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
i use awk to get last column
now my output is
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
---------- 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
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.
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.
This is the version of date that I used.
Hope that helps you with your script.
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
Hence if i execute a command it should give me count of APR 16 entries
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)
(As you wait for the experts) :-)
Hi Thanks Ni2, still need help
here is input data after using awk in your code
now when i use you code
i get Zero count, even though i added one entry with Apr 17
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
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