Sed to extract text between two strings

Solution 1:

sed -n '/^START=A$/,/^END$/p' data

The -n option means don't print by default; then the script says 'do print between the line containing START=A and the next END.

You can also do it with awk:

A pattern may consist of two patterns separated by a comma; in this case, the action is performed for all lines from an occurrence of the first pattern though an occurrence of the second.

(from man awk on Mac OS X).

awk '/^START=A$/,/^END$/ { print }' data

Given a modified form of the data file in the question:

START=A
  xxx01
  xxx02
END
START=A
  xxx03
  xxx04
END
START=A
  xxx05
  xxx06
END
START=B
  xxx07
  xxx08
END
START=A
  xxx09
  xxx10
END
START=C
  xxx11
  xxx12
END
START=A
  xxx13
  xxx14
END
START=D
  xxx15
  xxx16
END

The output using GNU sed or Mac OS X (BSD) sed, and using GNU awk or BSD awk, is the same:

START=A
  xxx01
  xxx02
END
START=A
  xxx03
  xxx04
END
START=A
  xxx05
  xxx06
END
START=A
  xxx09
  xxx10
END
START=A
  xxx13
  xxx14
END

Note how I modified the data file so it is easier to see where the various blocks of data printed came from in the file.

If you have a different output requirement (such as 'only the first block between START=A and END', or 'only the last ...'), then you need to articulate that more clearly in the question.

Solution 2:

Basic version ...

sed -n '/START=A/,/END/p' yourfile

More robust version...

sed -n '/^ *START=A *$/,/^ *END *$/p' yourfile

Solution 3:

Your sed expression has a space before end, i.e / ^END/. So sed gets the starting pattern, but does not get the ending pattern and keeps on printing till end. Use sed '/^START=A/, /^END/!d' input_file (notice /^END/)