Write a bash script to print a particular line from a file

Given a file, name file.txt, out task is to write a bash script which print particular line from a file.

Content of file.txt:

I
love
reading
articles
at
geeks
for
geeks

Syntax in Bash Script

1. awk :

   $>awk '{if(NR==LINE_NUMBER) print $0}' file.txt

2. sed :

   $>sed -n LINE_NUMBERp file.txt

3. head :

   $>head -n LINE_NUMBER file.txt | tail -n + LINE_NUMBER
   Here LINE_NUMBER is, 
   which line number you want to print

   Examples:

   Print a line from single file

   To print 4th line from the file then we will run following commands. The output of following lines will be “articles”.

   1. awk :

      $>awk '{if(NR==4) print $0}' file.txt

   2. sed :

      $>sed -n 4p file.txt 

   3. head :

      $>head -n 4 file.txt | tail -n + 4

   Print a line from multiple files

   Suppose we have two files, file1.txt and file2.txt, We can use the above commands and print particular line from multiple files by ‘&’.

   1. awk:

      $>awk '{if(NR==4) print $0}' file1.txt & awk '{if(NR==4) print $0}' file2.txt

   2. sed:

      $>sed -n 4p file1.txt & sed -n 4p file2.txt

   3. head:

      $>head -n 4 file1.txt | tail -n + 4 & head -n 4 file2.txt | tail -n + 4

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