I'm changing some notation in a few source code files.
In particular, variable names using the format
m_variable1
m_anothervariable
should be renamed and reformatted to
mVariable1
mAnotherVariable
That is, substitute m_ with m and make the next character uppercase.
I know how todo simple substitutions, like
%s/m_/m/gc
using vim, but not sure how to add syntax for changing a char to uppercase in a substitute statement?
You can make the first character of variable name uppercase, but I think you can hardly separate words from a consecutive string simply by built-in command.
I hope following command will help you:
:%s/\vm_(\w+)/m\u\1/g
Explaination
\v enables the 'very magic' mode
\u makes the first character of word after it uppercase
\1 references the first captured group
Result
mVariable1
mAnothervariable
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 have a file that is out-of-date and needs to be updated. The names have changed somewhat and I would like to clean them all up using a single substitution.
Here's what I'm trying to accomplish:
foo.foo_[single word] -> foo_bar.foo_[single word]_bar
where a single word is a string of n characters. In the file, they are always preceded by an underscore, but it needs to have "_bar" appended. There is always a "." after these instances, so I thought the following might work:
%s/foo\.foo_*\./foo_bar\.foo_*_bar\./g
Sadly, the first part doesn't even match what I want, so I'm back to square one.
I would first change:
foo_[word] -> foo_[word]_bar
and then
foo. -> foo_bar.
i.e.:
%s,\(foo_\w\+\),\1_bar,g|%s,foo\.,foo_bar\.,g
There are many ways to skin a cat but following should do the trick
%s/\vfoo.foo_(\w+)/foo_bar.foo_\1_bar/gc
what loosely translates to
\v Very Magic (:help magic)
foo.foo_ Search for exact string
(\w+) Search for a "word" and store in a backreference
/foo_bar.foo Replace search pattern with this exact string
\1 appended with backreference 1
_bar appended with _bar
or if you don't want to repeat the search in the replace part, you can go a bit nuts with backreferences and use
%s/\v(foo)\.foo_(\w+)/\1_bar.\1_\2_bar/gc
The most important parts you were missing were
using backreferences (:helpgrep backref)
using character classes (:h \w)
using repetition (_* is searching for 0 or more underscores. You probably meant _.*)
I have a pattern where there are double-quotes between numbers in a CSV file.
I can search for the pattern by [0-9]\"[0-9], but how do I retain value while removing the double quote. CSV format is like this:
"1234"5678","Text1","Text2"
"987654321","Text3","text4"
"7812891"3","Text5","Text6"
As you may notice there are double quotes between some numbers which I want to remove.
I have tried the following way, which is incorrect:
:%s/[0-9]\"[0-9]/[0-9][0-9]/g
Is it possible to execute a command at every search pattern, maybe go one character forward and delete it. How can "lx" be embedded in search and replace.
You need to capture groups. Try:
:%s/\(\d\)"\(\d\)/\1\2/g
[A digit can also be denoted by \d.]
I know that this question has been answered already, but here's another approach:
:%s/\d\zs"\ze\d
Explanation:
%s Substitute for the whole buffer
\d look up for a digit
\zs set the start of match here
" look up for a double-quote
\ze set the end of match here
\d look up for a digit
That makes the substitute command to match only the double-quote surrounded by digits.
Omitting the replacement string just deletes the match.
You need boundaries to use in regular expression.
Try this:
:%s/\([0-9]\)"\([0-9]\)/\1\2/g
A bit naive solution:
%s/^"/BEGINNING OF LINE QUOTE MARK/g
%s/\",\"/quote comma quote/g
%s/\"$/quota end of line/g
%s/\"//g
%s/quota end of line/"/g
%s/quote comma quote/","/g
%s/BEGINNING OF LINE QUOTE MARK/"/g
A macro can be created quite easy out of it and invoked as many times as needed.
I have several functions that start with get_ in my code:
get_num(...) , get_str(...)
