I am copy-pasting an example from a PDF to Vim and I have to replace all “ and ” with "
and all ‘ and ’ with ' so that the code works.
Well that will probably seem easier to understand:
I want to replace all foo and bar with foobar simultaneously.
Try this in vi:
:1,$s/[“”]/"/g
then
:1,$s/[‘’]/'/g
Use tr as a filter:
Unix way:
:%!tr “”‘’ \"\"\'\'
If you want to replace all "foo"s and all "bar"s with "foobar" you can use this:
%s/\v<(foo|bar)>/foobar/g
This will replace the "foo"s and the "bar"s but will leave any "foobar"s alone.
%s/ - substitute across the whole file
\v - use very magic regex syntax (see :help magic for more info)
< - match a left word boundary
(foo|bar) - foo or bar
> - match a right word boundary
/foobar/ - replacement string
g - globally (will happen for every occurrence, not just the first on the line)
Note that if you are just dealing with punctuation you'll probably want to remove the word boundary parts of this regex or it won't work.
Related
I am little new to Vim world. I am trying to substitute *=, ~=(actually [special char]=) in to [whatever is symbol]=(adding space both sides). Here is my substitute command:
:%s/[~,\*]=/ = /g
the problem in this case is that I am not able to add respective special symbol before the equal sign. Can you help me...
This is a classic capture and replace use case. Capture the symbol part by enclosing it in \(...\), and then reference it in the replacement part via \1. You'll find more details at :help s/\1 (or :help :substitute in general):
:%s/\([~,\*]\)=/ \1= /g
Alternatively, you can start the match only on the = with \zs. This asserts that the symbol part is there, but as it isn't included in the match, you don't need to reference it:
:%s/[~,\*]\zs=/ = /g
The same trick can be applied with \ze at the end. As you can see, this often results in shorter commands.
This is probably the simplest answer to your question:
:%s/[~,\*]=/ & /
An& in the replace segment means 'entire match'.
In vi (from cygwin), when I do searching:
:%s/something
It just replaces the something with empty string like
:%s/something// .
I've googled for a while but nothing really mentions this. Is there anything I should add to the .vimrc or .exrc to make this work?
Thanks!
In vi and vim, when you search for a pattern, you can search it again by simply typing /. It is understood that the previous pattern has to be used when no pattern is specified for searching.
(Though, you can press n for finding next occurence)
Same way, when you give a source (pattern) and leave the replacement in substitute command, it assumes that the replacement is empty and hence the given pattern is replaced with no characters (in other words, the pattern is removed)
In your case, you should understand that % stand for whole file(buffer) and s for substitute. To search, you can simply use /, followed by a pattern. To substitute , you will use :s. You need not confuse searching and substituting. Hence, no need for such settings in ~/.exrc. Also, remember that / is enough to search the whole buffer and % isnt necessary with /. / searches the entire buffer implicitly.
You may also want to look at :g/Pattern/. Learn more about it by searching :help global or :help :g in command line.
The format of a substitution in vim is as follows:
:[range]s[ubstitute]/{pattern}/{string}/[flags] [count]
In your case you have omitted the string from the substitution command and here what vim documentation stated about it:
If the {string} is omitted the substitute is done as if it's empty.
Thus the matched pattern is deleted. The separator after {pattern}
can also be left out then. Example: >
:%s/TESTING This deletes "TESTING" from all lines, but only one per line.
For compatibility with Vi these two exceptions are allowed:
"/{string}/" and "\?{string}?" do the same as "//{string}/r".
"\&{string}&" does the same as "//{string}/".
E146
Instead of the '/' which surrounds the pattern and replacement string, you can
use any other single-byte character, but not an alphanumeric
character, '\', '"' or '|'. This is useful if you want to include a
'/' in the search pattern or replacement string. Example: >
:s+/+//+
In other words :%s/something and :%s;something or :%s,something have all the same behavior because the / ; and , in the last examples are considered only as SIMPLE SEPARATOR
I'm using vim to make my programs in c/c++ and I would like to know how can I put "\n" (which represents a newline) in my code or use "%" using :%s.
For instance, sometimes I forget to put "%" in front of "d" or "f" in many lines or forget to put "\n" in some printf() calls.
printf("This is my d code.", x);
But the following command does not work, it puts a space in place of "\n"!
:%s/\<code.\>/code.\n/gc
or
:%s/\<d\>/%d/gc
How can I do what I want?
The :help s/\n has the answer:
\n insert a <NL> (<NUL> in the file)
(does NOT break the line) *s/\n*
You'll also find the solution there: to insert a literal backslash, escape it \\ by doubling; to split the line, a \r has to be used. Yes, this is inconsistent, and it works differently in similar tools like sed, but that's unfortunately how it is.
