Is it possible to write a script that takes 2 arguments: word for any word that is to be seacrched in the files of a directory and n for the position at which that word has to be in any line in that file, and finally print only the files that feature that word at that position in any line?
My code until now is:
word=$1
for files in .
do
grep -rl "$word"
done
This only prints the files with that word in them, however, and I'm not sure how to implement the rest.
Your loop seems a bit strange, you'd want to loop through all files, not through the set of things you provided (which is a set with a single entry, .). Not really useful at all.
So,
for file in * would make much more sense. Then, use that ${file} variable! You're not even doing anything with it in your for loop! That also makes no sense.
For example, you could
read to get lines from the file (thousands of examples on how to read lines from files using bash)cut to select the position[[ / ]] to test for the string in that position, and if successfulalternatively, learn your self a bit of regexes. Don't know which version of grep you have, but "from the beginning of the line, find the things that has N repititions the scheme "any repitititon of anything but a delimiter + one word delimiter followed by the word I'm looking for" isn't hard.
Something like, to look for "mustard" in the fifth word:
words_before=4
word="mustard"
# idea is to get the sed expression '/^\([^ ]\+ \)\{words_before\}word/!{qerror_code}'
sedtemplate_start='/^\([^ ]\+ \)'
sedtemplate_end='/!{q100}'
sedtemplate="${sedtemplate_start}\\{${words_before}\\}${word}${sedtemplate_end}"
#.... open all files, go through all lines
( echo "${this_line}" | sed -n "${sedtemplate}" ) && echo "${file}"
If I understand what you're trying to do correctly then the right approach is this (untested):
#!/usr/bin/env bash
awk -v word="$1" -v pos="$2" -v ORS='\0' '$pos == word { print FILENAME; nextfile }' * |
xargs -0 cat
That assumes your awk supports nextfile and printing of \0 (e.g. GNU awk).
