SED: addressing two lines before match

Question

Print line, which is situated 2 lines before the match(pattern).

I tried next:

sed -n \': loop
/.*/h
:x
{n;n;/cen/p;}
s/./c/p
t x
s/n/c/p
t loop
{g;p         


        

Answer 1

I've tested your sed command but the result is strange (and obviously wrong), and you didn't give any explanation. You will have to save three lines in a buffer (named hold space), do a pattern search with the newest line and print the oldest one if it matches:

sed -n '
    ## At the beginning read three lines.
    1 { N; N }
    ## Append them to "hold space". In following iterations it will append
    ## only one line.
    H 
    ## Get content of "hold space" to "pattern space" and check if the 
    ## pattern matches. If so, extract content of first line (until a 
    ## newline) and exit.
    g
    /^.*\nsix$/ { 
        s/^\n//
        P
        q 
    }
    ## Remove the old of the three lines saved and append the new one.
    s/^\n[^\n]*//
    h
' infile

Assuming and input file (infile) with following content:

one
two
three
four
five
six
seven
eight
nine
ten

It will search six and as output will yield:

four