Single variable but multiple values

Question

I want to write a program which will check if soft links exist or not

#!/bin/bash
file="/var/link1"
if [[ -L "$file" ]]; then
    echo "$file symlink is present";
    exit 0
else
    echo "$file symlink is not present";
    exit 1
fi

There will be link2, link3, link4, link5.

Do I have to write the same script n number of times for n number of links or can this be achieved in one script?

Also I want to have the exit 0 and exit 1 so that I can use for monitoring purpose.

You want to return 0 or 1 depending on whether the link exists, but also want handle multiple links. I don't see how you can have both features. You probably need to modify your script to accept a command line argument for file so that you can reuse it. stackoverflow.com/q/192249/397817 — Stephen Kennedy
– Stephen Kennedy, Commented Feb 18, 2018 at 16:37
Instead of writing the script n times, use a for loop to go over the links — alamoot
– alamoot, Commented Feb 18, 2018 at 17:01
This seems very, very much like an XY problem -- you're asking how to do X (store multiple values in a variable), but what you really want is to do Y (reuse some code with different values). — Charles Duffy
– Charles Duffy, Commented Feb 18, 2018 at 19:46
It's also notable that anything much more complicated than for link in /var/link{1,2,3,4,5}; do [[ -L "$link" ]] || exit; done is almost certainly more complicated than it needs to be. (exit exits with $? by default, so [[ -L non-existing-file ]] || exit exits with status 1; similarly, the implicit exit at the end of a script likewise uses $?, so the implicit exit from a script whose last command did not fail will always have exit status 0). — Charles Duffy
– Charles Duffy, Commented Feb 18, 2018 at 21:15

Walter A · Accepted Answer · 2018-02-18 20:04:44Z

1

You can use a function:

checklink() {
   if [[ -L "$1" ]]; then
      echo "$1 symlink is present";
      return 0
   else
      echo "$1 symlink is not present";
      return 1
   fi
}

file1="/var/link1"
file2="/var/link2"
file3="/var/link3"
file4="/var/link4"

for f in "${file1}" "${file2}" "${file3}" "${file4}"; do
   checklink "$f" || { echo "Exit in view of missing link"; exit 1; }
done
echo "All symlinks checked"
exit 0

edited Feb 18, 2018 at 20:04

answered Feb 18, 2018 at 17:36

Walter A

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2 Comments

Charles Duffy Over a year ago

{echo is not { echo. And there's a reference to $file in the function that needs to be changed to $1.

Premkumar Waghmare Over a year ago

It can be achieved by the below as well #!/bin/bash file=$1 if [[ -L "$file" ]]; then echo "$file symlink is present"; exit 0 else echo "$file symlink is not present"; exit 1 fi And call this script as test.sh /usr/bin/ls test.sh /bin/find

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