On 11/27/11 17:23, Bee wrote:
:let @a=''|g/^\(\%^\|\n\)\(.\+\n\)*.*Regexp/;'}y A
I tried to break out the pieces.
Does the ';' mean perform another search?
:let @a=''|g/^\(\%^\|\n\)\(.\+\n\)*.*foo/;'}y A
\%^ Matches start of the file
When matching with a string matches the start of the string
example, to find the first "VIM" in a file: /\%^\_.\{-}\zsVIM
\| A pattern is one or more branches
; Perform another search
'{ `{ To the start of the current paragraph
You're close...it could be clarified/expanded by putting a "." in
front of the ";" which delimits a range
:help :;
so the :g searches for a paragraph-break or the start of file,
followed by zero-or-more lines containing at least one character
(not a paragraph break), followed by a line containing the
"Regexp" you're searching for. It then, starting on the matching
line (the one with either the start-of-file or paragraph break),
a range is created through the next paragraph break ("'}") and
yanked-appending into register "a".
g/ On every line matching this pattern
^ a start-of-line
\( where either of these two branches match:
\%^ the start-of-file
\| or
\n a newline (thus this is a blank line)
\) (end of the branches)
\(.\+\n\)* followed by 0+ lines containing 1+ characters
.*foo followed by stuff matching up to "foo" (your RE)
/ Perform the following action
(implicit current line as the start-of-range)
; through
'} the next-paragraph mark (either a blank line
or the EOF)
y A Yank the range into register "a", appending
Hope that helps shed some light on the opacity of the expression.
-tim
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