Passing command line parameters through variables

Question

I'm trying to construct a command line by adding parameters with information from the "outside world" (namely, user input).

Say that I want to grep a file for a set of words that the user gives me (it is just to show you the use case, I'm really trying to run a bioinformatics code, but everybody have grep for testing purposes). The idea is that I will have a list of words in a bash array:

$ L=("foo" "bar")

And I want to use that list to end up with a line equivalent to

$ grep -e foo -e bar input.txt

I used an approach of generating the needes parameter string with

$ P=${L[@]/#/-e }

$ echo "$P"
-e foo -e bar

But I cannot use it as is (assuming a file that contains foo and bar):

$ grep "$P" input.txt
<No output>

Last command, actually searched for the string "-e foo -e bar".

If I run that command without the quotes, it works, but the it also triggers the SC2086 warning in shellcheck.net, which I'm trying to avoid.

I already tried several combinations of arrays, parenthesis, quotes... but I'm unable to nail the exact syntax for my needs. Anyone have that syntax?

PesaThe · Accepted Answer · 2018-08-20 13:05:36Z

4

If you quote the argument string, it will be interpreted as one long argument. If you don't quote it, then you rely on word-splitting which has many caveats and is not recommended.

You could build the arguments like this:

arg=()
arr=(foo bar)
for a in "${arr[@]}"; do arg+=(-e "$a"); done

grep "${arg[@]}" input

answered Aug 20, 2018 at 13:05

PesaThe

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