bash: variable outside function scope

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.

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.