Multiple command output into variable Bash script with $()

Question

I reviewed the multiple threads on this and am still having issues, here is the command I\'m trying to execute. Commands without the $() print the desired output t

Answer 1

Spaces Not Legal in Variable Assignments

Variable assignments must not have spaces between the variable name, the assignment operator, and the value. Your current line says:

MODEL3= $(/usr/sbin/getSystemId | grep "Product Name" | awk '{print $4}')

This actually means "run the following expression with an empty environment variable, where MODEL3 is set but empty."

What you want is an actual assignment:

MODEL3=$(/usr/sbin/getSystemId | grep "Product Name" | awk '{print $4}')

Answer 2

That is correct:

MODEL3=$(/usr/sbin/getSystemId | grep "Product Name" | awk '{print $4}')

But you can write the same without grep:

MODEL3=$(/usr/sbin/getSystemId | awk '/Product Name/{print $4}')

Now you have the result in the MODEL3 variable and you can use it further as $MODEL3:

echo "$MODEL3"