grep command for last nth line

Question

I have a grep command returns more than 20 lines. I need to display the 4th line from last line of the grep result,

lets say my file has below data,

test.txt

one  
two  
three  
four  
five  
six  
seven  
eight  
nine  
ten  

Task is :

grep "nine" test.txt // i want seven as result 

Can someone please help me with this ?

Answer 1

You can use -B option in grep to take any number of lines before your search line and then head -1 will get you the first line from output:

grep -B2 nine file | head -1

seven

Answer 2

Given:

$ cat file
one  
two  
three  
four  
five  
six  
seven  
eight  
nine  
ten  

You can use awk to keep n+1 number of lines in a buffer and then print the line that is n lines before the match:

$ awk -v n=2 'BEGIN{b=n+1} { buf[NR%b]=$0 } /nine/{ print buf[(NR-n)%b] }' file 
seven  

Answer 3

In awk:

$ awk -v s=nine '$0~s{print q}{q=p;p=$0}' file
seven  

Explained:

awk -v s='nine' '   # search string into s var
$0~s { print q }    # if match, print buffer q
{ q=p; p=$0 }       # q is previous of previous
' file

If there's going to be only one match you can add exit after the print q; .

Answer 4

If you want the nth line from the end, then clearly your input is limited so you don't need a streaming solution. So you might as well reverse the input and print the nth line from the beginning:

grep ... | tac | sed -n "${n}{p; q; }"