Filtering Filenames with bash - bash

Question

I have a directory full of log files in the form
${name}.log.${year}{month}${day}
such that they look like this:
logs/
production.log.20100314
production.log.20100321
production.log.20100328
production.log.20100403
production.log.20100410
...
production.log.20100314
production.log.old
I'd like to use a bash script to filter out all the logs older than x amount of month's and dump it into *.log.old
X=6 #months
LIST=*.log.*;
for file in LIST; do
is_older = file_is_older_than_months( ${file}, ${X} );
if is_older; then
cat ${c} >> production.log.old;
rm ${c};
fi
done;
How can I get all the files older than x months? and... How can I avoid that *.log.old file is included in the LIST attribute?

Answer 1

The following script expects GNU date to be installed. You can call it in the directory with your log files with the first parameter as the number of months.
#!/bin/sh
min_date=$(date -d "$1 months ago" "+%Y%m%d")
for log in *.log.*;do
[ "${log%.log.old}" "!=" "$log" ] && continue
[ "${log%.*}.$min_date" "<" "$log" ] && continue
cat "$log" >> "${log%.*}.old"
rm "$log"
done

Answer 2

Presumably as a log file, it won't have been modified since it was created?
Have you considered something like this...
find ./ -name "*.log.*" -mtime +60 -exec rm {} \;
to delete files that have not been modified for 60 days. If the files have been modified more recently then this is no good of course.

Answer 3

You'll have to compare the logfile date with the current date. Start with the year, multiply by 12 to get the difference in months. Do the same with months, and add them together. This gives you the age of the file in months (according to the file name).
For each filename, you can use an AWK filter to extract the year:
awk -F. '{ print substr($3,0,4) }'
You also need the current year:
date "+%Y"
To calculate the difference:
$(( current_year - file_year ))
Similarly for months.

Answer 4

assuming you have possibility of modifying the logs and the filename timestamp is the more accurate one. Here's an gawk script.
#!/bin/bash
awk 'BEGIN{
months=6
current=systime() #get current time in sec
sec=months*30*86400 #months in sec
output="old.production" #output file
}
{
m=split(FILENAME,fn,".")
yr=substr(fn[m],0,4)
mth=substr(fn[m],5,2)
day=substr(fn[m],7,2)
t=mktime(yr" "mth" "day" 00 00 00")
if ( (current-t) > sec){
print "file: "FILENAME" more than "months" month"
while( (getline line < FILENAME )>0 ){
print line > output
}
close(FILENAME)
cmd="rm \047"FILENAME"\047"
print cmd
#system(cmd) #uncomment to use
}
}' production*