Shell Programming and Scripting

HiCan someone show me how to print all lines from a file after a line matching a pattern using sed?Thanks

try

Code:

sed '1,/regex/d'

will not include the first pattern line.

example:

Code:

$ cat data
123
ABC
DEF
GHI
$ sed '1,/A/d' data
DEF
GHI

Thanks. Doesn't quite work perhaps because I am trying to pipe the lines through sed rather than read a file. What I am trying to do is lose the message of the day on stdout when running a command though ssh by preceding the command with an echo of a string "PARSE_FROM_HERE". The intention is to lose all of the MOTD stuff and just pick up the ls output. I have trouble posting code from here so I hope this is not too garbled, if so I will post it tonight: -

Code:

ssh [IP ADDRESS] "echo PARSE_FROM_HERE;ls -lrt" | sed '1,/PARSE_FROM_HERE/d'

What about:

Code:

ssh [IP ADDRESS] "ls -lrt"

Try:

Code:

sed '0,/PARSE_FROM_HERE/d'

, that's right! I take a mistake.

Code:

ssh [IP ADDRESS] "echo PARSE_FROM_HERE;ls -lrt " |sed '0,/PARSE_FROM_HERE/d'

This gives me the MOTD, the PARSE_FROM_HERE line and the ls output. If I substitute a 1 for the 0 I get the MOTD and nothing from the PARSE_FROM_HERE line. Which seems to be the opposite of what is expected from redraiment's results above.