Searching specific lines of files using GREP - linux

Question

I have a directory with many text files. I want to search a given string in specific lines in the files(like searching for 'abc' in only 2nd and 3rd line of each file). Then When I find A match I want to print line 1 of the matching file.
My Approach - I'm doing a grep search with -n option and storing the output in a different file and then searching that file for the line number. Then I'm trying to get the file name and then print out it's first line.
Using the approach I mentioned above I'm not able to get the file name of the right file and even if I get that this approach is very lengthy.
Is there a better and fast solution to this?
Eg.
1.txt
file 1
one
two
2.txt
file 2
two
three
I want to search for "two" in line 2 of each file using grep and then print the first line of the file with match. In this example that would be 2.txt and the output should be "file 2"
I know it is easier using sed/awk but is there any way to do this using grep?

Answer 1

Use sed instead (GNU sed):
parse.sed
1h # Save the first line to hold space
2,3 { # On lines 2 and 3
/my pattern/ { # Match `my pattern`
x # If there is a match bring back the first line
p # and print it
:a; n; ba # Loop to the end of the file
}
}
Run it like this:
sed -snf parse.sed file1 file2 ...
Or as a one-liner:
sed -sn '1h; 2,3 { /my pattern/ { x; p; :a; n; ba; } }' file1 file2 ...
You might want to emit the filename as well, e.g. with your example data:
parse2.sed
1h # Save the first line to hold space
2,3 { # On lines 2 and 3
/two/ { # Match `my pattern`
F # Output the filename of the file currently being processed
x # If there is a match bring back the first line
p # and print it
:a; n; ba # Loop to the end of the file
}
}
Run it like this:
sed -snf parse2.sed file1 file2 | paste -d: - -
Output:
file1:file 1
file2:file 2

Answer 2

$ awk 'FNR==2{if(/one/) print line; nextfile} FNR==1{line=$0}' 1.txt 2.txt
file 1
$ awk 'FNR==2{if(/two/) print line; nextfile} FNR==1{line=$0}' 1.txt 2.txt
file 2
FNR will have line number for the current file being read
use FNR>=2 && FNR<=3 if you need a range of lines
FNR==1{line=$0} will save the contents of first line for future use
nextfile should be supported by most implementations, but the solution will still work (slower though) if you need to remove it

Answer 3

With grep and bash:
# Grep for a pattern and print filename and line number
grep -Hn one file[12] |
# Loop over matches where f=filename, n=match-line-number and s=matched-line
while IFS=: read f n s; do
# If match was on line 2 or line 3
# print the first line of the file
(( n == 2 || n == 3 )) && head -n1 $f
done
Output:
file 1

Answer 4

Only using grep, cut and | (pipe):
grep -rnw pattern dir | grep ":line_num:" | cut -d':' -f 1
Explanation
grep -rnw pattern dir
It return name of the file(s) where the pattern was found along with the line number.
It's output will be somthing like this
path/to/file/file1(.txt):8:some pattern 1
path/to/file/file2(.txt):4:some pattern 2
path/to/file/file3(.txt):2:some pattern 3
Now I'm using another grep to get the file with the right line number (for e.g. file that contains the pattern in line 2)
grep -rnw pattern dir | grep ":2:"
It's output will be
path/to/file/file3(.txt):2:line
Now I'm using cut to get the filename
grep -rnw pattern dir | grep ":2:" | cut -d':' -f 1
It will output the file name like this
path/to/file/file3(.txt)
P.S. - If you want to remove the "path/to/file/" from the filename you can use rev then cut and again rev, you can try this yourself or see the code below.
grep -rnw pattern dir | grep ":2:" | cut -d':' -f 1 | rev | cut -d'/' -f 1 | rev