Printing asterisk ("*") in bash shell

Question

a=5
echo "*/$aMin * * * * bash /etc/init.d/ckDskCheck.sh"

When I try to run the following code, it displays properly

*/5 * * * * bash /etc/init.d/ckDskCheck.sh

But when I try to assign the result using the following code to the variable and print it out, it displays as this:

a=5
cronSen=`echo "*/$a * * * * bash /etc/init.d/ckDskCheck.sh"`
echo $cronSen

Result:

So I try to escape the asterisk by

cronSen=`echo "\*/$a \* \* \* \* bash /etc/init.d/ckDskCheck.sh"`

But it still doesn't work. Why? How can I fix this?

Answer 1

You have two problems:

1. Useless Use of Echo in Backticks

2. Always quote what you echo

So the fixed code is

a=5
cronSen="*/$a * * * * bash /etc/init.d/ckDskCheck.sh"
echo "$cronSen"

It appears you may also have a Useless Use of Variable, but perhaps cronSen is useful in a larger context.

In short, quote everything where you do not require the shell to perform token splitting and wildcard expansion.

Token splitting;

 words="foo bar baz"
 for word in $words; do
      :

(This loops three times. Quoting $words would only loop once over the literal token foo bar baz.)

Wildcard expansion:

pattern='file*.txt'
ls $pattern

(Quoting $pattern would attempt to list a single file whose name is literally file*.txt.)

In more concrete terms, anything containing a filename should usually be quoted.

A variable containing a list of tokens to loop over or a wildcard to expand is less frequently seen, so we sometimes abbreviate to "quote everything unless you know precisely what you are doing".