Hide comments mode

[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.