score 3 · Answer 1 · answered Jun 03 '16 at 13:42

3

Assign it to the variable using $(...) and make sure no space before and after = that must avoid split and provide proper result

Failover_check=$(traceroute 8.8.8.8 -n | grep 192.168.0.2 | awk '{print  $2;}')

answered Jun 03 '16 at 13:42

SriniV

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Chris Maes · Accepted Answer · 2016-06-03T13:58:24.833

2

because the output of your command is not assigned to Failover_check variable. Use:

Failover_check=$(traceroute 8.8.8.8 -n | grep 192.168.0.2 | awk '{print  $2;}')

also; you might consider quoting your variables; to avoid errors when the variable is empty:

if [ "$Failover_check" = "192.168.0.2" ]

edited Jun 03 '16 at 13:58

answered Jun 03 '16 at 13:41

Chris Maes

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You beat me in seconds :P – SriniV Jun 03 '16 at 13:43
@realspirituals: too slow :) – Chris Maes Jun 03 '16 at 13:43
No need for double brackets or quoting the variable `if [ $Failover_check == "192.168.0.2" ]` should work fine. – m0skit0 Jun 03 '16 at 13:54
@m0skit0 quoting is for safety; otherwise I get "unary operator expected" errors. You're right about the double brackets; I'll adapt my answer. – Chris Maes Jun 03 '16 at 13:58

Bash if statement not catching

2 Answers2