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Old 12-12-2011, 06:37 PM   #1
soph75
LQ Newbie
 
Registered: Jun 2011
Posts: 4

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Batch Log File Count number of "unknown user” algorithm


Hi,

How can I write a script that extracts number "Unknown User" from a log file?



Sample:

At 2011-12-07 03:04:21.902: 250 OK^M
At 2011-12-07 03:04:21.902: 550 "Unknown User"^M
At 2011-12-07 03:04:21.902: 503-All RCPT commands were rejected with this error:^M
503-"Unknown User"^M
503 Valid RCPT command must precede DATA^M
At 2011-12-07 03:04:21.902: Sent: RSET^M

At 2011-12-07 03:04:21.941: 250 Reset OK^M
Log-Entry-Offset/Size: 608761155/0000663^M
Content-ID: 1121995/20029_1 892879 HU^M
MailHost: verizon.net(0.0.0.0)^M
Addressee: edjjb4@verizon.net^M
Status-Attempt-Session-Error: 1:1:2d:0^M
Date: Wed Dec 07 03:04:21 EST 2011^M
At 2011-12-07 03:04:21.491: Sent: MAIL FROM:<news@go.intervalworld.com>^M
RCPT TO:<edjjb4@verizon.net>^M
DATA^M



Thanks
 
Old 12-12-2011, 09:18 PM   #2
corp769
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Registered: Apr 2005
Posts: 5,817

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Hello,

You can use "grep" to grep out what you need, and use it within a for loop, counting as you go with a variable. On a side note though, you should use the command "dos2unix" on your file to convert it, since I am seeing "^M" in your file.

Cheers,

Josh
 
Old 10-14-2014, 05:50 AM   #3
wje_lq
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Registered: Sep 2007
Location: Mariposa
Distribution: FreeBSD,Debian wheezy
Posts: 811

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grep '"Unknown User"' logfile.txt | wc -l
 
  


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