Q. I am getting "command not found" error when doing dynimic variable assignemet in Shell scripts. How can I resolve this issue?

 

Let use replicate this issue.

#!/bin/bash
VAR1=abc
i=0
VAR$i=$VAR1
echo "Value of VAR0 is $VAR0"

Save this file and execute this script.
bash abc.sh

abc.sh: line 4: VAR0=abc: command not found

Value of VAR0 is

If you see I am getting command not found error.

This is due to the line VAR$i=$VAR1, in which we are trying to substitute two variable at a time, one on RHS and other on LHS. This is not possible by default in Shell scripting. First we have to substitute "i" value then substitute $VAR1 value. To resolve this issue we have to use eval command which gives us second chance to execute a command.

Modified version of above script as below.

#!/bin/bash
VAR1=abc
i=0
eval VAR$i=$VAR1
echo "VAlue of VAR0 is $VAR0"

Output:

bash abc.sh

VAlue of VAR0 is abc

Hope this helps to resolve dynamic assignmet of variables.

 

Post Views: 8,638
The following two tabs change content below.
My Twitter profileMy Facebook profileMy Google+ profileMy YouTube channel

Surendra Anne

Mr Surendra Anne is from Vijayawada, Andhra Pradesh, India. He is a Linux/Open source supporter who believes in Hard work, A down to earth person, Likes to share knowledge with others, Loves dogs, Likes photography. He works as Devops Engineer with Taggle systems, an IOT automatic water metering company, Sydney . You can contact him at surendra (@) linuxnix dot com.