Concatenate arguments using a loop to form command in shell script

Question

I've tried different suggestions on this website and elsewhere but nothing has worked. I need to pass some arguments to a shell script and then concatenate them on a string to then launch it as a command. So I do this

command="perl perl_script.pl"

for arg
do

command+="$arg "

done

eval $command

But I get this error

bowtie_script_simple.sh: 43: bowtie_script_simple.sh: comando+=-n : not found
bowtie_script_simple.sh: 43: bowtie_script_simple.sh: comando+=3 : not found

According to other threads I've seen that should work. Any ideas?

Thanks

Interesting result because the example code given shows "command" the error shows "comando+=-n" which is not the same.... are you sure this is the exact code that can cause the error? — onaclov2000
– onaclov2000, Commented Nov 29, 2012 at 15:23
Oh yeah sorry I wrote it in english here but it's in spanish on my code. But the code is the same, it gives the error — Atirag
– Atirag, Commented Nov 29, 2012 at 15:25

Brian Agnew · Accepted Answer · 2012-11-29 15:24:50Z

2

command="$command $arg"

Do your args have (or are likely to have) whitespace ? You should probably quote further for safety's sake.

answered Nov 29, 2012 at 15:24

Brian Agnew

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1 Comment

Atirag Over a year ago

Yeah well some arguments go like this -n 3 but I figured shell will read them as separate arguments.

glenn jackman · Accepted Answer · 2012-11-29 16:12:24Z

2

Concatenating as a string introduces difficulties when handling args with spaces. With bash/zsh/ksh, using an array works better:

command=(perl perl_script.pl)
for arg; do
    command+=("$arg")
done

# or more simply
# command=(perl perl_script.pl "$@")

# now, execute the command with each arg properly quoted. 
"${command[@]}"

From you error message, it looks like you're using /bin/sh -- that shell does not have a var+=string construct -- have to use a shell with more features.

answered Nov 29, 2012 at 16:12

glenn jackman

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Comments

Faiz · Accepted Answer · 2012-11-29 15:35:30Z

1

I think this should work:

command="perl perl_script.pl"

for arg in $@
do
  command="$command $arg"
done

$command

answered Nov 29, 2012 at 15:35

Faiz

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