I want to write a script that displays the third line of a file, for example, filename.txt using only grep, without using backticks, &&, || or ; or sed and awk in only one line.
So I have tried this command: grep -w "3" filename.txt but it doesn't seem to work, any rectifications?
You don't need grep since you aren't searching for a pattern, you want line #3 (or line N).
To get line 3 you can output the first 3 lines from the file, then take the last line from those three:
head -3 filename.txt | tail -1
You mention backticks, so if you want to assign that, or output it, enclose it in $(...) instead of backticks.
linethree=$(head -3 filename.txt | tail -1)
If the file is very large and the line N you want is far into the file it would be more efficient to use sed, such as
sed 4321q;d filename.txt or
sed <n>q;d filename.txt
as outlined in this answer.
Using BASH version 4 or higher mapfile:
Use mapfile to read third line only -s 2 discards 2 lines and -n 1 copies only 1 line into array named third_line:
mapfile -s 2 -n 1 third_line < filename.txt
Output array contents:
printf "%s" "${third_line[@]}"
You can use grep -n to number the lines, then you can pipe the output to another grep to extract the third line:
grep -n ^ file | grep ^3:
If you need to remove the 3: from the beginning of the line, you can pipe it to
| cut -f2- -d:
or, if you insist on grep, you can use
| grep -Po '(?<=^3:).*'
i.e. only output the part preceded by 3: at the beginning of a line (but -P is not supported everywhere).
