Linux – SED: How to print every line after first instance of string using Sed

bashlinuxsedshell

I have a file with a similar format…

16:28 asdfasdf
16:29 4398upte
16:30 34liuthr
16:31 34tertio

How can I use SED to print out every line including and after the line with "16:30"?

The result would be…

16:30 34liuthr
16:31 34tertio

Right now, I am using sed as follows, but I have to manually find the first line's line number e.g. "562697":

sed -n '562697,$p'

Best Answer

Addresses in sed can be either line numbers or patterns. Try this:

sed -n '/16:30/,$p'

If the pattern contains a /, you can escape it with a \. For example, to search for 16/30 instead of 16:30, try this:

sed -n '/16\/30/,$p'