Properly escaping asterisk/glob in (z)sh script

Question

I have the following Problem, I need an environment variable that contains globbing characters as clear text, and I can't seem to escape them. Escaping with backslash works, but for some reason the backslash is still part of the string."

Disabling GLOB_SUBST resolves the problem, but I´d like to avoid that.

I used the following to narrow down the problem:

touch foo; chmod +x ./foo

# glob expansion i.E. $test contains all files in the directory.
echo 'export test="*"; echo $test' > foo; source ./foo 

# no glob expansion, but the escape character is added to the string.
echo 'export test="\*"; echo $test' > foo; source ./foo  
# output: \*

Any help would be appreciated.

Simply quote the `test` variable `"$test"`? – Inian Dec 12 '21 at 15:10 — Inian, Dec 12 '21 at 15:10
.. or use `noglob echo $test`. – Kusalananda Dec 12 '21 at 18:06 — Kusalananda, Dec 12 '21 at 18:06

score 1 · Accepted Answer · answered Dec 12 '21 at 17:27

In the first case, you assign the string * to the variable, and since globsubst is enabled, the unquoted expansion $test is subject to globbing (as in a POSIX shell). In the second case, you assign the string \*, but it's not expanded as a glob since the backslash escapes the asterisk. (Otherwise zsh would complain that there were no matching files found, unless you've disabled that, too. You probably don't have filenames starting with a backslash anyway, but the glob to match them would have to be \\*.)

To stop the globbing, quote the variable expansion, i.e. echo "$test".

See: When is double-quoting necessary?

Also, you don't need export if you're just going to use the variables within the same shell (even between the main shell and the sourced script).

Properly escaping asterisk/glob in (z)sh script

1 Answers1