extract data from log file in specified range of time with awk getline bash

问题

I was searching for parsing a log file and found what I need in this link extract data from log file in specified range of time

But the most useful answer (posted by @Kent):

# this variable you could customize, important is convert to seconds. 
# e.g 5days=$((5*24*3600))
x=$((5*60))   #here we take 5 mins as example

# this line get the timestamp in seconds of last line of your logfile
last=$(tail -n1 logFile|awk -F'[][]' '{ gsub(/\//," ",$2); sub(/:/," ",$2); "date +%s -d \""$2"\""|getline d; print d;}' )

#this awk will give you lines you needs:
awk -F'[][]' -v last=$last -v x=$x '{ gsub(/\//," ",$2); sub(/:/," ",$2); "date +%s -d \""$2"\""|getline d; if (last-d<=x)print $0 }' logFile

I think the error is in the "date +%s -d .... part

is giving the following error:

sh: -c: line 0: unexpected EOF while looking for matching `"'
sh: -c: line 1: syntax error: unexpected end of file
sh: -c: line 0: unexpected EOF while looking for matching `"'
sh: -c: line 1: syntax error: unexpected end of file

I spend lot of time trying to solve before I ask here but didn't find any solution.

The script will be called by the crontab to get the last 1 min log lines and count how many times an ip is listed in one minute so I can detect if it is an attack or not. and this is another task hope that an expert will help giving the needed code here in the same question.(I think it can be solved in 2 lines).

回答1:

Basing from your input here, you could use a script like this:

#!/bin/bash

LOGFILE=/path/to/logfile

X=$(( 60 * 60 )) ## 1 Hour

function get_ts {
    DATE="${1%%\]*}"; DATE="${DATE##*\[}"; DATE=${DATE/:/ }; DATE=${DATE//\// }
    TS=$(date -d "$DATE" '+%s')
}

get_ts "$(tail -n 1 "$LOGFILE")"
LAST=$TS

while read -r LINE; do
    get_ts "$LINE"
    (( (LAST - TS) <= X )) && echo "$LINE"
done < "$LOGFILE"

Save it to a file and change the value for LOGFILE, then run with bash script.sh.

Example output:

157.55.34.99 - - [06/Sep/2013:09:13:10 +0300] "GET /index.php HTTP/1.1" 200 16977 "-" "Mozilla/5.0 (compatible; bingbot/2.0; +http://www.bing.com/bingbot.htm)"
85.163.134.149 - - [06/Sep/2013:09:50:23 +0300] "GET /wap/wapicons/mnrwap.jpg HTTP/1.1" 200 1217 "http://mydomain.com/main.php" "Mozilla/5.0 (Linux; U; Android 4.1.2; en-gb; GT-I9082 Build/JZO54K) AppleWebKit/534.30 (KHTML, like Gecko) Version/4.0 Mobile Safari/534.30"

回答2:

The problem is probably just that you're not quoting your shell variables. Look:

$ foo='ab cd'

$ awk -v bar="$foo" 'BEGIN{print bar}'
ab cd

$ awk -v bar=$foo 'BEGIN{print bar}'
awk: fatal: cannot open file `BEGIN{print bar}' for reading (No such file or directory)

Yes, I know that's a different error message - what happens when you leave shell variables unquoted can by any number of things depending on the value of the variable, contents of your directory, etc., some of them VERY bad like removing every file in your filesystem.

So, quote your variables:

-v last="$last" -v x="$x"

then see if you still have the problem.

By the way, here's how to really solve your original problem using GNU awk with the input file http://pastebin.com/BXmS4zLn:

$ cat tst.awk
BEGIN {
    ARGV[ARGC++] = ARGV[ARGC-1]

    mths = "JanFebMarAprMayJunJulAugSepOctNovDec"

    if (days)  { hours = days * 24  }
    if (hours) { mins  = hours * 60 }
    if (mins)  { secs  = mins * 60  }
    deltaSecs = secs
}

NR==FNR {
    nr2secs[NR] = mktime($6" "(match(mths,$5)+2)/3" "$4" "gensub(/:/," ","g",$7))
    next
}

nr2secs[FNR] >= (nr2secs[NR-FNR] - deltaSecs)

$ awk -v hours=1 -f tst.awk file
157.55.34.99 - -  06 Sep 2013 09:13:10 +0300  "GET /index.php HTTP/1.1" 200 16977 "-" "Mozilla/5.0 (compatible; bingbot/2.0; +http://www.bing.com/bingbot.htm)"
85.163.134.149 - -  06 Sep 2013 09:50:23 +0300  "GET /wap/wapicons/mnrwap.jpg HTTP/1.1" 200 1217 "http://mydomain.com/main.php" "Mozilla/5.0 (Linux; U; Android 4.1.2; en-gb; GT-I9082 Build/JZO54K) AppleWebKit/534.30 (KHTML, like Gecko) Version/4.0 Mobile Safari/534.30"
83.113.48.218 - -  06 Sep 2013 10:13:07 +0300  "GET /english/nicons/word.gif HTTP/1.1" 200 803 "http://mydomain.com/french/details.php?eid=127928&cid=18&fromval=1&frid=18" "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.0; Trident/5.0)"

$ gawk -v mins=60 -f tst.awk file
157.55.34.99 - -  06 Sep 2013 09:13:10 +0300  "GET /index.php HTTP/1.1" 200 16977 "-" "Mozilla/5.0 (compatible; bingbot/2.0; +http://www.bing.com/bingbot.htm)"
85.163.134.149 - -  06 Sep 2013 09:50:23 +0300  "GET /wap/wapicons/mnrwap.jpg HTTP/1.1" 200 1217 "http://mydomain.com/main.php" "Mozilla/5.0 (Linux; U; Android 4.1.2; en-gb; GT-I9082 Build/JZO54K) AppleWebKit/534.30 (KHTML, like Gecko) Version/4.0 Mobile Safari/534.30"
83.113.48.218 - -  06 Sep 2013 10:13:07 +0300  "GET /english/nicons/word.gif HTTP/1.1" 200 803 "http://mydomain.com/french/details.php?eid=127928&cid=18&fromval=1&frid=18" "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.0; Trident/5.0)"

$ gawk -v mins=20 -f tst.awk file
83.113.48.218 - -  06 Sep 2013 10:13:07 +0300  "GET /english/nicons/word.gif HTTP/1.1" 200 803 "http://mydomain.com/french/details.php?eid=127928&cid=18&fromval=1&frid=18" "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.0; Trident/5.0)"

You can specify days= or hours= or mins= or secs= variables and it'll do the right thing.

If you only need a script to get the last 1 mins worth of log lines as your question states (now?), and want to see a one-liner to do it:

$ gawk 'NR==FNR {nr2secs[++nr] = mktime($6" "(match("JanFebMarAprMayJunJulAugSepOctNovDec",$5)+2)/3" "$4" "gensub(/:/," ","g",$7)); next} nr2secs[FNR] >= (nr2secs[nr] - 60)' file file
83.113.48.218 - -  06 Sep 2013 10:13:07 +0300  "GET /english/nicons/word.gif HTTP/1.1" 200 803 "http://mydomain.com/french/details.php?eid=127928&cid=18&fromval=1&frid=18" "Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.0; Trident/5.0)"

来源：https://stackoverflow.com/questions/18664086/extract-data-from-log-file-in-specified-range-of-time-with-awk-getline-bash

标签

bash

parsing

awk

logfiles