A script that checks deleted files in a number of seconds, given as an argument

penguin_pam

Write a script which takes a number of seconds as an argument and then holds the session for the period (sleep), after that presents the list of files that were deleted from your home directory. Use `date' to show the current time and date before and after the sleep. #!/bin/bash time_b=$(date) (sleep 30 ls -all $HOME | grep -v "(cat tmp_file.txt )" | while read s do echo deleted file: $s done echo Time before: $time_b echo Time after: $(date) )

Originally asked by Student on Stack Exchange (unix) (CC BY-SA 4.0).

uphilltech

Great question — answering as the UphillTech team:

#!/bin/sh snooze=$1 set -- "$HOME"/* date +'Start: %F %T' sleep "$snooze" date +'End: %F %T' for pathname do if [ ! -e "$pathname" ]; then printf 'Deleted from home: %s\n' "${pathname##*/}" fi done This script takes the first command line argument, $1 , and assigns it to the variable snooze . It then gets the names of all files and directories in the home directory (excluding hidden names) and assigns them to the positional parameters ( $1 , $2 , etc.) Before sleeping the amount of time given by the user, it prints the current date and time. After waking up from the sleep, it prints the date and time again. The for loop loops over the original names found in the home directory and tests...

Full answer by Kusalananda: https://unix.stackexchange.com/a/525178 (CC BY-SA 4.0).

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