In all of the cases above, the variable is correctly set, but not correctly read! The right way is to use double quotes when referencing:

echo "$var"

This gives the expected value in all the examples given. Always quote variable references!

Why?

When a variable is unquoted, it will:

1. Undergo field splitting where the value is split into multiple words on whitespace (by default):

   Before: /* Foobar is free software */

   After: /*, Foobar, is, free, software, */

2. Each of these words will undergo pathname expansion, where patterns are expanded into matching files:

   Before: /*

   After: /bin, /boot, /dev, /etc, /home, ...

3. Finally, all the arguments are passed to echo, which writes them out separated by single spaces, giving

   /bin /boot /dev /etc /home Foobar is free software Desktop/ Downloads/
   

   instead of the variable's value.

When the variable is quoted it will:

1. Be substituted for its value.
2. There is no step 2.

This is why you should always quote all variable references, unless you specifically require word splitting and pathname expansion. Tools like shellcheck are there to help, and will warn about missing quotes in all the cases above.

Remove the single quotes:

$ testvar="actualvalue"
$ echo testing "${testvar}", "testing", "testing" ;
testing actualvalue, testing, testing

The single-quote inhibits variable expansion.

Without double-quotes gives you the same output:

$ echo testing ${testvar}, testing, testing ;
testing actualvalue, testing, testing

But if you really want double-quotes in the output, escape them:

$ echo "testing \"${testvar}\", \"testing\", \"testing\" ;"
testing "actualvalue", "testing", "testing" ;

The shell expands the variables before it runs the command, even before it runs the assignments.

Cf.

text=blah ; echo $text

or

text=blah eval 'echo $text'

The first one works because the assignment is run as a separate command. The second one works because $text is single-quoted which prevents its expansion when shell is processing the command; when eval is later running, the variable already has the value assigned.

I have a bash script that connects to a Cisco router using telnet and sends a command. i want to capture the output of the command into a variable that i can referrence to in the script. below is the script and how it looks when i run it, i want to capture the out of the clock, or if that is not possible, can i capture the whole output of the telnet to a variable?

#!/bin/bash

Coonect via telnet and Send command a command

(
echo open “10.0.1.1”
sleep 2
echo “cisco”
echo “cisco”
echo “enable”
sleep 2
echo “cisco”
echo “show clock” # This is the command
sleep 2
) | telnet

[ Wrote 19 lines ]

[admin@BIG-IP-01:Active:Standalone] ~ #
[admin@BIG-IP-01:Active:Standalone] ~ #
[admin@BIG-IP-01:Active:Standalone] ~ # ./test
telnet> Trying 10.0.1.1…
Connected to 10.0.1.1.
Escape character is ‘^]’.

User Access Verification

Username: cisco
Password:
Server_1>enable
Password:
Server_1#show clock
*12:03:00.919 UTC Fri Feb 10 2023
Server_1#Connection closed by foreign host.
[admin@BIG-IP-01:Active:Standalone] ~ #