Since there seem to be still quite a few people who want to do this with sed...let's see how to select ranges of lines in the same way as with awk (explained here).
We should also avoid the same issue described there, that is, if other /BEGIN/ lines are found while we are inside a range, those lines should be printed. So with this input:
1 BEGIN 2 foo 3 bar 4 BEGIN 5 baz 6 END
at least lines 2 to 5 should be printed (line 1, or 6, or both may also be printed, depending on whether and which range endpoint we are including/excluding).
We're going to assume a sed with ERE (-E) support (as should be the norm these days anyway).
From BEGIN to END, inclusive
This is obviously the easy one:
# print lines from /BEGIN/ to /END/, inclusive $ sed '/BEGIN/,/END/!d' $ sed -n '/BEGIN/,/END/p'
No mysteries here. Let's get to the interesting cases.
From BEGIN to END, excluding END
# print lines from /BEGIN/ to /END/, excluding /END/ $ sed '/BEGIN/!d; :loop; n; /END/d; $!bloop'
We start a loop when we see a /BEGIN/, and keep looping until we see an /END/, at which point we delete the line so it's not printed.
From BEGIN to END, excluding BEGIN
# print lines from /BEGIN/ to /END/, excluding /BEGIN/
$ sed -E '/BEGIN/!d; :loop; N; /END/{ s/^[^\n]*\n//; p; d;}; $!bloop'
Same loop, but the lines are accumulated in the pattern space, and the first of them is removed before printing the whole block (note that the "D" command cannot be used for that purpose here, as it starts a new cycle).
From BEGIN to END, not inclusive
This is of course just a small variation on the preceding one, in that we delete both the first and the last line:
# print lines from /BEGIN/ to /END/, excluding both lines
$ sed -E '/BEGIN/!d; :loop; N; /END/{ s/^[^\n]*\n//; s/\n?[^\n]*$//; /./p; d;}; $!bloop'
Since we're excluding both the start and the end line, what's left after removing them may be empty, so we check that there's at least one character left and we only print the pattern space if that is the case.
For anything more complex, just use awk!
