Bash shell script to find out the missing file/newly added file in a directory

Question

This program should run everyday so that if any file is missed or added, I can get a list of all those.

Please some 1 suggest possible way.

Answer 1

write your own diff script like this

#!/bin/bash

#The first time you execute the script it create old_list file that contains your directory content
if [[ ! -f old_list ]] ; then
   ls -t1  > old_list ;
   echo "Create list of directory content" ;
   exit
fi
#Create new file 'new_list' that contains new directory content
ls -t1  > new_list

#Get a list of modified file (created and/or deleted)
MODIFIED=$(cat old_list  new_list | sort | uniq -u)

for i in $MODIFIED ;
do
    EXIST=$(echo $i | grep old_list)
    #if exist in old_list so its newly deleted
    if [[ ! -z "$EXIST" ]] ; then
       echo "file : $i deleted"
    else
       echo "file $i added"
    fi
done

#Update old_content content
ls -t1  > old_content ;
exit

Answer 2

Here you have a small script that will do what you want. It's dirty and has almost no checks, but it will do what you want:

#!/bin/bash

# Directory you want to watch
WATCHDIR=~/code
# Name of the file that will keep the list of the files when you last checked it
LAST=$WATCHDIR/last.log
# Name of the file that will keep the list of the files you are checking now
CURRENT=/tmp/last.log

# The first time we create the log file
touch $LAST

find $WATCHDIR -type f > $CURRENT

diff $LAST $CURRENT > /dev/null 2>&1

# If there is no difference exit
if [ $? -eq 0 ]
then
    echo "No changes"
else
    # Else, list the files that changed
    echo "List of new files"
    diff $LAST $CURRENT | grep '^>'
    echo "List of files removed"
    diff $LAST $CURRENT | grep '^<'

    # Lastly, move CURRENT to LAST
    mv $CURRENT $LAST
fi

Answer 3

Many ways to implement it..Psedo code:

ls -ltrh | wc -l  

gives you number of files/folders in the current directory

Check the number every day/as per your requirement and compare the value. Put the shell script in a cronjob as you need to run it every day

Answer 4

My first naive approach would be using a hidden file to track the list of files.

Suppose the directory contains three files at the beginning.

a
b
c

Save the listing in a hidden file:

ls > .trackedfiles

Now a file d is added:

$ ls
a b c d

Save the previous state and record the new state:

mv .trackedfiles .previousfiles
ls > .trackedfiles

To show the differences, you can use diff:

$ diff .previousfiles .trackedfiles
3a4
> d

Diff shows that d is only in the right file.

If files were both added and removed, diff would show something like this:

$ diff .previousfiles .trackedfiles
2d1
< b
3a3
> d

Here, b was removed and d was added.

Now you can grep the lines in the output to determine which files where added or removed.

It is probably a good idea to sort the output of ls above, to make sure the list of files is always in the same order. (Simply write ls | sort > .trackedfiles instead.)

One drawback of this approach is that hidden files are not covered. This could be solved by a) listing hidden files, too, and b) excluding the tracking file from the process.

Like mentioned in another answer, the script can be run by a cron job periodically.

Answer 5

You can setup a daily cron job that checks the birth time of all files in the directory against the current date output to see if it was created in the last 24 hours.

current=$(date +%s)
for f in *; do
    birth=$(stat -c %W "$f")
    if [ $((current-birth)) -lt 86400 ]; then
        echo "$f" >> new_files
    fi
done