script to eliminate left and right fields and to get the ouput.

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# 1  
Old 10-21-2009
Java script to eliminate left and right fields and to get the ouput.

Hi Experts,
I have a file as given below and want to filter out the filenames in it , by deleting left and right filds and to have the fllenames (There are spaces in the filename),

Sun Jan 11 11:20:10 2009 1 0 /home/output/file2311_recent.list user1 user2 0 done
Sun Jan 11 11:20:10 2009 1 0 /home/output/file2312 jan recent.list user1 user2 0 done
Sun Jan 11 11:20:10 2009 1 0 /home/output/Output2313 feb recent.text user1 user2 0 done

I want to eliminate first 7 field from left side , and 4 fields from right side , and to get the output . Is there anything available with scripting , sed awk ?

Output should be:

/home/output/file2312 jan recent.list 
/home/output/Output2313 feb recent.text

Thanks in advance,

Last edited by vgersh99; 10-21-2009 at 07:28 PM.. Reason: code tags, please.
# 2  
Old 10-21-2009
nawk '{gsub("^[^/]*", "");$(NF-3)=$(NF-2)=$(NF-1)=$NF="";gsub(" *$", "")}1' myFile

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# 3  
Old 10-21-2009
rev infile8|cut -d' ' -f5-|rev|cut -d' ' -f8-


If all the file names contain a . with an extension then this would do too:
grep -o '/[^.]*\.[^ ]*' infile

# 4  
Old 10-21-2009
Many Thanks for the answers:

Scrutinizer , the rev and grep code worked nicely, However couldnot understand the grep code properly, appreciate if you could explain the grep code a bit.

The nawk code looks nice but I could not use it, as nwak not available in redhat linux.
# 5  
Old 10-21-2009
if you have Python
# python -c "for line in open('file'): print ' '.join(line.split()[7:-4])"
/home/output/file2312 jan recent.list
/home/output/Output2313 feb recent.text

# 6  
Old 10-22-2009
Sure, rveri
grep -o '/[^.]*\.[^ ]*' infile

means look for a match that consists of
/      the character / ,
[^.]*  followed by any number of characters that are not a dot 
\.     followed by a dot
[^ ]*  followed by any number of characters that are not a space

and then from man grep:
-o     Print  only  the  matched  (non-empty)  parts  of  a  matching line, 
       with each such part on a separate output line.

Actually it is good you asked because now I realize that \. is not necessary, so the expression becomes:
grep -o '/[^.]*[^ ]*' infile

Last edited by Scrutinizer; 10-22-2009 at 01:06 AM..
# 7  
Old 10-22-2009
although OP's sample doesn't have other fields (including if filenames contains dots as well) containing dots in them, its prudent to include checks for them as well if using the above grep command. otherwise, eliminating by column is a more flexible way to go
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