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Top Forums Programming Open Source Splitting files using awk and reading filename value from input data Post 302976078 by rbatte1 on Thursday 23rd of June 2016 01:35:45 PM
Old 06-23-2016
Splitting files using awk and reading filename value from input data

I have a process that requires me to read data from huge log files and find the most recent entry on a per-user basis. The number of users may fluctuate wildly month to month, so I can't code for it with names or a set number of variables to capture the data, and the files are large so I don't want to read the it several times.

The entries of interest have a particular string so I can extract just them from the overall log file and I have a way to split the output into separate files on a per-user basis, my plan being to then just read the last line of each files created with tail -1 and the filename giving me the user account in question.

My boss, however, worries about false-positive data matches for my expression (by chance or maliciously) that might try to overwrite a critical file.


My data has a syslog-type date in it which means doing a sort -u is proving tricky too. I've got this far with splitting the data out to files under /tmp/logs as splitlog.rbatte1 or similar but if field 11 were ever */../../etc/passwd then potentially I would be in trouble.

The date is the first three fields and 'as far as I am aware' a valid user name would be in field 11, but ........

A simplified part of the code would be:-
Code:
grep "Active transaction started" /var/log/qapplog | awk "{print \$1, \$2, \$3, \$11> \"/tmp/logs/splitlog.\"\$11}"
for userfile in /tmp/logs/splitlog.*
do
   lastrecord=$(tail -1 $userfile)
   printf "User %s last record is %s\n" "$userfile" "$lastrecord"
   .... whatever else here ....
done

I have considered adding tr -d "\/" to strip out the characters, but now that it's been raised, I'm concerned that there may be other things I'm not considering.

Is there a better way to work here, potentially with awk getting the equivalent of basename "$11" or variable substitution in the shell of "${{11}##*/}"?


Any suggestions welcome. Perhaps there is a better design overall that will find the last entry on a per-user basis. The log is thankfully written in time order, so the last in the file by user name is the last by time already.

Kind regards,
Robin
 

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sulog(4)							   File Formats 							  sulog(4)

NAME
sulog - su command log file SYNOPSIS
/var/adm/sulog DESCRIPTION
The sulog file is a record of all attempts by users on the system to execute the su(1M) command. Each time su(1M) is executed, an entry is added to the sulog file. Each entry in the sulog file is a single line of the form: SU date time result port user-newuser where date The month and date su(1M) was executed. date is displayed in the form mm/dd where mm is the month number and dd is the day number in the month. time The time su(1M) was executed. time is displayed in the form HH/MM where HH is the hour number (24 hour system) and MM is the minute number. result The result of the su(1M) command. A ` + ' sign is displayed in this field if the su attempt was successful; otherwise a ` - ' sign is displayed. port The name of the terminal device from which su(1M) was executed. user The user id of the user executing the su(1M) command. newuser The user id being switched to with su(1M). EXAMPLES
Example 1 A sample sulog file. Here is a sample sulog file: SU 02/25 09:29 + console root-sys SU 02/25 09:32 + pts/3 user1-root SU 03/02 08:03 + pts/5 user1-root SU 03/03 08:19 + pts/5 user1-root SU 03/09 14:24 - pts/5 guest3-root SU 03/09 14:24 - pts/5 guest3-root SU 03/14 08:31 + pts/4 user1-root FILES
/var/adm/sulog su log file /etc/default/su contains the default location of sulog SEE ALSO
su(1M) SunOS 5.11 6 Jun 1994 sulog(4)
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