sed and remove string between two patterns

As @steeldriver points out, it is easy if you have non-greedy regexps. If not, you can do it with a loop, like this:

sed ':a;s/PATTERN-2/\n/;s/PATTERN-1.*\n//;ta' test.txt

This works because we know there are no newlines in the middle of any line. It would also work with any other character that does not occur in any line, e.g. §.

If you want to use only sed try like below

sed 's/PATTERN-1[^P]*PATTERN-2//g' test.txt

In your example, the .* matches stuff that you want to keep.

You can capture that stuff and replace it back by using:

sed 's/PATTERN-1\(.*\)PATTERN-2/\1/g' test.txt

Everything between the brackets gets stored in the first capture buffer and \1 substitute in the value of that buffer.

It's easy in a regex implementations that support a non-greedy qualifier, such as perl's ? For example:

perl -pe 's/PATTERN-1.*?PATTERN-2//g' test.txt

"Closest" is not really a sed term. But if there is a reasonable limit to the number of repeats of the sequence PATTERN-1.*PATTERN-2 you can hardcode for that number as follows:

     $ sed -E 's/(PATTERN-1).*(PATTERN-2)(.*)\1.*\2/\3/g;s/PATTERN-1.*PATTERN-2//g' <<"end"
     PATTERN-1xxxxPATTERN-2aaa
     PATTERN-1xxxxPATTERN-2fffPATTERN-1zzzzPATTERN-2gggPATTERN-1zzzzPATTERN-2
     PATTERN-1xxxxPATTERN-2bbb
     end

     aaa
     ggg
     bbb

Note that I use the -E option for extended regex syntax. Also note that in the search expression I use backreferences for the PATTERN-1 and -2 strings, just for your comfort.