use bash $HOME in shell script

Question

How to made bash to execute variable value. For example, we have this code, where variable value was set in single quotes(!).

#!/bin/bash
V_MY_PATH='$HOME'
echo "$V_MY_PATH"
ls $V_MY_PATH

The output is

$HOME
ls: $HOME: No such file or directory

How to made bash to translate shell variable insto its value if there is some.

I want to add some code after V_MY_PATH='$HOME' to make output like echo $HOME.

It's something simple, but i'm stuck. (NB: I know that with V_MY_PATH="$HOME", it works fine.)

EDIT PART: I just wanted to make it simple, but I feel that some details are needed.

I'm getting parameter from a file. This part works good. I don't want to rewite it. The problem is that when my V_MY_PATH contains a predefined variable (like $home) it's not treated like its value.

Answer 1

Remove the single quotes

V_MY_PATH='$HOME'

should be

V_MY_PATH=$HOME

you want to use $HOME as a variable

you can't have variables in single quotes.

Complete script:

#!/bin/bash
V_MY_PATH=$HOME
echo "$V_MY_PATH" 
ls "$V_MY_PATH"  #; Add double quotes here in case you get weird filenames

Output:

/home/myuser
0
05430142.pdf
4
aiSearchFramework-7_10_2007-v0.1.rar

etc.

Answer 2

use variable indirect reference so:

pete.mccabe@jackfrog$ p='HOME'
pete.mccabe@jackfrog$ echo $p
HOME
pete.mccabe@jackfrog$ ls ${p}
ls: cannot access HOME: No such file or directory
pete.mccabe@jackfrog$ ls ${!p}
bash                        libpng-1.2.44-1.el6      python-hwdata           squid
...
pete.mccabe@jackfrog$ 

The ${!p} means take the value of $p and that value is the name of the variable who's contents I wish to reference