I'm trying to replace extra blank after every line which matched like this
[\s]+\n
the problem is empty lines will also be matched
Base on my understanding, "\s" is equal to "[ \f\n\r\t\v]", so I changed the expression into
[ \f\r\t\v]+\n
to rule out "\n", but this time I got nothing...
Anybody knows why?
to find white space at ends of lines I would use
[ \t]+(?=\r?\n)
It will not catch the last line with white space if it has no \n. For that you need to use $ that is not affected by \n
[ \t]+$
Background
First of all, Visual Studio Code regex does not work in the usual way sometimes as it was "tweaked" to work with a text editor.
Here, note that \s only matches line break chars (vertical whitespace) if there is \n or \r in the pattern. So, [\s]+\n matches across lines, it matches 1 or more any whitespace chars, and then stops at the last newline in this chunk of whitespaces.
So, if you need to match horizontal whitespace at the end of a line, I thought you could use a combination of \s+ (one or more whitespace chars) and $ (end of a line).
However, \s+$ only matches whitespace at the end of a line if there are at least two occurrences of the whitespace chars.
Solution
\s*$
Demo screenshot:
