[SOLVED] searching for file unix script

Issue

My script is as shown: it searches for directories and provides info on that directory, however I am having trouble setting exceptions.

if [ -d "$1" ];
then
  directories=$(find "$1" -type d | wc -l)
  files=$(find "$1" -type f | wc -l)
  sym=$(find "$1" -type l | wc -l)

  printf "%s %'d\n" "Directories" $directories
  printf "%s %'d\n" "Files" $files
  printf "%s %'d\n" "Sym links" $sym
  exit 0
else
  echo "Must provide one argument"
  exit 1
fi

How do I make it so that if I search for a file it tells me that a directory needs to be inputted? I'm stuck on it, I've tried test commands but I don't know what to do.

Solution

You're missing your shebang in the first line of your script:

#!/bin/bash

I get correct results from your script if I add it:

Directories 1,991
Files 13,363
Sym links 0

You may have to set the correct execution permissions also chmod +x scriptname.sh?

Entire script looks like this:

#!/bin/bash

if [ -z "$1" ];
        then
                echo "Please provide at least one argument!"
                exit 1
elif [ -d "$1" ];
        then
                directories=$(find "$1" -type d | wc -l)
                files=$(find "$1" -type f | wc -l)
                sym=$(find "$1" -type l | wc -l)
                printf "%s %'d\n" "Directories" $directories
                printf "%s %'d\n" "Files" $files
                printf "%s %'d\n" "Sym links" $sym
                exit 0
        else
                echo "This is a file, not a directory"
                exit 1
fi

Answered By - user14757127