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

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:

Answer 2

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.

Answer 3

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!

Answer 4

sed:

sed '5!d' file

awk:

awk 'NR==5' file

Answer 5

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

Answer 6

You could use sed -n 5p file.

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