Print Multipul Fiels as One shot

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# 1  
Old 04-10-2002
Question Print Multipul Fiels as One shot

Hi there

This is my first post. I just want to know the best way to print N number of files as one shot. Let's simulate the case ...

> ls


What is the best way to print all this files.. what I'm doing now is

/ > print file1.txt ; print file2.txt ; print file3.txt ; print file4.txt

is there's faster way to do that...... thaanks
# 2  
Old 04-10-2002

try this

find /yourDirectory -name "*.txt" -exec print {} \;

the {} is for all files found by the find command

for more information try

man find Smilie


# 3  
Old 04-10-2002
Thanks But......

Thanks but....

actually u didn't get what I ment, ,I just give an example *.txt files,, actually I have dirctory contain 100000 (OR MORE) files (some old reports and logs ). When I want to get the files I just grep it " ls -lrt | grep "date" | grep "another Parameter" then I will get the result. Now I want to know whatis the best way to print all the files with out manually give print command for each file. I hope u get what I mean
# 4  
Old 04-10-2002
If you need to list only top 10 files in an directory, then -

ls -l | head -11

you can also use - "ls -lt | head -11" for viewing 10 recently modified files.

Hope this is useful.
# 5  
Old 04-10-2002
Well, first, I don't see why you need "ls -lrt". Do they need to print in date order?

But I still agree with isacs:
find /your/dir -type f -name "*date*[0-9][0-9]*" -exec lp -dmy_printer {} \;

That will find only files (not directories) in /your/dir that have the word "date" before some numbers, then send each one to the line printer in the order in which it finds them.

Look at the name page for find for all finds of options, including ones to keep the search in the /your/dir directory, instead of trying to keep searching below it.
# 6  
Old 04-10-2002
Tools I'm Confuse!!!!!!!

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 "

ls -lrt | grep "Apr 10" | grep 1998

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


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 -printername
I will get the printes result not the file it self...Pleassse try to understand me and execuse me for my "GOOD" english Smilie

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

# 7  
Old 04-10-2002
Use backtilt

lp printername `ls -ltr|grep "your_param"|grep "your_param2"`


lp printername `ls -ltr | grep "your_param1.*your_param2"` - If you are using the param as per your example.
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