Get specific line from text file using just shell script

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). I need to do this only using a basic shell script.

cat file | while read line
	do
	   #do something
	done

I know how to iterate through lines, as shown above, but what if I just need to get the contents of a particular line

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:

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.)

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]}"

In parallel with William Pursell's answer, here is a simple construct which should work even in the original v7 Bourne shell (and thus also places where Bash is not available).

i=0
while read line; do
    i=`expr "$i" + 1`
    case $i in 5) echo "$line"; break;; esac
done <file

Notice also the optimization to break out of the loop when we have obtained the line we were looking for.

Assuming the question was asked for bash, here is the fastest simplest way to do this.

readarray -t a <file ; echo ${a[5-1]}

You may may discard array a when not needed anymore.

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

line=5; prep=`grep -ne ^ file.txt | grep -e ^$line:`; echo "${prep#$line:}"

#!/bin/bash
for i in {1..50}
do
 line=$(sed "${i}q;d" file.txt)
 echo $line
done