topic Re: awk issue in Operating System - HP-UX

awk issue

NDO — Tue, 22 Mar 2011 07:52:25 GMT

Hi All

I´m not very familiar with scripting, Í am still learning.
Please can someone tell me what could be wrong with that command:
root@nrtrde1 # ls -lrt | awk '{if ($6 == "Jan") && ($7 == "1")print $9}'
awk: syntax error near line 1
awk: illegal statement near line 1

The idea is to list all files of 1st of january 2011. If I am able to do this I could write a script to transfer files from january 2011.
Your help is appreciated

F.R.

Re: awk issue

Victor Fridyev — Tue, 22 Mar 2011 08:01:32 GMT

ls -lrt | awk '{if($7 == "1")if ($7==1) print $9}'

Re: awk issue

NDO — Tue, 22 Mar 2011 08:06:45 GMT

>>ls -lrt | awk '{if($7 == "1")if ($7==1) print $9}'

Sorry the first if shouldnÂ´t be "Jan" as opposed to what you wrote which is "1"?

F.R.

Re: awk issue

Victor Fridyev — Tue, 22 Mar 2011 08:21:55 GMT

Sorry, 8(((
ls -lrt | awk '{if($6 == "Jan")if ($7==1) print $9}'

Re: awk issue

NDO — Tue, 22 Mar 2011 11:00:46 GMT

Hi

It does work now, but how if I want to exclude from that list all january 2010 files, because now is listing me all january 1st 2010 files, but what I want is January 2011.

F.R.

Re: awk issue

Hein van den Heuvel — Tue, 22 Mar 2011 11:15:59 GMT

It is still not 'proper' awk.

Proper awk IMHO is a series of conditions follows by code blocks.

The example unconditionally goes into a codeblock and then starts a condition.

Try this: $ ls -ltr | awk '($6=="Jan") && ($7==1) {print $9}'

Not for the real problem you do NOT want to use awk.
You want to use FIND.

Check out the -ctime option, or the usage of 'reference files' through the -newer and -older options.
Google and the MAN PAGES are your friend in this learning exercise.

Cheers,
Hein

Re: awk issue

Victor Fridyev — Tue, 22 Mar 2011 11:47:36 GMT

So please define your request properly.
If you need some date filter, use find:
find /home/usersname -type f -mtime +5 -exec ls -ls {} \;
or something similar.

Good luck !

Re: awk issue

James R. Ferguson — Tue, 22 Mar 2011 13:20:56 GMT

Hi:

# touch -amt 201012312359.59 /tmp/ref1
# touch -amt 201101012359.59 /tmp/ref2
# find . -type f -newer /tmp/ref1 -a ! -newer /tmp/ref2

...will list the files *last_modified* on January 1, 2011

See the manpages for 'touch(1)' and for 'find(1)'.

Regards!

...JRF...

Re: awk issue

Dennis Handly — Wed, 23 Mar 2011 04:05:59 GMT

>Hein: Proper awk IMHO is a series of conditions follow by code blocks.

True but you can do C style programming:
... | awk '
{
if ($6 == "Jan" && $7 == 1)
print $9
}'

Re: awk issue

NDO — Wed, 23 Mar 2011 05:50:45 GMT

Hi
Dennis:

Your suggestion give me this:
-rw-r--r-- 1 taprap other 60414 Jan 1 2010 CDNAM01MOZ0105105
-rw-r--r-- 1 taprap other 123 Jan 1 2010 CDMWICPMOZ0103249
-rw-r--r-- 1 taprap other 1698 Jan 1 2010 CDISRCLMOZ0103103
-rw-r--r-- 1 taprap other 129 Jan 1 01:34 CDINDA9MOZ0101336
-rw-r--r-- 1 taprap other 3516 Jan 1 01:34 CDZMBCEMOZ0103029
-rw-r--r-- 1 taprap other 129 Jan 1 01:35 CDINDJBMOZ0102144
-rw-r--r-- 1 taprap other 647 Jan 1 01:35 CDINDA2MOZ0101581
-rw-r--r-- 1 taprap other 123 Jan 1 01:36 CDGRCPFMOZ0105837
-rw-r--r-- 1 taprap other 123 Jan 1 01:36 CDPRYHTMOZ0101023

BUT I do not want the 2010 files, I just want the other files. How can I get rid of the Jan 1 2010 files?

F.R.

Re: awk issue

Shibin_2 — Wed, 23 Mar 2011 06:37:54 GMT

try this.

... | awk '
{
if ($6 == "Jan" && $7 == 1 && $8 != 2010)
print $9
}'

Re: awk issue

NDO — Wed, 23 Mar 2011 09:04:33 GMT

Hi
Shibin!

Your suggestion worked perfectely, I havwe modified slightly to list all January files of current year as this:

ls -lrt | awk '{if ($6 == "Jan" && $8 != 2010)print $0}'

F.R.

Re: awk issue

NDO — Tue, 29 Mar 2011 13:34:38 GMT

Thank you for your contributions. Already assign points

F.R.