Print The ouput From ls | grep "!!!"

guys forget the find command coz with find command u can't get condational output like grep. I will give small example :-

Apr 10 09:12 aacbl222_12aug1998.lqc
Apr 10 09:12 sscbl4534_4sep2001.lqc
Apr 10 09:12 ah66fmi_5jan1997.lqc
Apr 10 09:12 y313h1_7sep1998.lqc
May 11 09:12 aalike_9aug2000.lqc

all this files and thosands more in /test/test2
to get my file I should give some parameter " in this example I give 2 gre some time it will take more than 6 grep's to get what I want " forget it "
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ls -lrt | grep "Apr 10" | grep 1998

Apr 10 09:12 y313h1_7sep1998.lqc
Apr 10 09:12 aacbl222_12aug1998.lqc

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now when I want to print what I got from my command I will do this : -

print y313h1_7sep1998.lqc ; print aacbl222_12aug1998.lqc

when I type

ls -lrt | grep "Apr 10" | grep 1998 | lp
I will get the printe of result not the file it self...Pleassse try to understand me and execuse me for my "GOOD" english

Better to read all what I wrote above or just read this

[ HOW I CAN PRINT THE RESULTED FILES AS A FILE NOT AS OUTPUT ]

Havent I already answered the question?..

Or Was the command unsuccessful?

The answer is...

pipe the output in a file

ls > FILE.txt

creates or if exists overwrites the file FILE.txt and stores the output of the ls command in it.

ls >> FILE.txt appends the output of the command if the file exists or create a new one if not

hope this solves your problem

see the other thread as well

Frank

Please try to understand me

ls | grep "bla bla bla" >> result.txt

ok I will get the result in result.txt as u said then I want to print the files, don't tell me to print result.txt coz I will get in name of the file in the print out and I want to print the information inside the file.It's very simple please don't make me feel my english toooo BAD & unix As well

what I want is the fastest way to print multipul files

what I'm using right now is

print file1 ; print file2 ; print file3 ; print file4

Please please read this carefully and try it out before deciding it won't work, because as far as I can tell, it will!

I think that if you can figure out exactly what you are needing to do (which files to print, and when), you could use a very simple find command to do it.
But if you really really really want to use grep a bunch of times, you could do this:

ls -ltr | grep "Apr 10*1998.lqc" > /tmp/lp_file
print `cat /tmp/lp_file`

or use:

for each in `cat /tmp/lp_file`; do print $each; done

Note: I used backticks ` not single quotes ' above. The backticks are usually near the top left on a standard PC keyboard.

Or like this (if you can narrow what you want down by the filename):

for each in *sep2001.lcq; do; print $each; done

Or what about :

ls -lrt | grep "Apr 10*1998.lqc" | xargs print

Some of these may complain of an error with too many arguments, or possibly that the jobs may not all print, so you may have to use the first/second idea of:

ls -lrt | grep "Apr 10*1998.lcq" > /tmp/lp_list
for each in `cat /tmp/lp_list; do print $each; done

These examples all assume you are using a Bourne shell variant, such as ksh, sh, bash, etc...

Please try these before posting back saying that none of them work. And please let us know why they don't work.

Or even the basic step of taking the file.txt and adding print in the beginning of each line in the file and making it exe.

I know this is horrible scripting, but it works.

I agree that one of Livinfree's suggestions will work. see below his code.

code:
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for each in `cat /tmp/lp_file`; do print $each; done
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