Bash script variable dropping output

Question

I run a command at the command line and it works fine. I have added the same command to a bash script and it cuts the output. I have made sure the quotes are correct and I can't seem to see where I am going wrong.

This is the command when I run it manually:

./ciscocmd -u username -p password -s passwowrd -t 192.168.1.1 -c "show mac address-table address 34e6.xxxx.xxxx"

and it shows

Processing... 192.168.1.1
switch#show mac address-table address 34e6.xxxx.xxxx
Mac Address Table
-------------------------------------------

Vlan    Mac Address       Type        Ports
----    -----------       --------    -----
10    34e6.xxxx.xxxx    DYNAMIC     Gi0/2
Total Mac Addresses for this criterion: 1

This is in the script:

portis=`./ciscocmd -u username -p password -s passwowrd -t 192.168.1.1 -c "show mac address-table address 34e6.xxxx.xxxx"`
echo $portis

but yet it outputs this

[root@server]# ./test.sh
switch#ddresses for this criterion: 1--4e6.xxxx.xxxx
[root@server]#

I must be missing something. I have tried double "" and '' to surround the mac address but its not working.

Any suggestions?

Answer 1

1. don't use echo for arbitrary data
2. quote your variables

So:

printf '%s\n' "$portis"

3. Avoid `...`. Use $(...) instead.