Get specific line from text file using just shell script

Question

I am trying to get a specific line from a text file.

So far, online I have only seen stuff like sed, (I can only use the sh -not bash or sed or anything like that).

Answer 1

Easy with perl! If you want to get line 1, 3 and 5 from a file, say /etc/passwd:

perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd

Answer 2

The standard way to do this sort of thing is to use external tools. Disallowing the use of external tools while writing a shell script is absurd. However, if you really don't want to use external tools, you can print line 5 with:

i=0; while read line; do test $((++i)) = 5 && echo "$line"; done < input-file

Note that this will print logical line 5. That is, if input-file contains line continuations, they will be counted as a single line. You can change this behavior by adding -r to the read command. (Which is probably the desired behavior.)

Answer 3

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

Answer 4

I didn't particularly like any of the answers.

Here is how I did it.

# Convert the file into an array of strings
lines=(`cat "foo.txt"`)

# Print out the lines via array index
echo "${lines[0]}"
echo "${lines[1]}"
echo "${lines[5]}"