Get specific line from text file using just shell script

Question

sed:

sed '5!d' file

awk:

awk 'NR==5' file

Assuming line is a variable which holds your required line number, if you can use head and tail, then it is quite simple:

head -n $line file | tail -1

If not, this should work:

x=0
want=5
cat lines | while read line; do
  x=$(( x+1 ))
  if [ $x -eq "$want" ]; then
    echo $line
    break
  fi
done

You could use sed -n 5p file.

You can also get a range, e.g., sed -n 5,10p file.

Best performance method

sed '5q;d' file

Because sed stops reading any lines after the 5th one

Update experiment from Mr. Roger Dueck

I installed wcanadian-insane (6.6MB) and compared sed -n 1p /usr/share/dict/words and sed '1q;d' /usr/share/dict/words using the time command; the first took 0.043s, the second only 0.002s, so using 'q' is definitely a performance improvement!

If for example you want to get the lines 10 to 20 of a file you can use each of these two methods:

head -n 20 york.txt | tail -11

or

sed -n '10,20p' york.txt 

p in above command stands for printing.

Here's what you'll see: