Data chopping

[IMAUser]

Hi ,

I m running the following command

who -u | nawk ' { if ( substr($5,1,2) > 9 ) print $0 }'

This is just to get the list of users who have logged on after 9. But this line comes back with nothing. At the same time if I just say

who -u | nawk ' { print substr($5,1,2) }'

It shows me the substr of the time and some of the records have time more than 9. So I fail to understand what its doing.

Any pointers please.

Thnx

 

[Ygor]

You are comapring a string to a number. Try..

' { if ( int(substr($5,1,2)) > 9 ) print $0 }'

 

[IMAUser]

Thnx, Should have noticed that.

 

[johngiggs]

IMAUser,

If you want to find out who has logged in after 9:00, you should probably use:

who -u | awk ' { if ( int(substr($5,1,2)) >= 9 ) print $0 }'

because > 9 will only print lines that are greater than 9. Also keep in mind that this does not take into consideration the login date. Someone could have logged in the previous day after 9:00 and still be displayed on your listing.

John

 

[IMAUser]

Actually yes.

I have a few issues with the way I m checking. I need to add some flexibility. So the immediuate requirement is to find out who has logged in say in the last tow hours. so the "9" is not relevant but I need to somehow get the current time from the command date. I tried using date like you wud use it in a shell script
var=`date +%H` within the awk commmand but awk dosent like it. So I m stuck. Any ideas please..

Thnx,

 

[CaKiwi]

Try

"date +%H" | getline var

CaKiwi

 

[IMAUser]

Hi CaKiwi,

Where do I use it. If I had this on the command line

who -u | awk ' { if ( int(substr($5,1,2)) >= 9 ) print $0 }'

Where wud the getline var come.

Thanks

 

[CaKiwi]

Something like

who -u | awk ' BEGIN{"date +%H" | getline var} { if ( int(substr($5,1,2)) >= var ) print $0 }'



CaKiwi