Sed : print all lines after match

Question

I got my research result after using sed :

zcat file* | sed -e \'s/.*text=\\(.*\\)status=[^/]*/\\1/\' | cut -f 1 - | grep \"pattern\"
         


        

Answer 1

The seldom used branch command will do this for you. Until you match, use n for next then branch to beginning. After match, use n to skip the matching line, then a loop copying the remaining lines.

cat file | sed -n -e ':start; /pattern/b match;n; b start; :match n; :copy; p; n ; b copy'