bash arguments in a variable and then loop through

Question

I want DIR to loop through all the arguments (with space and ""). But this code do not recognize, "cd ef" as one. Dividing them into two. How can I do that?

#test.sh
echo Number of arguments: $#

if [ $# -eq 0 ]
then
  echo "No arguments"
  DIR=.
else 
  DIR="$*"
fi

echo DIR =$DIR
for arg in $DIR;
do 
  echo $arg;
done

Command

#bash test.sh ab "cd ef"

Output, Number of arguments: 2

But, enumerating 3 elements in the loop as ab, cd, and ef.

I know, I can do,

for arg in "$*";

But that not the solution in this case. Can I use DIR for this purpose? The platform is OSX Mavericks shell.

$* is almost never appropriate for use -- it concatenates arguments with the first character of IFS, meaning that if any argument happens to contain the first character of IFS, it's impossible to distinguish between the characters delimiting the arguments and the characters within the arguments. — Charles Duffy
– Charles Duffy, Commented Sep 3, 2014 at 15:08
Also, non-exported, non-builtin variables should have at least one lower-case variable to avoid potential namespace conflicts with environment variables and builtins. — Charles Duffy
– Charles Duffy, Commented Sep 3, 2014 at 15:09
BashFAQ #50, mywiki.wooledge.org/BashFAQ/050 is (tangentially) relevant, inasmuch as it addresses many of the same misunderstandings that result in using scalar variables where they aren't appropriate. — Charles Duffy
– Charles Duffy, Commented Sep 3, 2014 at 15:11

Charles Duffy · Accepted Answer · 2014-09-03 15:12:22Z

4

You need to use an array, not a regular scalar variable, for this purpose -- and "$@", not $*:

if (( $# )); then
  dirs=( "$@" )
else
  dirs=( . )
fi

for dir in "${dirs[@]}"; do
  ...
done

Comments

chepner · Accepted Answer · 2014-09-03 15:52:48Z

1

If you don't actually need a variable DIR, you can use a form of parameter expansion:

echo "DIR=${@:-.}"
for dir in "${@:-.}"; do
   ...
done

answered Sep 3, 2014 at 15:52

chepner

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