I want to change them to get_*_struct(...).
Can I somehow match the get_* regex and then replace according to the pattern so that:
get_num(...) becomes get_num_struct(...),
get_str(...) becomes get_str_struct(...)
Can you also explain some logic behind it, because the theoretical regex aren't like the ones used in UNIX (or vi, are they different?) and I'm always struggling to figure them out.
This has to be done in the vi editor as this is main work tool.
Thanks!
To transform get_num(...) to get_num_struct(...), you need to capture the correct text in the input. And, you can't put the parentheses in the regular expression because you may need to match pointers to functions too, as in &get_distance, and uses in comments. However, and this depends partially on the fact that you are using vim and partially on how you need to keep the entire input together, I have checked that this works:
%s/get_\w\+/&_struct/g
On every line, find every expression starting with get_ and continuing with at least one letter, number, or underscore, and replace it with the entire matched string followed by _struct.
Darn it; I shouldn't answer these things on spec. Note that other regex engines might use \& instead of &. This depends on having magic set, which is default in vim.
For an alternate way to do it:
%s/get_\(\w*\)(/get_\1_struct(/g
What this does:
\w matches to any "word character"; \w* matches 0 or more word characters.
\(...\) tells vim to remember whatever matches .... So, \(w*\) means "match any number of word characters, and remember what you matched. You can then access it in the replacement with \1 (or \2 for the second, etc.)
So, the overall pattern get_\(\w*\)( looks for get_, followed by any number of word chars, followed by (.
The replacement then just does exactly what you want.
(Sorry if that was too verbose - not sure how comfortable you are with vim regex.)
I have a file containing string like this one :
print $hash_xml->{'div'}{'div'}{'div'}[1]...
I want to replace {'div'}{'div'}{'div'}[1] by something else.
So I tried
%s/{'div'}{'div'}{'div'}[1]/by something else/gc
The strings were not found. I though I had to escape the {,},[ and ]
Still string not found.
So I tried to search a single { and it found them.
Then I tried to search {'div'}{'div'}{'div'} and it found it again.
Then {'div'}{'div'}{'div'}[1 was still found.
To find {'div'}{'div'}{'div'}[1]
I had to use %s/{'div'}{'div'}{'div'}[1\]
Why ?
vim 7.3 on Linux
The [] are used in regular expressions to wrap a range of acceptable characters.
When both are supplied unescaped, vim is treating the search string as a regex.
So when you leave it out, or escape the final character, vim cannot interpret a single bracket in a regex context, so does a literal search (basically the best it can do given the search string).
Personally, I would escape the opening and closing square brace to ensure that the meaning is clear.
That's because the [ and ] characters are used to build the search pattern.
See :h pattern and use the help file pattern.txt to try the following experiment:
Searching for the "[9-0]" pattern (without quotes) using /[0-9] will match every digit from 0 to 9 individually (see :h \[)
Now, if you try /\[0-9] or /[0-9\] you will match the whole pattern: a zero, an hyphen and a nine inside square brackets. That's because when you escape one of [ or ] the operator [*] ceases to exist.
Using your search pattern, /{'div'}{'div'}{'div'}[1\] and /{'div'}{'div'}{'div'}\[1] should match the same pattern which is the one you want, while /{'div'}{'div'}{'div'}[1] matches the string {'div'}{'div'}{'div'}1.
In order to avoid being caught by these special characters in regular expressions, you can try using the very magic flag.
E.g.:
:%s/\V{'div'}[1]/replacement/
Notice the \V flag at the beginning of the line.
Because the square brackets mean that vim thinks you're looking for any of the characters inside. This is known as a 'character class'. By escaping either of the square brackets it lets vim know that you're looking for the literal square string ending with '[1]'.
Ideally you should write your expression as:
%s/{'div'}{'div'}{'div'}\[1\]/replacement string/
to ensure that the meaning is completely clear.