[SOLVED] problem creating an array from a function in bash

mia_tech · 03-20-2014, 04:39 PM

I have an array and a function which list all files in a directory and initializes the array; however, that method is not working. Aparently b/c the variable is local to the function. How can I make it global, or what would be the correct way to pass it to the function without using return

Code:

list_files() {
 files=($(ls | grep txt 2>/dev/null))
 if [ $? -ne 0 ]; then
  echo "error...."
  exit 1
}

files=""
list_files

by the way, I know this will work just fine

Code:

files=($(ls | grep txt 2>/dev/null))

I wanted to use a function which output an error in case of an empty directory

grail · 03-20-2014, 08:43 PM

Quote:

I have an array and a function which list all files in a directory and initializes the array; however, that method is not working. Aparently b/c the variable is local to the function. How can I make it global, or what would be the correct way to pass it to the function without using return

Please explain what you are getting and how it differs from what you want?
Also, if you place set -xv in your script it will show you what the interpreter is doing.

Lastly, there is no need for grep and ls should not be used for parsing (see here).
Just use simple globbing:

Code:

files=( *.txt )

ntubski · 03-21-2014, 09:17 AM

Quote:

Originally Posted by grail

Just use simple globbing:

Code:

files=( *.txt )

+1

And use shopt -s failglob to get an error if there are no .txt files.