Question

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):

1. Using the commands who and cut, extract the list of usernames currently logged in the system
2. To check after a minute, the sleep command can be used: sleep 60
3. Get the new list of users logged in after a minute
4. For each user in the list;
   1. check if they are 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
   2. If the user is not in the second list, print that the user has logged out
5. For each user in the second list,
   1. If he is in the first list, print that the user has logged in
6. Rename the second list as the first list
7. Repeat the process from step 2

Answer 1

1). ANSWER :

GIVENTHAT :

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

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 scconds(10)

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

4. saw current "who" output into file "A"(overwrite)

5. detect change "diff" of files and output to console in formatted

6. ctrl-c 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

<b0users01ttys010

<b0user02 ttys011

User(s) LoggedOut

>b93033 organ90