Hi 2 everyone.. i need some assista

[jrb23]

Hi 2 everyone.. i need some assistance on reporting AIX user that have not logged for previous 3 months.
I happen to have more than 5000 entries on the passwd file.
I am devising a script that will interrogate the /etc/passwd(to get all the first column of the passwd file) then relate it to the /var/adm/wtmp using the last command.
Any script that has been posted on this topic will be much appreciated.
Or any other ways of doing it?

Thank you in advance

 

[Gloups]

Use that to get the fisrt collumn of /etc/passwd:

more /etc/passwd | cut -d ':' -f 1

 

[ogniemi]

it is not tested - have no access to AIX now - should check last output column/fields

for i in `awk -F\: '{print $1}' /etc/passwd`;do [ `last $i|head -1|awk '{print $4" "$5}'` = "Jan 21" ] && echo $i last logged on Jan 21;done

or

for i in `awk -F\: '{print $1}' /etc/passwd`;do [ x$(last $i|head -1|awk '{print $4" "$5}') = x"Jan 21" ] && echo $i last logged on Jan 21;done

 

[Gloups]

Woooops i read a little bit to fast.

Sorry for my poor answer )

 

[ogniemi]

it looks much complex.

do the following:

1. last > last.output
2. vi last.output
(find first line containing "22 Jan" and press "Esc-d-G" , save and exit)

3. Use this to determine which user was not logged within last 3 months:

for i in `awk -F\: '{print $1}' /etc/passwd`;do [ `grep -w -c ^$i last.output` -eq 0 ] && echo $i not logged more then 3 months;done

 

[theycallmetim]

No way, the

lsuser -a time_last_login ALL

command is by far the easiest. It will give you the time in julian date.

 

[ogniemi]

Bingo! you whom they call tim.

 

[jrb23]

Thank you -- much appreciated guys..
ogniemi - even not tested but good source that i can modify
Gloups - have not thought bout using last as output to interrogate - might be a good try..
theycallmetim - good idea i will surely explore this

once i have completed i will post it here..