How to recognize a single line via regexs
Hi,
how can I identify a single line no longer than e.g. 60 characters
preceded and followed by a blank line via regexs. This way I want to
identify section headings. What I did was mark every blank line with
%s/^$// and than chomp the CR %s/\n//g and if the text
between the 's isn't longer than 60 characters put it into
\section{}, and replace every with \r\r. But in larger files
this takes a while. Is there a smarter solution to the problem?
Thanks in advance
Chris
Re: How to recognize a single line via regexs
Informationen wrote:
Hi,
how can I identify a single line no longer than e.g. 60 characters
preceded and followed by a blank line via regexs. This way I want to
identify section headings. What I did was mark every blank line with
%s/^$// and than chomp the CR %s/\n//g and if the text
between the 's isn't longer than 60 characters put it into
\section{}, and replace every with \r\r. But in larger files
this takes a while. Is there a smarter solution to the problem?
The following regexp will do the match:
^\n\zs.\{1,60}\ze\n$
^ beginning of line
\n newline
\zs pattern really starts here (but must be precededed by the foregoing)
.\{1,60} one to sixty characters
\ze pattern really ends here (but must be followed by the trailing pattern)
\n newline
$ end-of-line
Please read: :help regexp for (much) more.
Regards,
Chip Campbell
Re: How to recognize a single line via regexs
how can I identify a single line no longer than e.g. 60 characters
preceded and followed by a blank line via regexs. This way I want to
identify section headings. What I did was mark every blank line with
%s/^$// and than chomp the CR %s/\n//g and if the text
between the 's isn't longer than 60 characters put it into
\section{}, and replace every with \r\r. But in larger files
this takes a while. Is there a smarter solution to the problem?
You might try:
/^\s*\n\zs.\{1,60}\ze\n\s*$
/^\n\zs.\{1,60}\ze\n$
which should find what you describe (the former allows a blank
line to contain whitespace while the latter looks for a truely
blank line with zero whitespace).
It breaks down as
^ the start of line
\s*(the optional whitespace, as described)
\n a newline
\zstreat this as the beginning of the match
.\{1,60} 1-60 non-newline characters
\zeyou could optionally use this here
to treat this as the end of the match
\n another newline
\s*(more optional whitespace, as described)
$ the end of a line
The above regexp(s) (with or without the \s* or the \ze) can
also be used in :s commands to do things like
:%s/^\s*\n\zs.\{1,60}\ze\n\s*$/\U
to uppercase all your titles, or
:%s/^\s*\n\zs.\{1,60}\ze\n\s*$/\=submatch(0).\n.substitute(submatch(0),
'.', '=', 'g')
which will underline all your hits with a row of = the same
length as your section-title.
Just a few ideas,
-tim
