Linux script help needed - linux

Question

I am trying to create a script that will use the /etc/passwd file to create a listing like this:
Full Name: Blah Blah User Account: bblah User ID: 5000 Last logged in: Sun Feb 21 18:13
Full Name: Mike Mike User Account: mmike User ID: 5001 Last logged in: Sun Feb 21 18:13
With my script, I get it to show
Full Name: Blah Blah User Account: bblah User ID: 1000
because I grep the 1000 accounts in my passwd file. As soon as I do the 5000 series, I get all the full names, then all the user accounts, and then all the user ids listed. How do I go about fixing this? How do I go about adding the "Last logged in" bit?
Is there a way to do this without using awk?
Script so far is:
#!/bin/bash
passfile=/etc/passwd
for i in $(grep 5000 ${passfile} | cut -d : -f 5)
do
account=$(grep 5000 /etc/passwd | cut -d : -f 3)
username=$(grep 5000 /etc/passwd | cut -d : -f 1)
echo Full Name:${i} Username: ${username} User ID: ${account} Last login: ${lastlogin)
done

Answer 1

Last logged in is done by last command:
last username

Answer 2

Use this:
#!/bin/bash
echo -en "BEGIN{\n}{\n system(var)\n}\n" > h.awk
while read in; \
do \
uname=`echo $in | sed -r 's/:/\ /g' | awk '{print $1}'`; \
p1=`echo $in | sed -r 's/:/\ /g' | awk '{print "name: "$5, "username: "$1, "id: "$3}'`; \
p2=`echo | awk -f ./h.awk var="last $uname| head -n 1"|awk '{print $5 $6}'`; \
echo $p1 last login: $p2; \
done < /etc/passwd
If you need more detail of last login to be prompted, use this one:
echo -en "BEGIN{\n}{\n system(var)\n}\n" > h.awk
while read in; \
do \
uname=`echo $in | sed -r 's/:/\ /g' | awk '{print $1}'`; \
p1=`echo $in | sed -r 's/:/\ /g' | awk '{print "name: "$5, "username: "$1, "id: "$3}'`; \
p2=`echo | awk -f ./h.awk var="last $uname| head -n 1"|awk '{print $4" "$5 $6" "$7}'`; \
echo $p1 last login: $p2; \
done < /etc/passwd