check if variable is set and grep

Question

i have three variables $a, $b, $c. I don't know whether the three variables are set. I want the variable in the GREP query only if the variables are set. How do i do this?

find . -iname "*.txt" -type f | xargs grep -inw "$a" -sl | xargs grep -inw "$b" -sl |  xargs grep -inw "$c" -sl
find .* -iname "*.txt" -type f | xargs grep -iw "$a|$b|$c" -sl

Please edit your question to describe your overall goal. Perhaps you're going about it all wrong. — ooga
– ooga, Commented Jun 4, 2014 at 16:51
the second line is a query that i need to use in my script somewhere else. — corbett
– corbett, Commented Jun 4, 2014 at 17:00

konsolebox · Accepted Answer · 2014-06-04 17:14:12Z

1

You can prepare multiple -e arguments on an array:

args=()
for x in "$a" "$b" "$c"; do
    [[ -n $x ]] && args+=(-e "$x")
done
[[ ${#args[@]} -gt 0 ]] && find . -iname "*.txt" -type f | xargs grep -iw "${args[@]}" -sl

Note: Having -e "$a" -e "$b" -e "$c" is practically synonymous to "($a|$b|$c)" and might be even safer. Also if you don't intend "$a", "$b", and "$c" to be parsed as regex, you can just use fgrep or add the option -F; that which can't be done with "($a|$b|$c)".

edited Jun 4, 2014 at 17:14

answered Jun 4, 2014 at 16:54

konsolebox

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2 Comments

ooga Over a year ago

-n makes no sense in conjunction with -l.

ooga Over a year ago

Nope. You copied it exactly as he had it, even splitting the options into two groups for no reason.

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