bash: variable outside function scope

Question

Is there a possible way to initialize a global variable in bash and assign it a value in a function, then use it out side the function scope?

Function example:

globla_var=""

_DBINFO()
{
  curl  -su $AUTH https://<balla bla >/databases | jq -c 'map(select(.plan.name != "Sandbox")) | .[] | {id, name}'| \
  while  read db
  do
    idb=$(echo "$db" | jq -r '.id')
    name=$(echo "$db" | jq -r '.name')
    if [[ $name = '<bla>' ]]; then
        $global_var_her = $(<bla value>)
     fi
  done
}

then use it outside the function:

 echo $global_var

the result

   $0: line 16: =<bla bla>: command not found

I tried using declare:

declare -r global_var

same results

In fact, variables in a function are global by default; local variables must be explicitly declared using either the local or declare builtins. — chepner
– chepner, Commented Oct 27, 2015 at 19:45
there is a typo at the beginning globla_var which should be global_var as I believe. — Ömer A.
– Ömer A., Commented Jul 24, 2021 at 8:48
@ÖmerAn Furthermore, _DBINFO() uses $global_var_her, not $global_var nor $globla_var. — avpaderno
– avpaderno, Commented Dec 31, 2024 at 12:41

that other guy · Accepted Answer · 2015-10-27 20:01:09Z

8

Yes you can, but you have to be careful about subshells that limit scope in unexpected ways. Piping to a while read loop like you are doing is a common pitfall.

Instead of piping to a while read loop, use redirection and process substitution:

_DBINFO()
{     
  while read db
  do
    idb=$(echo "$db" | jq -r '.id')
    name=$(echo "$db" | jq -r '.name')
    if [[ $name = '<bla>' ]]; then
        global_var=value
     fi
  done  <  <(curl  -su "$AUTH" "https://$host/databases" |
               jq -c 'map(select(.plan.name != "Sandbox")) | .[] | {id, name}')
}

AUTH="user:password"
host="example.com"
_DBINFO
echo "The global variable is $global_var"

You also need to make sure your assignment is syntactically valid. $var = value is not a valid bash assignment, while var=value is. shellcheck can point out many things like that.

edited Oct 27, 2015 at 20:01

answered Oct 27, 2015 at 19:54

that other guy

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Comments

whoan · Accepted Answer · 2024-07-05 06:32:23Z

Maybe you're confusing bash with php. Just remove the $, the _her part, and the space:

global_var=$(<bla value>) # or maybe: global_var=`<bla value>`

instead of:

$global_var_her = $(<bla value>)

With that, the error goes away.

On the other hand, check that other guy's answer regard to the pipelines.

And yes, it is possible to define a global variable and it is done the way you're doing it.

Maybe is useful pointing something about this:

$(<bla value>)

If <bla value> is a command, the sentence is ok, because that't the way to capture the output of a command with Command Substitution.

If, instead, is a literal value, just remove the $() leaving just '<bla value>' (global_var='<bla value>'). This is the same to do $(echo '<bla value>') but that would be an unnecessary waste of resources.

D. Schreier · Accepted Answer · 2020-06-18 22:51:09Z

3

This is already answered, but I'd give it a shorter solution

test() { declare x="Important output"; }

test && echo $x

Will leave us with no output, while the answer has a workaround, here's what I suggest to use

test() { export x="Important output"; }

test && echo $x

Will give us an output

edited Jun 18, 2020 at 22:51

D. Schreier

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answered Jun 18, 2020 at 11:08

Tigran

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1 Comment

Leonardo Dagnino Over a year ago

This does not address the problem at hand though - while you can read it, you still can't write to it in a way that propagates to the parent shell.

user2350426 · Accepted Answer · 2015-10-27 20:06:08Z

All variables have global scope, unless declared local inside some function.

So, all the vars in your program are valid outside the function.
But there are several problems:

You are sending the output of curl ... with a pipe |. That creates a sub-shell. Variables changed inside a sub-shell are not translated to global.
The name of the var was globla_var="". Is that a typo? Should it be global_var?.
There is an space before and after the = in: $global_var_her = "bla". That is incorrect syntax in bash, and in shell in general.
There is a $ on the left side of an equality. That does not assign to the variable global_var_her.
There is an additional _her which seems UN-needed.

All corrected, this should work:

global_var=""

_DBINFO(){

    content=$(
        curl  -su $AUTH https://<balla bla >/databases |
        jq -c 'map(select(.plan.name != "Sandbox")) | .[] | {id, name}'
    )

    while  read -r db
    do
        idb=$(echo "$db" | jq -r '.id')
        name=$(echo "$db" | jq -r '.name')
        if [[ $name = '<bla>' ]]; then
            global_var=$(echo "<bla value>")
        fi
    done <<<"$content"

}

Jan Święcki · Accepted Answer · 2023-01-22 14:07:38Z

0

Try declare -g. From declare help page:

-g  create global variables when used in a shell function; otherwise
    command.  The `-g' option suppresses this behavior.

f() {
    declare -g newvar="It works!";
}
f
echo $newvar;  # It works!

answered Jan 22, 2023 at 14:07

Jan Święcki

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