What am I doing wrong here?

UNIX for Advanced & Expert Users
1

I am working on a simple login ID check shell script that should prompt for a user ID then check to see if this user is logged on. Trying to get the hang of this stuff so I am thinking of my own little projects.

#! /bin/sh
echo "please enter a user name"
read user
if user=$user
    then
 do
User=`who |grep user1,user2,user3|cut -d " " -f1`
echo User ID = $User
fi
 done
2

It is hard to guess what a script was intended to do when it contain many syntax and logic errors.

The functioning lines appear to boil down to displaying the given user from a "who" report:

#!/bin/sh
echo "please enter a user name"
read user
who -u | grep "^${user} " | awk '{print $1}'

Btw. If you get an error message, please always post the error message as well as the script.

3

The script worked as you have it. Thanks for the help... but what I was looking for is a way to tell if the person is currently logged in and give an echo to when they logged in. Sorry I should have been more concise!

4

One idea is to count the number of times a user appears in the "who -u" list by using the "wc -l" command (which just counts the number of lines it sees).
The backticks ` ` are important and cause the commands to be executed. Note the correct syntax for an "if" statement. Not clear where the "do" and "done" figure in Post #1.

#!/bin/sh
echo "please enter a user name"
read user
user_login_count=`who -u | grep "^${user} " | wc -l`
if [ ${user_login_count} -gt 0 ]
then
       echo "User ${user} is logged in ${user_login_count} times"
       # Show some evidence
       who -u | grep "^${user} " | awk '{print $1}'
else
       echo "User ${user} is not logged in"
fi

If you have a modern POSIX shell, the line containing the backticks can be replaced with the more eloquent:

user_login_count=$(who -u | grep "^${user} " | wc -l)

When posting it always helps to say what Operating System and Shell you have. There is much variation.

1 Like