Wildcard in variable, Shell bash, how to do ? Variable=$variable2*

Question

For me it worked:

diretorio=$(echo 'test 123'*)

but not worked when i used variable in quotes

Var2="test 123" 
diretorio=$(echo '$Var2'*)

How to solve it?

why on earth are you putting a concatenated list of filenames into a variable in the first place? — ormaaj, Mar 15 '13 at 00:00
Another interesting mess: `a="*"; echo $a; a="'*'"; echo $a` -- no easy way to disable wildcard expansion without introducing other issues, think `if [ -n "$1" ]; then a="-name $1";fi; find /dir $a` (Arrays also works for that case...) — Gert van den Berg, Jun 17 '16 at 07:03

score 6 · Answer 1 · answered Mar 14 '13 at 22:53

The mistake in your glob is that

diretorio=$(echo '$Var2'*)

is a shot in /dev/null, because the shell don't expand variables in single quotes.

So :

diretorio=$(echo "$Var2"*)

Learn the difference between ' and " and `. See http://mywiki.wooledge.org/Quotes and http://wiki.bash-hackers.org/syntax/words

score 4 · Accepted Answer · answered Mar 15 '13 at 02:39

May I suggest an alternate approach? Instead of making a space-separated list of filenames (which will cause horrible confusion if any of the filenames contain spaces, e.g. "test 123"), use an array:

diretorio=("${Var2}"*)
doSomethingWithAllFiles "${diretorio[@]}"
for umDiretorio in "${diretorio[@]}"; do
    doSomethingWithASingleFile "$umDiretorio"
done

score 1 · Answer 3 · answered Mar 14 '13 at 22:53

1

Use double quotes:

diretorio=$(echo "$Var2"*)

Single ones prevent variable substitution

answered Mar 14 '13 at 22:53

jlliagre

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@sputnick indeed, although not yet propagated to my browser when I was writing my reply. – jlliagre Mar 14 '13 at 23:01

Wildcard in variable, Shell bash, how to do ? Variable=$variable2*

3 Answers3