How to enumerate and print out the variables in bash?

Question

Is there a way to enumerate all the variables in a bash script and print out their values?

If that is not possible is there a way to print out the value of avariable give its name, ie from a string like 'ABC' print the value of $ABC if it has been assigned?

All the variables directly used by a script or all the variables available in the shell session? — Etan Reisner
– Etan Reisner, Commented Mar 24, 2015 at 11:37
Those used in the script, but if those available in the shell session can also be listed that will also be good. — vfclists
– vfclists, Commented Mar 24, 2015 at 11:39
The latter is easier. The former is somewhat impossible. That being said there isn't a way to "inspect" a script for the variables it will use (or expects to have set before being run or anything like that). — Etan Reisner
– Etan Reisner, Commented Mar 24, 2015 at 11:43

anubhava · Accepted Answer · 2015-03-24 11:43:30Z

2

You can use compgen -v builtin:

while read -r line; do echo "$line=${!line}"; done < <(compgen -v)

Fore more details use:

help compgen

To list all variable created by your scripts (.bashrc etc):

{ set -o posix; set; }

edited Mar 24, 2015 at 11:43

answered Mar 24, 2015 at 11:34

anubhava

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3 Comments

anubhava Over a year ago

Yes both snippets will work from inside the script also.

vfclists Over a year ago

Is there a way to restrict the output to only the variables in the script?

anubhava Over a year ago

Did you try: { set -o posix; set; } from your script?

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