How to print lines of text with date older than two days - awk

Question

I have the following text file that I am working with and must be able to parse only the object name value when the creationdatetime is older than two days.
objectname ...........................: \Path\to\file\hpvss-LUN-22May12 22.24.11\hpVSS-LUN-29Aug12 22.39.15
creationdatetime .....................: 01-Sep-2012 02:17:43
objectname ...........................: \Path\to\file\hpVSS-LUN-22May12 22.24.11\hpVSS-LUN-28Aug12 22.16.19
creationdatetime .....................: 03-Sep-2012 10:18:09
objectname ...........................: \Path\to\file\hpVSS-LUN-22May-12 22.24.11\hpVSS-LUN-27Aug12 22.01.52
creationdatetime .....................: 03-Sep-2012 10:18:33
An output of the command for the above would be:
\Path\to\file\hpvss-LUN-22May12 22.24.11\hpVSS-LUN-29Aug12 22.39.15
Any help will be greatly appreciated.
Prem

Answer 1

Date parsing in awk is a bit tricky but it can be done using mktime. To convert the month name to numeric, an associative translation array is defined.
The path names have space in them so the best choice for field separator is probably : (colon followed by space). Here's a working awk script:
older_than_two_days.awk
BEGIN {
months2num["Jan"] = 1; months2num["Feb"] = 2; months2num["Mar"] = 3; months2num["Apr"] = 4;
months2num["May"] = 5; months2num["Jun"] = 6; months2num["Jul"] = 7; months2num["Aug"] = 8;
months2num["Sep"] = 9; months2num["Oct"] = 10; months2num["Nov"] = 11; months2num["Dec"] = 12;
now = systime()
two_days = 2 * 24 * 3600
FS = ": "
}
$1 ~ /objectname/ {
path = $2
}
$1 ~ /creationdatetime/ {
split($2, ds, " ")
split(ds[1], d, "-")
split(ds[2], t, ":")
date = d[3] " " months2num[d[2]] " " d[1] " " t[1] " " t[2] " " t[3]
age_in_seconds = mktime(date)
if(now - age_in_seconds > two_days)
print path
}
All the splitting in the last block is to pick out the date bits and convert them into a format that mktime accepts.
Run it like this:
awk -f older_than_two_days.awk infile
Output:
\Path\to\file\hpvss-LUN-22May12 22.24.11\hpVSS-LUN-29Aug12 22.39.15

Answer 2

I would do it in 2 phases:
1) reformat you input file
awk '/objectname/{$1=$2="";file=$0;getline;$1=$2="";print $0" |"file}' inputfile > inputfile2
This way you would deal with
01-Sep-2012 02:17:43 | \Path\to\file\hpvss-LUN-22May12 22.24.11\hpVSS-LUN-29Aug12 22.39.15
03-Sep-2012 10:18:09 | \Path\to\file\hpVSS-LUN-22May12 22.24.11\hpVSS-LUN-28Aug12 22.16.19
03-Sep-2012 10:18:33 | \Path\to\file\hpVSS-LUN-22May-12 22.24.11\hpVSS-LUN-27Aug12 22.01.52
2) filter on dates:
COMPARDATE=$(($(date +%s)-2*24*3600)) # 2 days off
IFS='|'
while read d f
do
[[ $(date -d "$d" +%s) < $COMPARDATE ]] && printf "%s\n" "$f"
done < inputfile2