I am trying to control multi-line comments for the Lua language in Vim.
If, in insert mode, I input the sequence:
--[[<CR>hello<CR>goodbye<CR>]]
Then I want the final text to look like this:
--[[
hello
goodbye
]]
So far, I have not been able to find a setting of 'comments' to accomplish this.
In particular, the m (middle of 3-part comment) does not seem to accept a space (let alone 2 spaces) as its string.
I tried this:
set com=s:--[[,m:\ \ ,e:]],:--
But there are no leading spaces whatsoever. I get:
--[[
hello
goodbye
]]
Note, if I add a non-blank character, it does work (kind of):
set com=s2:--[[,m:.,e:]],:--
--[[
. hello
. goodbye
. ]]
I am especially confused because the Vim help page says:
Any blank space in the text before and after the {string} is part of
the {string}, so do not include leading or trailing blanks unless the
blanks are a required part of the comment string.
They way I read that, it says that any blanks I put in comment= are taken literally as blanks, and not ignored. But in my case, they seem to be ignored...
Any advice, here?
Related
I'm trying to convert multiple instances of Unicode codes to their corresponding characters.
I have some text with this format:
U+00A9
And I want to generate the following next to it:
©
I have tried to select the code in visual mode and use the selection range '<,'> in command mode as input for i_CTRL_V but I don't know how to use special keys on a command.
I haven't found anything useful in the manual with :help command-mode . I could solve this problem using other tools but I want to improve my vim knowledge. Any hint is appreciated.
Edit:
As #m_mlvx has pointed out my goal is to visually select, then run some command that looks up the Unicode and does the substitution. Manually input a substitution like :s/U+00A9/U+00A9 ©/g is not what I'm interested in as it would require manually typing each of the special characters on every substitution.
Any hint is appreciated.
Here are a whole lot of them…
:help i_ctrl-v is about insert mode and ranges matter in command-line mode so :help command-mode is totally irrelevant.
When they work on text, Ex commands only work on lines, not arbitrary text. This makes ranges like '<,'> irrelevant in this case.
After carefully reading :help i_ctrl-v_digit, linked from :help i_ctrl-v, we can conclude that it is supposed to be used:
with a lowercase u,
without the +,
without worrying about the case of the value.
So both of these should be correct:
<C-v>u00a9
<C-v>u00A9
But your input is U+00A9 so, even if you somehow manage to "capture" that U+00A9, you won't be able to use it as-is: it must be sanitized first. I would go with a substitution but, depending on how you want to use that value in the end, there are probably dozens of methods:
substitute('U+00A9', '\(\a\)+\(.*\)', '\L\1\2', '')
Explanation:
\(\a\) captures an alphabetic character.
+ matches a literal +.
\(.*\) captures the rest.
\L lowercases everything that comes after it.
\1\2 reuses the two capture groups above.
From there, we can imagine a substitution-based method. Assuming "And I want to generate the following next to it" means that you want to obtain:
U+00A9©
you could do:
v<motion>
y
:call feedkeys("'>a\<C-v>" . substitute(#", '\(\a\)+\(.*\)', '\L\1\2', '') . "\<Esc>")<CR>
Explanation:
v<motion> visually selects the text covered by <motion>.
y yanks it to the "unnamed register" #".
:help feedkeys() is used as low-level way to send a complex series of characters to Vim's input queue. It allows us to build the macro programatically before executing it.
'> moves the cursor to the end of the visual selection.
a starts insert mode after the cursor.
<C-v> + the output of the substitution inserts the appropriate character.
That snippet begs for being turned into a mapping, though.
In case you would like to just convert unicodes to corresponding characters, you could use such nr2char function:
:%s/U+\(\x\{4\}\)/\=nr2char('0x'.submatch(1))/g
Brief explanation
U+\(\x\{4\}\) - search for a specific pattern (U+ and four hexadecimal characters which are stored in group 1)
\= - substitute with result of expression
'0x'.submatch(1) - append 0x to our group (U+00A9 -> 0x00A9)
In case you would like to have unicode character next to text you need to modify slightly right side (use submatch(0) to get full match and . to append)
In case someone wonders how to compose the substitution command:
'<,'>s/\<[uU]+\(\x\+\)\>/\=submatch(0)..' '..nr2char(str2nr(submatch(1), 16), 1)/g
The regex is:
word start
Letter "U" or "u"
Literal "plus"
One or more hex digits (put into "capture group")
word end
Then substituted by (:h sub-replace-expression) concatenation of:
the whole matched string
single space
character by UTF-8 hex code taken from "capture group"
This is to be executed in Visual/command mode and works over selected line range.
I'm trying to complete an exercise from https://learnvimscriptthehardway.stevelosh.com/chapters/16.html
The sample text to be worked on is:
Topic One
=========
This is some text about topic one.
