On Wed, Apr 19, 2023 at 09:16:26PM +0200, lejeczek via CentOS wrote:
On 19/04/2023 08:46, wwp wrote:
Hello lejeczek,
...
Surround ${_Val} with double quotes (as you should) and things will be different:
$ unset _Val; test -n "${_Val}"; echo $?
1
Now you get it? :-)
I don't know, am not sure, I remembered it differently, did not think enclosing quotes were necessary(always?) for that were {}
{} does not prevent this (at least not in bash):
$ FOO="a b"
$ test -z $FOO
bash: test: a: binary operator expected
$ test -z ${FOO}
bash: test: a: binary operator expected
Because after $FOO or ${FOO} variable expansion, bash parsed:
test -z a b
'b' is unexpected, from a grammar point of view.
Quoting is expected, here:
$ test -z "$FOO"
<no error>
When FOO is unset, apparently it's a different matter, where you end up
with $?=0 in all unquoted -n/-z cases, interestingly. I could not find
this specific case in the bash documentation. That may not be portable
to other shells, BTW. I only use {} when necessary (because of what
bash allows to do between {}, plenty!, or when inserting $FOO into a
literal string that may lead the parser to take the whole string for a
variable name: echo $FOObar != echo ${FOO}bar).
Regards,
There is a several ways to run tests in shell, but 'test'
which is own binary as I understand, defeats me..
Yes, there is a binary for test (and its alternate '[').
But most shells, including bash have incorporated code for test
(and other commands) into the shell code itself for efficiency.
$ type test
test is a shell builtin
I'd expect a consistency, like with what I usually do to
test for empty var:
-> $ export _Val=some; [[ -v _Val ]]; echo $?
0
-> $ unset _Val; [[ -v _Val ]]; echo $?
1
I do hope you don't use -v to test for empty variables as
it tests for "set" variables and valid name syntax.
Set variables can be "empty" ( name= ).
But in your last example _Val is "un"set, it does not
exist. Thus it can neither be empty nor occupied.
--
Jon H. LaBadie jcu@xxxxxxxxxx
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