bash passing arguments/parameters?

Question

#!/bin/bash

traverse() {
local x=$1
if [ -d $x ]
then
lst=(`ls $x`)
for((i=${#lst[@]}; --i;)); do
    echo "${lst[i]}"
    done
else echo "not a directory"
fi
}

traverse

I want to pass a parameter such as "/path/to/this/directory/" when executing program but only works if I'm running the program in the same directory as my bash script file and any other parameter I pass is completely ignored.

the script is supposed to take a parameter and check if it's a directory and if it's a directory then list all the files/folders in descending order. If not display error message.

What is wrong with code, thanks!

that other guy · Accepted Answer · 2014-02-13 18:21:33Z

3

This happens because $1 in the function refers to traverse's parameters, not your script's parameters.

To run your function once with each argument, use

for arg in "$@"   # "$@" is also the default, so you can drop the 'in ..'
do
  traverse "$arg"
done

If you in the future want to pass all the script's parameters to a function, use

myfunc "$@"

This is just the problem at hand, though. Other problems include not quoting your variables and using command expansion of ls, lst=(`ls $x`), instead of globs, lst=( "$x"/* )

answered Feb 13, 2014 at 18:21

that other guy

125k12 gold badges187 silver badges214 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user3016065 Over a year ago

@[that other guy] thanks very helpful information! now runs as I expected!

anubhava · Accepted Answer · 2014-02-13 18:22:15Z

3

You don't need to call ls for that. You can use this code:

traverse() {
    local x="$1"
    if [ -d "$x" ]; then
        arr=( "$x/"* )
        for ((i=${#arr[@]}; i>0; i--)); do
           echo "${arr[$i]}"
        done
    else
        echo "not a directory"
    fi
}

answered Feb 13, 2014 at 18:22

anubhava

790k67 gold badges603 silver badges671 bronze badges

Comments

glenn jackman · Accepted Answer · 2014-02-13 18:34:00Z

0

"That other guy" has the right answer. The reason it always looks at the current directory:

you invoke traverse with no arguments
the $1 in the traverse function is empty, therefore $x is empty
the test is therefore [ -d ], and when [ is given 1 argument, it returns success if the argument is not empty. Your if command always executes the "true" block and ls $x is just ls when x is empty

Use [[ ... ]] with bash: it is smarter about empty arguments. Otherwise, quote your variables:

$ x=; [ -d $x ] && echo always true || echo not a directory
always true
$ x=; [[ -d $x ]] && echo always true || echo not a directory
not a directory
$ x=; [ -d "$x" ] && echo always true || echo not a directory
not a directory

answered Feb 13, 2014 at 18:34

glenn jackman

249k42 gold badges233 silver badges362 bronze badges

Collectives™ on Stack Overflow

bash passing arguments/parameters?

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related