It has multiple paragraphs.
Topic Two
=========
This is some text about topic two. It has only one paragraph.
The mapping to delete the heading of Topic One or Topic Two (depending on which body the cursor is placed in) and enter insert mode is:
:onoremap ih :<c-u>execute "normal! ?^==\\+$\r:nohlsearch\rkvg_"<cr>
Enter 'cih' in the body of either text below the headings and respective heading will be erased and the cursor will be placed there ready to go, in insert mode. Great mapping--but, I'm trying to understand what's happening with \+$.
When I omit \+$ and use this mapping:
:onoremap ih :<c-u>execute "normal! ?^==\r:nohlsearch\rkvg_"<cr>
it works fine, seemingly identically to the other mapping. So what is the use of the \+$?
Here is how Mr. Losh explains it:
The first piece,
?^==\+$
performs a search backwards for any line that consists of two
or more equal signs and nothing else. This will leave our cursor on
the first character of the line of equal signs."
But what does \+$ accomplish? I've tried to enter it manually in command but I just get an error sound. It works as intended as part of the full function, though. but like I said, when I remove it and run the full command without, it works fine.
There's something I'm missing about the necessity of that '+$'... Maybe it has to do with the "two or more equal signs and nothing else"?
The author's command:
?^==\+$
searches backward for a line consisting exclusively of 2 or more equal signs:
^ anchors the pattern to the beginning of the line,
= matches a literal equal sign,
^= thus matches a literal equal sign at the beginning of the line,
= matches a second equal sign,
\+ matches one or more of the preceding atom, as many as possible,
=\+ thus matches one or more equal sign, as many as possible,
$ anchors the pattern to the end of the line,
so the pattern above is going to match any of the following lines:
==
===
=============
etc.
but not lines like:
==foo
== <- six spaces
etc.
which is exactly the goal of that exercice.
Your command, on the other hand:
?^==
searches backward for a sequence of two equal signs at the beginning of a line:
^ anchors the pattern to the beginning of the line,
== matches two literal equal signs,
so your pattern is going to match the same lines as above:
==
===
=============
etc.
but also lines like:
==foo
== <- six spaces
etc.
because it is not strict enough.
Your pattern would definitely be good enough if used manually to jump to one of those underlines because it gets the job done with minimal typing. But the goal, here, is to make a mapping. Those things have to be generalised to be reliable, which pretty much requires a level of explicitness and precision your pattern lacks.
In short, Steve's pattern checks all the boxes while yours doesn't: it is explicit and precise while yours is implicit and imprecise.
The \+$ is part of the regular expression matching a line of only equals signs. Without it, your mapping would recognize, for example,
This is not a heading
=This is not an underline
as a heading.
The \+ means "At least two of the previous character (=)". The $ means End of line, so there cannot be anything after the equals signs.
I'm new into vim, I have hug text file as follow:
ZK792.6,ZK792.6(let-60),cel-miR-62(18),0.239
UTR3,IV:11688688-11688716,0.0670782
ZC449.3b,ZC449.3(ZC449.3),cel-miR-62(18),0.514
UTR3,X:5020692-5020720,0.355907
First, I would like to get delete all rows with even numbers (2,4,6...).
Second, I would like to remove (18) from entire file. as a example:
cel-miR-62(18) would be cel-miR-62.
Third: How can I get delete all parentheses including it's inside?
Would someone help me with this?
For the first one:
:g/[02468]\>/d
where :g matches all lines by the regex between the slashes and runs d (delete line) on the matching lines. The regex is quite easy to read, the only interesting symbol there is perhaps the \>, which matches end of a word.
For the second question:
:%s/\V(18)//g
where % is the specification meaning "all lines of the file", s is the substitute command, \V sets the "very nomagic" mode of regexes (not sure what your default is, you might not need this) and the final g makes vim substitute all occurrences on each line (with an empty string, the one between slashes). Make sure that :set gdefault? prints nogdefault (the default setting of gdefault), otherwise, drop the final g from the substitute command.
To remove every even line (or every other line):
:g/^/+d
To remove every instance of (18):
:%s/(18)//g
Remove all the parenthetical content:
:%s/(.\\{-})//g
Note: the pattern in third answer is a non-greedy match.
I'm trying to figure out how to dt or df the last occurrence of a character in a string.
For example, let's say I have the following line:
foo not.relevant.text.bar
If I f df. I expectedly get foo relevant.text.bar but I would like to get foo bar. Using f 3df. is not an option as I don't know how many of that character will be in the string. Additionally, I may want to get foo .bar (f 3dt.), or if the line ends with a dot, I may want to get foo .. I want to always find the last one regardless of how many there are.
