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how do you supply a literal asterisk * when specifying a filename to a command substitution in bash? Consider the following directory:

host:/tmp/backtick-test# ls -l
total 0
-rw-r--r--  1 root  wheel  0 29 Jun 23:18 test
-rw-r--r--  1 root  wheel  0 29 Jun 23:18 test*

I can't work out how to specify the second file test* with $().

List both files:

host:/tmp/backtick-test# ls test*
test  test*

List only the file with the trailing *:

host:/tmp/backtick-test# ls test\*
test*

Now with commmand expansion:

host:/tmp/backtick-test# echo $(ls test*)
test test test*

Returns three filenames.

/tmp/backtick-test# echo $(ls -al "test*")
-rw-r--r-- 1 root wheel 0 29 Jun 23:18 test test*

Better, returns two but I only want the second one.

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  • 7
    Your $(ls -al "test*") may be just an example, but if you were thinking of using the output of ls to construct a command line, read this first: Don't Parse ls - in short, use find instead. Also look into using arrays to store any arguments you want to pass to programs, it avoids all issues with word-splitting. e.g. readarray -d '' -t myarray < <(find ... -print0) Commented Jun 29 at 15:02
  • The answer here depends on if the command within the command substitution prints only a single filename (in that case, see unix.stackexchange.com/questions/68694/…), or if it prints multiple filenames (meaning the output would need to be NUL-separated (as cas showed above) to deal with arbitrary filenames, or at least you'd need to modify IFS and disable globbing to deal with a newline-separated list) Commented Jun 29 at 16:28
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    Obviously using echo to print a command substitution is quite redundant, and using ls with a glob is also unnecessary if you only want the filenames (and none of any possible metadata), so this smells of an XY problem. What is it you actually want to achieve? Commented Jun 29 at 17:20

1 Answer 1

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You need to escape or quote the asterisk in the command substitution, and quote the expansion itself (with double quotes which still allow expansions to be performed within but prevent split+glob to be performed over the result):

echo "$(ls -d 'test*')"

Thus ls will only list test* (and with -d, not its contents if that file happens to be of type directory), and the result won’t undergo word splitting + pathname expansion (where test* is split on characters of $IFS, here staying test* with the default value of $IFS and each resulting word processed as a glob, resulting in test and test* passed to echo which it prints space separated).

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