Calculate N days from a file output

Shell Programming and Scripting
1

OK, here is the output from a cron I have here:

FULL OUTPUT:

acoxxx Lastlogin= 2010/07/15 13:10
db2t Lastlogin= 2010/07/16 13:09
db2tadm Lastlogin= 2010/07/20 13:09
eisuser Lastlogin= 2010/07/20 11:53
israel Lastlogin= 2010/07/10 11:42
nmon Lastlogin= 2010/07/05 12:55
norbac Lastlogin= 2010/07/04 14:21
oracle Lastlogin= 2010/07/03 08:05
root Lastlogin= 2010/07/03 13:01
s03josa0 Lastlogin= 2010/07/01 14:10

I'm looking for a script that 'grep' ONLY the lines older than N days.
Example: "Lastlogin older than 10 days":
OUPUT:

nmon Lastlogin= 2010/07/05 12:55
norbac Lastlogin= 2010/07/04 14:21
oracle Lastlogin= 2010/07/03 08:05
root Lastlogin= 2010/07/03 13:01
s03josa0 Lastlogin= 2010/07/01 14:10

How can I do this?

thanks
ISrael.

2

lets just say your cron output is stored in a file named file1

then you can use something like this:-

 awk -v f=0 '/2010/07/10/ {f=0} { print > "file_"f }'  file1

this will place all the lines before the date 2010/07/10 into file0. but in this case you have to enter the date manually. i mean the date 10days earlier date. or you can take it as a parameter. or pass it via giving output of find command.

3

Hi daz,

it gives me an error:

israel@bsades2: /home/israel # awk -v f=0 '/2010/07/10/ {f=0} { print > "file_"f }' /tmp/prueba1
 Syntax Error The source line is 1.
 The error context is
                /2010/07/10/ >>>  { <<<
 awk: 0602-500 Quitting The source line is 1.

What's wrong? Thanks

---------- Post updated at 09:44 AM ---------- Previous update was at 09:17 AM ----------

daz,

What if I the date is 36 days?

thanks
Israel.

4

With GNU date, or you give the date directly to $DATE by format as 20100710

DATE=$(date -d "36 days ago" +%Y%m%d)
awk -v d="$DATE" '{t=$3;gsub(/\//,"",t)}{if (t<d) print} ' urfile
5

Hi rdcwayx,

I think I don't have gnu-date because I'm on AIX6.1. Thanks.