Is this possible without a regex? I suppose I could use a regex but I was hoping there was a simple vim command that I'm missing. I find myself trying to do something like this often.
one way without using regex, without counting "dot" (could be other letters)... see if others have better way..
foo[I]not.relevant.text.bar ([I] is cursor)
you could try:
lmm$T.d`m
or in this format, may look better?
lmm$T.d`m
this will do the job. you could create a mapping if you use that often.
EDIT
I add a GIF animation to show it works. :)
note
I typed #= in normal mode after moving my cursor to the right starting point (by f(space)), to display the keys I pressed in command line.
You can use my JumpToLastOccurrence plugin. It provides ,f / ,F / ,t / ,T commands that do just that.
I would use f df...
It is not necessarily shorter to type, but I find it easier to use "repeat last command" than counting in advance the number of word/sentence I want to delete.
Then you can adjust the number of . you type to adjust the length of the string you want to delete.
For your example: ET.dB
foo not.relevant.text.bar
And it works, as long as the cursor is anywhere within the text following "foo".
Strip Path from Path+Filename: ET/dB
I use it for stripping a pathname of all but the trailing filename.
Strip the path from /some/long/path/filename.ext leaving only the filename.
Just as long as:
The cursor is anywhere within the bold word
There are no spaces in that word
E Go to the end (since there are no spaces - also works if not the last thing on the line)
T/ Find the last / (stop just after it, so it will be deleted, as well)
dB Delete to the beginning of the word
In visual mode:
$F.d^
The $ goes to the end of the current line, F searches backward for a period and d^ deletes till the beginning of the line.
Is there an easy way to flip code around an equal sign in vi/vim?
Eg: I want to turn this:
value._1 = return_val.delta_clear_flags;
value._2._1 = return_val.delta_inactive_time_ts.tv_sec;
value._2._2 = return_val.delta_inactive_time_ts.tv_nsec;
value._3 = return_val.delta_inactive_distance_km;
(...)
into this:
return_val.delta_clear_flags = value._1;
return_val.delta_inactive_time_ts.tv_sec = value._2._1;
return_val.delta_inactive_time_ts.tv_nsec = value._2._2;
return_val.delta_inactive_distance_km = value._3;
(...)
on A LOT of lines in a file.
I know this seems a little trivial, but I've been running into lots of cases when coding where I've needed to do this in the past, and I've never had a good idea/way to do it that didn't require a lot of typing in vim, or writing a awk script. I would think this would be possible via a one liner in vi.
Explanations of the one-liners is very welcome and will be looked upon highly when I select my accepted answer. :)
Something like this:
:%s/\([^=]*\)\s\+=\s\+\([^;]*\)/\2 = \1
You might have to fiddle with it a bit if you have more complex code than what you have shown in the example.
EDIT: Explanation
We use the s/find/replace comand. The find part gets us this:
longest possible string consisting of anything-but-equal-signs, expressed by [^=]* ...
... followed by one or more spaces, \s\+ (the extra \ in front of + is a vim oddity)
... followed by = and again any number of spaces, =\s\+
... followed by the longest possible string of non-semicolon characters, [^;]*
Then we throw in a couple of capturing parentheses to save the stuff we'll need to construct the replacement string, that's the \(stuff\) syntax
And finally, we use the captured strings in the replace part of the s/find/replace command: that's \1 and \2.
For interest's sake, here's how I did it as a recorded macro:
qq0"+df=xA<BACKSPACE> = <ESC>"+pxi;<ESC>jq
Peforming that on the first line sets the "q" macro to do what's required. Then on each subsequent line you can execute the macro by typing:
#q
or, say you want to apply the macro to the next 10 lines:
10#q
I always find macros easier for a quick switch like this than figuring out the regex, because they're essentially an extension of how I would do it by hand.
Edit: Dan Olson points out in his comment that if you want to then apply the macro to a range of lines, for instance lines 6-100, you can enter the following. I don't know if there's a more concise syntax that doesn't require the ".*" pattern match.
:6,100g/.*/normal #q
Explanation of the macro
qq
start recording in register q
0
go to beginning of line
"+df=
delete up to the '=' and put the text into the '+' register
x
delete extra space
A
go to end of line and enter insert mode
<BACKSPACE> = <ESC>
Delete the semicolon, insert an equals sign and a space
"+p
insert the test copied earlier into register '+'
xi;<ESC>
reinsert the trailing semicolon
j
move down to the next line, ready to reapply the macro
q
stop recording
:%s/^\s*\(.\{-}\)\s*=\s*\(.\{-}\)\s*;\s*$/\2 = \1;/
should work nicely.
:%s/\([^ =]*\)\s*=\s*\([^;]*\);/\2 = \1;/