How to read n-th line from a text file in bash?
bash, file, linux
Solution
Using `head` and `tail`
$ head -2 inputFile | tail -1
5 6 7 8
OR
a generalized version
$ line=2
$ head -"$line" input | tail -1
5 6 7 8
Using sed
$ sed -n '2 p' input
5 6 7 8
$ sed -n "$line p" input
5 6 7 8
What it does?
`-n` suppresses normal printing of pattern space.
`'2 p'` specifies the line number, `2` or (`$line` for more general), `p` commands to print the current patternspace
`input` input file
Edit
To get the output to some variable use some command substitution techniques.
$ content=`sed -n "$line p" input`
$ echo $content
5 6 7 8
OR
$ content=$(sed -n "$line p" input)
$ echo $content
5 6 7 8
To obtain the output to a bash array
$ content= ( $(sed -n "$line p" input) )
$ echo ${content[0]}
5
$ echo ${content[1]}
6
Using awk
Perhaps an `awk` solution might look like
$ awk -v line=$line 'NR==line' input
5 6 7 8
Thanks to Fredrik Pihl for the suggestion.
Problem
Say I have a text file called "demo.txt" who looks like this: ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ``` Now I want to read a certain line, say line 2, with a command which will look something like this: ``` Line2 = read 2 "demo.txt" ``` So when I'll print it: ``` echo "$Line2" ``` I'll get: ``` 5 6 7 8 ``` I know how to use 'sed' command in order to print a n-th line from a file, but not how to read it. I also know the 'read' command but dont know how to use it in order a certain line. Thanks in advance for the help.