Running shell commands within AWK - shell

Question

I'm trying to work on a logfile, and I need to be able to specify the range of dates. So far (before any processing), I'm converting a date/time string to timestamp using date --date "monday" +%s.
Now, I want to be able to iterate over each line in a file, but check if the date (in a human readable format) is within the allowed range. To do this, I'd like to do something like the following:
echo `awk '{if(`date --date "$3 $4 $5 $6 $7" +%s` > $START && `date --date "" +%s` <= $END){/*processing code here*/}}' myfile`
I don't even know if thats possible... I've tried a lot of variations, plus I couldn't find anything understandable/usable online.
Thanks
Update:
Example of myfile is as follows. Its logging IPs and access times:
123.80.114.20 Sun May 01 11:52:28 GMT 2011
144.124.67.139 Sun May 01 16:11:31 GMT 2011
178.221.138.12 Mon May 02 08:59:23 GMT 2011

Answer 1

Given what you have to do, its really not that hard AND it is much more efficient to do your date processing by converting to strings and comparing.
Here's a partial solution that uses associative arrays to convert the month value to a number. Then you rely on the %02d format specifier to ensure 2 digits. You can reformat the dateTime value with '.', etc or leave the colons in the hr:min:sec if you really need the human readability.
The YYYYMMDD format is a big help in these sort of problems, as LT, GT, EQ all work without any further formatting.
echo "178.221.138.12 Mon May 02 08:59:23 GMT 2011" \
| awk 'BEGIN {
mons["Jan"]=1 ; mons["Feb"]=2; mons["Mar"]=3
mons["Apr"]=4 ; mons["May"]=5; mons["Jun"]=6
mons["Jul"]=7 ; mons["Aug"]=8; mons["Sep"]=9
mons["Oct"]=10 ; mons["Nov"]=11; mons["Dec"]=12
}
{
# 178.221.138.12 Mon May 02 08:59:23 GMT 2011
printf("dateTime=%04d%02d%02d%02d%02d%02d\n",
$NF, mons[$3], $4, substr($5,1,2), substr($5,4,2), substr($5,7,2) )
} ' -v StartTime=20110105235959
The -v StartTime is ilustrative of how to pass in (and the matching format) your starTime value.
I hope this helps.

Answer 2

Here's an alternative approach using awk's built-in mktime() function. I've never bothered with the month parsing until now - thanks to shelter for that part (see accepted answer). It always feels time to switch language around that point.
#!/bin/bash
# input format:
#(1 2 3 4 5 6 7)
#123.80.114.20 Sun May 01 11:52:28 GMT 2011
awk -v startTime=1304252691 -v endTime=1306000000 '
BEGIN {
mons["Jan"]=1 ; mons["Feb"]=2; mons["Mar"]=3
mons["Apr"]=4 ; mons["May"]=5; mons["Jun"]=6
mons["Jul"]=7 ; mons["Aug"]=8; mons["Sep"]=9
mons["Oct"]=10 ; mons["Nov"]=11; mons["Dec"]=12;
}
{
hmsSpaced=$5; gsub(":"," ",hmsSpaced);
timeInSec=mktime($7" "mons[$3]" "$4" "hmsSpaced);
if (timeInSec > startTime && timeInSec <= endTime) print $0
}' myfile
(I've chosen example time thresholds to select only the last two log lines.)
Note that if the mktime() function were a bit smarter this whole thing would reduce to:
awk -v startTime=1304252691 -v endTime=1306000000 't=mktime($7" "$3" "$4" "$5); if (t > startTime && t <= endTime) print $0}' myfile

Answer 3

I'm not sure of the format of the data you're parsing, but I do know that you can't use the backticks within single quotes. You'll have to use double quotes. If there are too many quotes being nested, and it's confusing you, you can also just save the output of your date command to a variable beforehand.