BASH Wild-card : Select all variables (not the content)

Question

I have a bunch of variables that I want to check, and if they contain the value "None" then I want to empty them.

    var1=(check for some value, sometimes it returns "none")
    var2=(check for some value, sometimes it returns "none")
    var3=(check for some value, sometimes it returns "none")
    someBizzareName=(check for some value, sometimes it returns "none")

    if [[ "${var1}" == "None" ]] ; then
        var1=""
    fi
    if [[ "${var2}" == "None" ]] ; then
        var2=""
    fi

And this is all working fine and dandy, only since I have a lot of varN, I will end up with a ton of if [[ "${varN}" == "None" ]] and I have to know their names ; so I was wondering, since it's in BASH very nature to search and match everything, if there is a wild-card for variables, inside a for loop, that will match all vars, something like ${*} (I tried that, does'nt work) ? I have done all kinds of searches but always find something about matching variable content, not the var itself..?

score 8 · Accepted Answer · answered May 12 '11 at 10:15

8

All, no. But you can match a pattern (but not *).

$ echo "${!B*}"
BASH BASHOPTS BASHPID BASH_ALIASES BASH_ARGC BASH_ARGV BASH_CMDS BASH_COMMAND BASH_LINENO BASH_SOURCE BASH_SUBSHELL BASH_VERSINFO BASH_VERSION

answered May 12 '11 at 10:15

Ignacio Vazquez-Abrams

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Hey this works :) It looks like regular [regexp](http://www.regular-expressions.info/) syntax, so something like `[A-Z][0-9]` should work ; **and** it's local to my for loop ; Thanks ! – yPhil May 12 '11 at 10:48
Ah, thank you Ignacio. Makes my life easier on searching how can I match the end of the string instead of the beginning. – yPhil May 12 '11 at 11:03
Ah, [found it](http://www.gnu.org/software/bash/manual/bash.html#Shell-Parameter-Expansion) :) – yPhil May 12 '11 at 11:07
What's the purpose of the ! (3rd char, before the B) in the quoted expression? Also how do you replace * with numbers... say single digit 1-9 doesn't work with "{!B[1-9]}" – niken May 03 '17 at 14:46
@Nik: The "!" tells it to look for variables with that prefix. The "*" is not negotiable. – Ignacio Vazquez-Abrams May 03 '17 at 15:04
1

How to get suffix expansion? I can only find `${!prefix*}` – Bhargav Katkam Nov 21 '17 at 08:14
@BhargavKatkam: You don't. – Ignacio Vazquez-Abrams Nov 21 '17 at 08:15
1

Any other solution to get variables with a given suffix? – Bhargav Katkam Nov 21 '17 at 08:22

score 2 · Answer 2 · answered May 12 '11 at 11:28

2

You may also use the builtin compgen:

man bash | less -p 'compgen .option. .word.'
compgen -A variable B

answered May 12 '11 at 11:28

jim

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1

shellter · Answer 3 · 2011-05-12T11:11:58.520

1

All, yes ;-)

Most Unix/Linux support either the env or printenv which produce output like

 var=value

The export command, without arguments, will list all exported variables in your environment.

for varAndVal in $( env ) ; do
   case ${varAndVal} in
     *=none ) 
      eval \$${varAndVal}=
      #OR eval unset \$${varAndVal}
     ;;
    esac
 done

I hope this helps.

P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, and/or give it a + (or -) as a useful answer.

edited May 12 '11 at 11:11

answered May 12 '11 at 10:28

shellter

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This does not quite do what I want ; It matches (really just lists them) environment vars, when I want not only to match vars local to my script (obviously) but if possible, local to my for loop, just to be safe – yPhil May 12 '11 at 10:40
Yes, sorry, missed the part about for loop. Good luck! – shellter May 12 '11 at 11:15

BASH Wild-card : Select all variables (not the content)

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