Is there a way to grep for something and then print out 10 lines after it.
for example if I want to grep for a word, then output the following 10 or whatever number of lines after the word.

Maybe this...
sed -n '/word/{N;N;N;N;N;N;N;N;N;p;}'

I created a different script you can use. Although Pederabo's is waaay nicer looking and faster, you can use this if you wanted to print out more lines than just 10 (like, say 100)..

Replace the 4 on line 8 with the number of lines you want to print after the search string is found:

 1  for l in `sed -n '/someString/{=;}/g' file`
 2  do
 3  a=1;b=0
 4    while read LINE
 5    do
 6      if [ $a -eq `expr $l + 1` ]
 7      then
 8        b=4
 9      fi
10      if [ $b -gt 0 ]
11      then
12        echo $LINE
13      fi
14      b=`expr $b - 1`
15      a=`expr $a + 1`
16    done < file
17  done

The above is a possibility. However you'll only get the occurences where there are exactly the number of requested lines. If the requested string would be placed at the end of the file you would not see it.

I should do the trick with a little script

lnnr=10;
num_of_lines=10;
while read line
do
  if (( $lnnr < $num_of_lines ))
  then
    echo $line
    (( lnnr=$lnnr + 1 ))
  else
    echo $line | grep "string or whatever" >/dev/null 2>&1
    if (( $?==0 ))
    then 
      lnnr=0
      echo $line
    fi
  fi
done < inputfile

why not just use head or tail. grep "word" file/location | tail -10 or head -10

tail -10 and head -10 print out the first 10 lines and last 10 lines, respectively, of a file. They cannot be used to pull lines from the middle of a file. Plus what you've done will output one or more lines containing the word "word" and then try to feed those lines into the tail or head command, which actually needs a filename.