Sorting user by last login an count


 
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# 1  
Old 04-17-2012
Sorting user by last login an count

Experts need your help

i need to create a script which will give count of users logged between yesterday and today (Or if run exacle 12:00AM midnight, it will give count of of users logged on that day
here is how the data looks
Code:
Pappu, Paul|22-Feb-201|0|30-Jun-201
Aloor, Shiva|10-Apr-201|0|18-Aug-201
Singh, Anil|16-Mar-201|0|10-Nov-201
Sun, Henry|09-Apr-201|0|05-Sep-201
Proxy,ETS|10-Apr-201|0|08-Mar-201
Patel, Rishi|11-Feb-201|0|27-Jun-201
Dubey, Anand|16-Apr-201|0|13-Mar-201
Tiwari, Shirish|19-Feb-201|0|13-Dec-201
Vinayagam, Deiva|15-Feb-201|0|04-Jul-201
infodba_ie|28-Feb-201|0|02-Dec-201
Murugesan, Gokul|03-Apr-201|0|07-Dec-201
cntestuser|21-Mar-201|0|08-Nov-201
Gupta, Rajiv|02-Apr-201|0|27-Jun-201
Airmal, Shiva Kumar|16-Apr-201|0|09-Dec-201
Joshi, Deepak|16-Apr-201|0|17-Sep-201
Wang,Charles|10-Apr-201|0|08-Nov-201


i use awk to get last column
Code:
cat user.txt |awk -F"|" '{print $2}'`

now my output is
Code:
22-Feb-201
10-Apr-2012
16-Mar-2012
09-Apr-2012
10-Apr-2012
11-Feb-2012
16-Apr-2012
19-Feb-2012
15-Feb-2012
28-Feb-2012
03-Apr-2012
21-Mar-2012
02-Apr-2012
16-Apr-2012
16-Apr-2012
10-Apr-20122
24-Jan-2012
09-Mar-2012
16-Apr-2012
27-Feb-2012
21-Mar-2012
13-Feb-2012
03-Apr-2012
06-Mar-201

My challenge is
1- I need to sort based on month and date
2- Get a list of user logged (system time -1), mean if execute and 12-05 midnight it will give user logged yesterday
3. get the count of the rows in column 2

Please help


Moderator's Comments:
Mod Comment Please use code tags, thanks!


---------- Post updated at 05:39 AM ---------- Previous update was at 04:27 AM ----------

I was able to sort it now using sort -t '-' -k 3.1n,3.2 -k 2.1M,2.3 -k 1.1n,1.2
Code:
24-Jan-2012
11-Feb-2012
13-Feb-2012
15-Feb-2012
19-Feb-2012
22-Feb-2012
27-Feb-2012
28-Feb-2012
06-Mar-2012
09-Mar-2012
16-Mar-2012
21-Mar-2012
21-Mar-2012
02-Apr-2012
03-Apr-2012
03-Apr-2012
09-Apr-2012
10-Apr-2012
10-Apr-2012
10-Apr-2012
16-Apr-2012
16-Apr-2012
16-Apr-2012

If i get tip to get user count of yesterday's assuming i execute this 1 minute after midnight

Last edited by zaxxon; 04-17-2012 at 06:39 AM.. Reason: code tags, not html tags and also tag the code please, ty. See PM.
# 2  
Old 04-17-2012
Hi,

I am not quite sure I understand what you want. But if you just want to count the number of times 16-Apr-2012 appears you could do the following. From your list of dates.

Code:
echo "24-Jan-2012
11-Feb-2012
13-Feb-2012
15-Feb-2012
19-Feb-2012
22-Feb-2012
27-Feb-2012
28-Feb-2012
06-Mar-2012
09-Mar-2012
16-Mar-2012
21-Mar-2012
21-Mar-2012
02-Apr-2012
03-Apr-2012
03-Apr-2012
09-Apr-2012
10-Apr-2012
10-Apr-2012
10-Apr-2012
16-Apr-2012
16-Apr-2012
16-Apr-2012" | grep "$(date +%d-%b-%Y --date="1 day ago")" | wc -l

This is the version of date that I used.
Code:
date --version
date (GNU coreutils) 8.5
Copyright (C) 2010 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>.
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by David MacKenzie.

Hope that helps you with your script.
# 3  
Old 04-17-2012
Hi i need to check system date and list the count of entries which are one day older than the system date
My system date is in this format
Code:
infodba-ie10ux014:/home/infodba/execute\n\r-> date
Tue Apr 17 17:05:16 GMT 2012

Hence if i execute a command it should give me count of APR 16 entries

Moderator's Comments:
Mod Comment How to use code tags

Last edited by Scrutinizer; 04-17-2012 at 08:48 AM..
# 4  
Old 04-17-2012
Ok. So you will not use the list of dates that you got from awk?
If you pass the results thru grep it will give you 3. Which is a count of yesterdays dates?

So taking what you have done. (skipping the sort)
Code:
awk -F"|" '{print $2}' user.txt | grep "$(date +%d-%b-%Y --date="1 day ago")" | wc -l

(As you wait for the experts) :-)
# 5  
Old 04-18-2012
Hi Thanks Ni2, still need help
here is input data after using awk in your code
Code:
22-Feb-2012
10-Apr-2012
16-Mar-2012
09-Apr-2012
17-Apr-2012
11-Feb-2012
16-Apr-2012
19-Feb-2012
15-Feb-2012
28-Feb-2012
03-Apr-2012
21-Mar-2012
02-Apr-2012
16-Apr-2012
10-Apr-2012
24-Jan-2012
09-Mar-2012
16-Apr-2012
27-Feb-2012
21-Mar-2012
13-Feb-2012
03-Apr-2012
06-Mar-2012

now when i use you code
Code:
awk -F"|" '{print $2}' lastlogin_user.txt | grep "$(date +%d-%b-%Y --date="1 day ago")" | wc -l

i get Zero count, even though i added one entry with Apr 17
Code:
infodba-ie10ux014:/home/infodba/execute\n\r-> awk -F"|" '{print $2}' lastlogin_user.txt | grep "$(date +%d-%b-%Y --date="1 day ago")" | wc -l
       0

I mean i want to get count of entries with yesterday's date, in my case there one "17-Apr-2012" so the output should be 1

Please confirm where i am wrong, my system date format is
Code:
Wed Apr 18 11:14:14 GMT 2012

# 6  
Old 04-18-2012
What does this give you

Code:
date +%d-%b-%Y --date="1 day ago"

# 7  
Old 04-18-2012
Bug

Use below command to get result for yesterdays and today users..
Code:
awk -F "|" '{print $2}' lastlogin_user.txt | grep -E "$(date +%d-%b-%Y --date="1 day ago")|$(date +%d-%b-%Y)"


Last edited by Franklin52; 04-18-2012 at 04:07 AM.. Reason: Please use code tags for code and data samples, thank you
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