view a particular line of a file denoted by a number

Question

Okay, this is probably an evident thing but it escapes me, as in it could probably be done in a much simpler way that I'm not aware of, so far.. Say there's a "file" and I want to view only what's on line number "X" of that file, what would be the solution?

here's what i can think of:

head -X < file | tail -1  
 sed -n Xp < file

is there anything else (or any other way) from the standard set of unix/gnu/linux text-tools/utils?

Solution 1

sed -n 'Xp' theFile, where X is your line number and theFile is your file.

Solution 2

in vi:

vi +X filename

in EMACS:

emacs +X filename

in the shell:

nl -ba -nln filename| grep '^X '

you can use context grep cgrep instead of grep to see some lines above and below the matching line..

---

EXAMPLES:

print just that one line:

$ nl -ba  -nln  active_record.rb  | grep '^111 '
111       module ConnectionAdapters

with context:

$ nl -ba  -nln  active_record.rb  | grep -C 2 '^111 '
109       end
110     
111       module ConnectionAdapters
112         extend ActiveSupport::Autoload
113     

for context control in grep check man grep :

   Context Line Control
       -A NUM, --after-context=NUM
              Print NUM lines of trailing context after matching lines.  Places a line containing a group separator (--) between contiguous groups of matches.  With the -o or --only-matching option, this has no effect and a warning is given.

       -B NUM, --before-context=NUM
              Print NUM lines of leading context before matching lines.  Places a line containing a group separator (--) between contiguous groups of matches.  With the -o or --only-matching option, this has no effect and a warning is given.

       -C NUM, -NUM, --context=NUM
              Print NUM lines of output context.  Places a line containing a group separator (--) between contiguous groups of matches.  With the -o or --only-matching option, this has no effect and a warning is given.

Solution 3

awk one-liner:

awk "NR==$X" file

bash loop:

for ((i=1; i<=X; i++)); do
  read l
done < file
echo "$l"

Solution 4

Just use vi

vi file

When in the file type

:X

where X is the line number you want to see

However, the sed -n Xp file is a good way if you really only want to see the one line

View more solutions

Comments

[A. Rex]

Essentially equivalent to the sed one: echo Xp | ed -s file

[Admin]

you can go directly to the line - as the other answer suggests

[matja]

why was this downvoted? this is the most portable, simplest and most efficient way. one doesn't even need the quotes.

[fig]

@XXL - so? The point is that it's the best answer, so it should be voted as such.

[XXL]

@fig what are you talking about? how is it the best answer? the poster didn't bother reading the OP's post which already had this exact part. What's the use of posting the same thing?!

[Dan Passaro]

@XXL to reinforce that this is a simple and efficient way to do the task at hand

[Daniel Fortunov]

I rarely look for the answer in the question, so it's useful to have it as a separate answer, that can be voted up or down relative the other answers. +1