Notes
The \n doesn't insert a space, but the special <NL> character, which usually is shown as ^#.
The \<code.\> isn't right; to match a literal period, you have to escape it: \.. Else, it matches any character. Likewise, the . usually isn't a keyword character, so the \> boundary wouldn't match.
You don't need to repeat the match text in the replacement, you can use & for it. Also read up on capture groups (:help /\() and the submatch references :help s/\1. This is a better way:
:%s/\<code\./&\\n/gc
(I don't see a problem with the second substitution.)
You want to insert the two-character sequence \n, not a literal newline (the latter would create a syntax error).
A sample line to be changed is:
printf("This is my d code.", x);
One problem with your attempt:
:%s/\<code.\>/code.\n/gc
is that there is no word boundary between the . and the " following the word code. The other problem is that \ in the target is used to escape special characters (for example you can refer to a / character as \/), so the \ must itself be escaped.
This should do the job:
:%s/\<code\."/code.\\n"/gc
A more general solution might be:
:g/printf/s/"/\\n"/egc
which offers to replace " by \n" on each line that contains printf -- but that will miss any printf calls that span more than one line.
As for replacing the d by %d, the command you have in your question:
:%s/\<d\>/%d/gc
is correct.
I always wanted to know, how you can substitute within given parameters.
If you have a line like this:
123,Hello,World,(I am, here), unknown
and you wnat to replace World with Foobar then this is an easy task: :%s/World/Foobar/
Now I wonder how I can get rid of a , which is wihtin the ( ).
Theoretically I just have to find the first occurance of ( then substitute the , with a blank until ).
Try lookahead and lookbehind assertions:
%s/([^)]*\zs,\ze.*)//
(\zs and \ze tell where pattern starts and end)
or
%s/\(([^)]*\)\#<=,\(.*)\)\#=//
The first one is more readable, the second one uses \( ... \) groupings with parentheses inside groups which makes it look like obfuscated, and \#<= which apart from being a nice ASCII-art duck is the lookbehind operator, and \#= that is the lookahead operator.
References: :help pattern (more detail at :help /\#=, :help /\ze, etc.)
You use the GUI and want to try those commands? Copy them into the clipboard and run :#+ inside Gvim.
Modifying slightly the answer of #Tom can give you a quite good and "a bit" more readable result :
%s/\(.*\)(\(.*\),\(.*\))\(.*\)/\1(\2\3)\4/
That way you will have : in \1 will store what is at the left outside of the parenthesis, \4 what is at the right outside of the parenthesis and \2 and \3 what is inside the parenthesis, respectively on the left (\2) and on the right (\3).
With that you can easily swap your elements if your file is organised as column.
You can also select the text you want to change (either with visual or visual-block modes) and enter the : to start the replace command. vi will automatically start the command with :'<,'> which applies the command to the selected area.
Replacing a , can be done with:
:'<,'>s/,/ /g
For your example, this is the same thing as suggested by #ubuntuguy
%s/\(.*\)(\(.*\),\(.*\)/\1(\2\3
This will do the exact replacement you want.
Yet another approach, based on the fact that actually you want to substitute only the first occurrence of , inside the parenthesis:
:%s#\((.\{-}\),#\1 #
Explanation:
:%s for substitution in the whole file (don't use % if you want to work only with the current line)
we can use # or : as a delimiter to make the command more readable
in (.\{-} we ask to find any symbol (dot) after the left parenthesis and the rest stands for 0 or more occurrence (as few as possible) of this symbol. This expression is put inside \(...\) to be able to refer to this group as \1 in the future.
I know for searching a whole word I should use /\<mypattern\>. But this is not true for dash (+U002d) character and /\<-\> always fails. I also try /\<\%d45\> and it fails too. anyone know the reason?
Edit2: As #bobbogo mentioned dash is not in 'iskeyword' so I add :set isk+=- and /\<-\> works!
Edit1: I think in Vim /\<word\> only is valid for alphanumeric characters and we shouldn't use it for punctuation characters (see Edit2). I should change my question and ask how we can search punctuation character as a whole world for example I want my search found the question mark in "a ? b" and patterns like "??" and "abc?" shouldn't be valid.
\< matches the zero-width boundary between a non-word character and a word character. What is a word character? It's specified by the isk option (:help isk).
Since - is not in your isk option, then - can never start a word, thus \<- will never match.
I don't know what you want, but /\>-\< will match the dash in hello-word.
Could always search for the regex \byourwordhere\b
As OP said. In order to include dash - into search just execute:
:set isk+=-
Thats all.
Example: When you press * over letter c of color-primary it will search for entire variable name not just for color.