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awk regardless positions

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# 1  
Old 09-20-2011
awk regardless positions
PHP Code:
brw-------    1 oracle   dba          49, 21 Apr 05 11:45 dprod_0000018
brw-------    1 oracle   dba          49, 26 Apr 05 11:45 dprod_0000019
brw-------    1 oracle   dba          43, 93 Feb 02 2011  dprod_000002
brw-------    1 oracle   dba          49, 27 Apr 05 11:45 dprod_0000020
brw-------    1 oracle   dba          49, 29 Apr 05 11:45 dprod_0000021
brw-------    1 oracle   dba          49, 30 Apr 05 11:46 dprod_0000022
brw-------    1 oracle   dba          49, 35 Apr 05 11:46 dprod_0000023 
I need to grep the file names (dprod_xxxxx), but
PHP Code:
ls -al |grep prod | awk '{ print $10 }' 
ls -al |grep prod | awk '{ print $9 }' 
don't let me get all of them because of positions due to the different date and time format (i.e. Feb 02 2011 and Apr 05 11:46)

Please advise how to get all of the file names (dprod_xxxxx).

Thank you
# 2  
Old 09-20-2011
$10 prints the filename for me for every file.

What's wrong with ls -al dprod_*?

For AWK, even though I had no problems, for filenames without spaces, you can use $NF instead of $9, $10, etc.
This User Gave Thanks to Scott For This Post:
Daniel Gate
# 3  
Old 09-20-2011
Great! $NF works fine.
# 4  
Old 09-20-2011
I saw your another post. I think you just need ls dprod* instead of all the piped commands. -l in ls command will give the long listing. Give it a try!

--ahamed
This User Gave Thanks to ahamed101 For This Post:
Scott
# 5  
Old 09-20-2011
Of course! The point I was trying to make (ls) for just the filenames. Doh! Smilie
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