Output text from file after certain string

Question

I am looking for a methode to return all text from after a certain phrase in the file.

So I have text file, lets say it called README, which contains the text below. I need a command to only output the text that comes after the line UserMode instructions. The text can vary in lengt so using a fixed line cut is not an option. It must be based on a specific string.

I've looked a the sed command but could not quite figure out how to achieve what I want.

Linux is easily portable to most general-purpose 32- or 64-bit architectures as long as they have a paged memory management unit (PMMU) and a port of the GNU C compiler (gcc) (part of The GNU Compiler Collection, GCC). Linux has also been ported to a number of architectures without a PMMU, although functionality is then obviously somewhat limited. Linux has also been ported to itself. You can now run the kernel as a userspace application - this is called UserMode Linux (UML).

UserMode instructions
In UML environments, host and guest kernel versions don't need to match, so it is entirely possible to test a "bleeding edge" version of Linux in User-mode on a system running a much older kernel. UML also allows kernel debugging to be performed on one machine, where other kernel debugging tools (such as kgdb) require two machines connected with a null modem cable.

Answer 1

Using range with awk:

awk '/UserMode instructions/,0' file

The end range condition 0 never match, so the rest of the file is printed.

---

Equivalent sed approach:

sed -n '/UserMode instructions/,$p' file

-n disables the default print.
p prints the line if the range matches