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Help required on grep command(Skip the first few lines from printing in the output)

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# 1  
Old 09-24-2009
Help required on grep command(Skip the first few lines from printing in the output)
Hi experts

I want the proper argument to the grep command so that I need to skip the first few lines(say first 10 lines) and print all the remaining instances of the grep output.
I tried to use grep -m 10 "search text" file*. But this gives the first 10 instances(lines) of the search string. But I want to skip the first 10 lines(instances) and print the remaining lines in the output.
I need to run this grep command on a group of files(100s) at once.
I hope my problem statement is clear.

Please help me in this regard.
Many thanks in advance.
# 2  
Old 09-24-2009
Code:
grep "search text" file | tail -n +11

# 3  
Old 09-24-2009
Problem with grep command..Very urgent
Quote:
Originally Posted by varontron
Code:
grep "search text" file | tail -n +11

Hi Varontron and all

It looks like you did not understood my question .
Suppose if I am grepping some text from three files, my output should be
file1:11th line onwards to last line from file1
file2:11th line onwards to last line from file2
file3:11th line onwards to last line from file3
I am not expecting the 11th line onwards in the cumulative file.

I hope problem statement is clear at this time.
Regards
Sidda
# 4  
Old 09-25-2009
so... if you want to grep a whole bunch of files with one command, you'll get one big set of output. You should get the filename at the front of each line in the output so you could then process the whole pig.

or...you could put the grep in a loop like this

Code:
for each in `ls dir_with_files`
do
  grep "searchtext" $each | tail -n +11 > dir_with_processed_files/$each
done

# 5  
Old 09-25-2009
Quote:
Originally Posted by varontron
so... if you want to grep a whole bunch of files with one command, you'll get one big set of output. You should get the filename at the front of each line in the output so you could then process the whole pig.

or...you could put the grep in a loop like this

Code:
for each in `ls dir_with_files`
do
  grep "searchtext" $each | tail -n +11 > dir_with_processed_files/$each
done

Thank you Varantron for your timely help.

I prefer the first method where one single command will get the required output from each with file identity.
For time being I am already using the second method of modifying the files and copying to other directory and then cat them together and process further. As I need work on roughly 2GB of input files the second method requires many file transfers/creating new directories etc. So I will be grateful if I can get single command to do this.
# 6  
Old 09-25-2009
well, even a single command is going to require some work, as I assume you'll need to output the results of each individual file to a separate file.

furthermore, your loop will need to be a little more complex because you'll have to iterate over the whole grep result and keep track of the file on the current line, the line number, etc.

I think it would look something like this...

Code:
cd mydir
grep "searchtext" * > results.txt
COUNT=1
OLD_FNAME=""
NEW_FNAME=""
while read line 
do
  NEW_FNAME=`echo $line|cut -f1 -d':'`
  if [ $OLD_FNAME = $NEW_FNAME ]
  then
    if [ $COUNT -gt 10 ]
    then
        # output the line, presumably without the filename
    fi 
    COUNT=$COUNT+1
  else
    OLD_FNAME = $NEW_FNAME
    COUNT=1
  fi
done < results.txt

# 7  
Old 09-25-2009
How about this:
Code:
sed '1,10d' file*

This User Gave Thanks to edidataguy For This Post:
javabuddy
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