How to store command output to variable which has asterisk(*)?

Question

I am trying capture command output to a variable but asterisk(*) characters automatically gets converted into directory files.

Command I want to run

secrets=$(vault read kv/data/secrets)

Expression to recreate the scenario

cronExpression=$(echo "33 * * *") && echo $cronExpression

Expected

cronExpression="33 * * *"

Actual

33 LICENSE README.md test.json LICENSE README.md test.json LICENSE README.md test.json

I have LICENSE README.md test.json files in a folder where I am trying to execute this command so each * gets converted into list files in the same directory

Gilles Quénot · Accepted Answer · 2023-03-05T07:46:23.357

1

Like this:

cronExpression="$(echo "33 * * *")" && echo "$cronExpression"
33 * * *

Learn how to quote properly in shell, it's very important :

"Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[@]}", "a & b". Use 'single quotes' for code or literal $'s: 'Costs $5 US', ssh host 'echo "$HOSTNAME"'. See
http://mywiki.wooledge.org/Quotes
http://mywiki.wooledge.org/Arguments
http://wiki.bash-hackers.org/syntax/words
when-is-double-quoting-necessary

edited Mar 05 '23 at 07:46

answered Mar 05 '23 at 07:30

Gilles Quénot

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Thanks for your answer. I have added `GLOBIGNORE=*` to fix the issue. – Thamaraiselvam Mar 05 '23 at 07:56
1

There's no need of this variable with proper quoting – Gilles Quénot Mar 05 '23 at 07:59
1

@Thamaraiselvam That breaks places where you want wildcard expansion to work, and doesn't avoid problems with word splitting. Double-quoting variable references *works*. – Gordon Davisson Mar 05 '23 at 08:10
Consider accepting answer if it fits your needs – Gilles Quénot Mar 05 '23 at 16:56

How to store command output to variable which has asterisk(*)?

1 Answers1