Easy enough, with regular expressions :-))

• Open the Replace dialog ( CTRL + H )

• SEARCH .(?=\d\d:\d\d:\d\d:)

• REPLACE $0\r\n

• Check the Regular expression search mode

• Leave the . matches newline option UNchecked

• if necessary, check the Wrap around option

• Click on the Replace All button

Et voilà !

Notes :

• The dot ( . ) represents any standard character ( different from \r, \n and \f )

• The syntax (?=\d\d:\d\d:\d\d:) is called a look-ahead, that looks, after the dot, for 3 sets of two digits, each of them followed with a colon. However, although this condition must be true, it’s NEVER part of the regex to be replaced. Therefore, the search regex is, only, the character just before the time syntax HH:MM:SS.

• The $0, in the replacement part, represents the entire regex matched, that is to say, the dot, which must be separated, from the HH:MM:SS form, with a line break ( \r\n ). If you use UNIX files, just change \r\n into \n.

So, the last line of your post :

17:30:24:THE SUN DEVILS OF ARIZONA STATE17:30:26:UNIVERSITY.17:30:27:ASU GETTING SET TO TAKE ON17:30:29:KENNESAW STATE.17:30:30:PART OF THE LEGENDS

will be replaced by the five lines below :

17:30:24:THE SUN DEVILS OF ARIZONA STATE
17:30:26:UNIVERSITY.
17:30:27:ASU GETTING SET TO TAKE ON
17:30:29:KENNESAW STATE.
17:30:30:PART OF THE LEGENDS

Best Regards,

guy038

P.S. :

You’ll find good documentation, about the new Boost C++ Regex library ( similar to the PERL Regular Common Expressions ) used by Notepad++, since the 6.0 version, at the TWO addresses below :

http://www.boost.org/doc/libs/1_48_0/libs/regex/doc/html/boost_regex/syntax/perl_syntax.html

http://www.boost.org/doc/libs/1_48_0/libs/regex/doc/html/boost_regex/format/boost_format_syntax.html

• The FIRST link explains the syntax, of regular expressions, in the SEARCH part

• The SECOND link explains the syntax, of regular expressions, in the REPLACEMENT part