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PeterJones

That feature doesn’t exist natively, as far as I know.

You can add a feature request, per the instructions in the FAQ – make sure you search the existing issues, to make sure you aren’t duplicating an existing request.

There might be a workaround, using PythonScript or similar automation tool, but I cannot immediately see how to easily do that. If you’d like a workaround like that, speak up, and maybe someone here will find the time to ponder that.

Alan Kilborn

@Александр-Хренников

Notepad++/Scintilla has a hide/unhide lines feature, but it doesn’t work very well (has inconsistencies). @Claudia-Frank was working on a Pythonscript-based replacement a while ago, but she seems to have departed us for greener pastures. :)

Still, @PeterJones is on the right track–something could be done to make this a reality. Note that as you’ve defined it, it would only work for full line comments, not partial line comments.

guy038

Hello, @александр-хренников, @peterjones, @alan-kilborn and All,

I’m thinking of two possible work-arounds :

First solution :

You could create a simple Python or Lua script which pastes your code in a new tab, then executes, in the new tab’s text, the regex search/replacement, below :

SEARCH (?-s)^(\h*[#;].*|\h*)\R

REPLACE Leave EMPTY

which just gives your expected text :

port 1194
proto udp
dev tun
ca ca.crt
cert server.crt
key server.key  # This file should be kept secret
dh dh2048.pem
server 10.8.0.0 255.255.255.0
ifconfig-pool-persist ipp.txt
keepalive 10 120
cipher AES-256-CBC
persist-key
persist-tun
status openvpn-status.log
verb 3
explicit-exit-notify 1

Obviously, no code modification could be done, in the new tab ,only visualization !

Second solution :

Run the regex search/replacement, below, in your real code file :

SEARCH (?-s)^(\h*[#;].*|\h*)\R

REPLACE Leave EMPTY

In this condensed code, bookmark some parts of your code, that you’re interesting in, for later modifications
Performs a Ctrl+Z operation to restore the original contents of your code
From beginning of file, press the F2 or Shift + F2 shortcuts to navigate throughout these bookmarks ( which are kept ! ) and modify your code just as you want to ;-))

Best Regards,

guy038

Eko palypse

@guy038 said:

(?-s)^(\h*[#;].|\h)\R

if you could find a regex which results in a block instead of returning each line then
we could use something like editor.research and editor.hideLines to make this work.
What do you think?

Eko palypse

the reason for asking matching a block is because this

def hideLine(match):
    line = editor.lineFromPosition(match.span()[0])
    editor.hideLines(line,line)
    
editor.research(r'(?-s)^(\h*[#;].*|\h*)\R', hideLine)

does work but is somehow slow.

guy038

Hi, @александр-хренников, @peterjones, @alan-kilborn, @eko-palypse and All,

Ah, OK, I understand ! I should have found it, directly :-(

So, my second attempt would be :

SEARCH (?-s)(^\h*(?:[#;].*|)\R)(?1)*

REPLACE Leave EMPTY

This new formulation decrease, drastically, the number of replacements ;-)) From 301 to 13, in our example !

REMARK : Do note the difference between :

(?1) which is a routine call to the referenced group 1 itself ( ^\h*(?:[#;].*|)\R )
\1 which is a back-reference to the present value of group 1

For instance :

The regex (\d\d\d)(?1)* would match any range of digits, which is a multiple of 3, where as :
The regex (\d\d\d)\1* would just match any range of 3 digits, possibly repeated

Just test these two regexes, against text below :

123123123          # 123, repeated 3 times
123456789          # digits from 1 to 9
123456456          # 123, followed with 456, twice
123123123123123    # 123, repeated 5 times
12345601234567890  # '123456' + '1234567890' ( 17 digits )

As you can see, the regex (\d\d\d)(?1)* is strictly equivalent to the regex (\d\d\d)(\d\d\d)*

I’ve, even, found out a shorter regex, for our problem :

SEARCH (?-s)^\h*(?:[#;].*|)\R(?0)*

REPLACE Leave EMPTY

Where the (?0) syntax is a call routine to the entire regex. But, in that case, from my quick explanations, in the remark section, I don’t exactly understand WHY that regex does work too !!

Indeed :

The regex (?-s)(^\h*(?:[#;].*|)\R)(?1)* is equivalent to the regex (?-s)(^\h*(?:[#;].*|)\R)(^\h*(?:[#;].*|)\R)* Right !
So the regex (?-s)^\h*(?:[#;].*|)\R(?0)* should be equivalent to the regex (?-s)^\h*(?:[#;].*|)\R(?-s)^\h*(?:[#;].*|)\R(?0)**, which has, again, a (?0)* syntax, leading to a second development, and so on… ??

Anyway, it works fine ;-))

BR,

guy038

Eko palypse

brilliant, and a script like

regex = r'(?-s)(^\h*(?:[#;].*|)\R)(?1)*'

def hideLine(match):
    start, end = match.span()
    _start = editor.lineFromPosition(start)
    if _start == 0:
        _start = 1
    _end = editor.lineFromPosition(end) - 1
    editor.hideLines(_start,_end)
    
editor.research(regex, hideLine)

could hide the lines.
Note, scintilla prevents hiding the first line.

Alan Kilborn

@Eko-palypse

Anybody else notice that if you hide lines in this manner, then switch the tab you are editing in N++, then switch back, the lines are no longer hidden??

Eko palypse

@Alan-Kilborn

yes, theory - looks like npp is explicitly unfolding code if it can find its own fold marker
when activating a buffer. Haven’t checked source code yet. If this is the case, then by
either using npps fold marker or buffer activated callback to rehide the lines might solve
the issue.

Alan Kilborn

There’s a possible solution for this thread’s topic, HERE.