Question

In UNIX

Write a script that checks each minute and reports on who logs in and who logs out.

You can follow the steps below (not the only way):

Using the commands who and cut, extract the list of usernames currently logged in the system

To check after a minute, the sleep command can be used: sleep 60

Get the new list of users logged in after a minute

For each user in the list;

check if he is in the second list using if and grep. You can dump the output of grep to /dev/null when using it in the as a test for the if statement

If the user is not in the second list, print that the user has logged out

For each user in the second list,

If he is in the first list, print that the user has logged in

Rename the second list as the first list

Repeat the process from step 2

Answer 1

Answer:

Below is algo used for bash script:

Steps:

0 - Infinite while loop :

1 - save current "who" output into a file "A"

2 - wait for configured seconds (10)

3 - move file "A" to "B" (new file)

4 - save current "who" output into file "A" (overwrite)

5 - detect changes "diff" of files and output to console in formatted

6 - ctrl-c to quit

CODE:

#################################################################################################

fileOne="/tmp/a.md"

fileTwo="/tmp/b.md"

clear

while true; do

echo "\n\tWaiting..\n"

for i in $(seq 1 10);

do

# echo $i

sleep 1

done

echo "Diff..\n"

mv $fileOne $fileTwo

who > $fileOne

echo "User(s) LoggedIn"

diff $fileOne $fileTwo | grep "^< " | cut -d' ' -f 1,2,3

echo "\nUser(s) LoggedOut"

diff $fileOne $fileTwo | grep "^> " | cut -d' ' -f 1,2,3

sleep 1

who > $fileOne

done

#################################################################################################

Sample output;

Waiting..

Diff..

User(s) LoggedIn
< b0user01 ttys010
< b0user02 ttys011

User(s) LoggedOut

> b93033 organ90

Waiting..