Create variable from string/nameonly parameter to extract data in bash?

Question

I want to save the variable name and its contents easily from my script.

Currently :-

LOGFILE=/root/log.txt
TEST=/file/path
echo \"TEST : ${TEST}\" &         


        

Answer 1

Consider using the printenv function. It does exactly what it says on the tin, prints your environment. It can also take parameters

$ printenv
SSH_AGENT_PID=2068
TERM=xterm
SHELL=/bin/bash
LANG=en_US.UTF-8
HISTCONTROL=ignoreboth
...etc

You could do printenv and then grep for any vars you know you have defined and be done in two lines, such as:

$printenv | grep "VARNAME1\|VARNAME2"
VARNAME1=foo
VARNAME2=bar

Answer 2

Yes, using indirect expansion:

echo "$1 : ${!1}"

Quoting from Bash reference manual:

The basic form of parameter expansion is ${parameter} [...] If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion

Answer 3

You want to use Variable Indirection. Also, don't use the function keyword, it is not POSIX and also not necessary as long as you have () at the end of your function name.

LOGFILE=/root/log.txt

save()
{
    echo "$1 : ${!1}" >> ${LOGFILE}
}

TEST=/file/path

save TEST

Proof of Concept

$ TEST=foo; save(){ echo "$1 : ${!1}"; }; save TEST
TEST : foo