Script in UNIX (Bash) to determine if the user is the owner of a file

bashscriptunix

Something that would take any number of arguments, where each argument would be a path to a file or directory. If the user does own a particular path then it should check it to see if the path represents a normal file AND if that file is executable. If it is, then your script should execute/run the file.

Thanks for any help I've been trying different things for too long and I'm frustrated.

Best Answer

magic() {
    for p in "$@"; do
        [ -O "$p" -a -x "$p" ] && /bin/sh "$p"
    done
}

read 'man test' to see what the checks do.

Related Solutions

How to write a script to accept any number of command line arguments? (UNIX, BASH)

One possible way to do what you want involves $@, which is an array of all the arguments passed in.

for item in "$@"; do
  if [ -O "$item" ]; then
    echo "you are the owner of $item"
  else
    echo "you are not the owner of $item"
  fi
done

Breaking a file down of strings, into separate files each based on the first letter. BASH

Try this

OLDIFS=$IFS
IFS='
'
typeset -a file
file=($(cat list.txt))
for i in "${file[@]}"; do
    echo $i >> ${i:0:1}.txt
done
IFS=$OLDIFS

Note, the IFS part is not usually necessary. Also, I tested it on Zsh 4.3.17 on linux and on Bash 4.2.37.

What it does is it declares an array, assigns the contents of the file to that array, then loops over each element of the array, hence each line and echo's that element into the file with the name of the first lettes plus '.txt' appended to it.

Best Answer

Related Solutions

How to write a script to accept any number of command line arguments? (UNIX, BASH)

Breaking a file down of strings, into separate files each based on the first letter. BASH

Related Question