How to delete all lines before the second match of a pattern with sed?

Question

I have the following file:

first
second
third
fourth
third
fifth
sixth

Using cat file | sed -n '/third/,$p' cat file | sed -n '/third/,$p' I can print starting from the first match, in order to get:

third
fourth
third
fifth
sixth

Is it possible to modify the sed command such that it essentially ignores the first match and prints from the second match? That would be:

third
fifth
sixth

Answer 1

Here is an awk that does that by keeping a buffer to store all lines from occurrence of third in a line and resetting the buffer when third is found again:

awk '/third/{p=$0 RS; next} p{p=p $0 RS} END{printf "%s", p}' file

third
fifth
sixth

Alternatively, you may use this tac + awk :

tac file | awk '1; /third/{exit}' | tac

third
fifth
sixth

Answer 2

With sed:

sed '1,/third/d' file | sed -n '/third/,$p'

Output:

third
fifth
sixth

Answer 3

与gnu sed

sed '/third/!d;:A;N;/\nthird/!{s/[^\n]*\n//;bA };s/[^\n]*\n//;:B;N;bB' infile