Extract Nth line after matching pattern

Question

I want to extract the Nth line after a matching pattern using grep, awk or sed.

For example I have this piece of text:

Answer 1

Printing the lnb^th line after first blank/empty line:

Index of line to print (in bash shell):

lnb=2

• Using sed:

  sed -ne '/^\s*$/{:a;n;0~'"$lnb"'!ba;p;q}' my_file`
  

• Using perl:

  perl -ne '/^\s+$/ && $k++;$k!=0 && $k++ && $k=='"$lnb"'+2 && (print,last)' my_file`
  

Printing the lnb^th line after regular-expression match:

• Using sed:

  sed -ne '/regex/{:a;n;0~'"$lnb"'!ba;p;q}' my_file
  

• Using perl:

  perl -ne '/regex/ && $k++;$k!=0 && $k++ && $k=='"$lnb"'+2 && (print,last)' my_file
  

  Bonus 1, Windows PowerShell (install Perl first) :

  $lnb=2
  
  perl -ne "/regex/ && `$k++;`$k!=0 && `$k++ && `$k==$lnb+2 && (print,last)" my_file
  

  Bonus 2, Windows DOS command line :

  set lnb=2
  
  perl -ne "/regex/ && $k++;$k!=0 && $k++ && $k==%lnb%+2 && (print,last)" my_file
  

Printing ALL lnb^th lines after regular-expression match:

• Using perl(bash example):

  perl -ne '/regex/ && $k++;$k!=0 && $k++ && $k=='"$lnb"'+2 && (print,$k=0)' my_file