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.
Related
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 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.)
In PDP11/40 assembling language a number ends with dot is interpreted as a decimal number.
I use the following pattern but fail to match that notation, for example, 8.:
syn match asmpdp11DecNumber /\<[0-9]\+\.\>/
When I replace \. with D the pattern can match 8D without any problem. Could anyone tell me what is wrong with my "end-with-dot" pattern? Thanks.
Your regular expression syntax is fine (well, you can use \d instead of [0-9]), but your 'iskeyword' value does not include the period ., so you cannot match the end-of-word (\>) after it.
It looks like you're writing a syntax for a custom filetype. One option is to
:setlocal filetype+=.
in a corresponding ~/.vim/ftplugin/asmpdp11.vim filetype plugin. Do this when the period character is considered a keyword character in your syntax.
Otherwise, drop the \> to make the regular expression match. If you want to ensure that there's no non-whitespace character after the period, you can assert that condition after the match, e.g. like this:
:syn match asmpdp11DecNumber /\<\d\+\.\S\#!/
Note that a word is defined by vim as:
A word consists of a sequence of letters, digits and underscores, or a
sequence of other non-blank characters, separated with white space
(spaces, tabs, ). This can be changed with the 'iskeyword'
option. An empty line is also considered to be a word.
so your pattern works fine if whitespace follows the number. You may want to skip the \>.
I think the problem is your end-of-word boundary marker. Try this:
syn match asmpdp11DecNumber /\<[0-9]\+\./
Note that I have removed the \> end-of-word boundary. I'm not sure what that was in there for, but it appears to work if you remove it. A . is not considered part of a word, which is why your version fails.
I'm trying to search and replace $data['user'] for $data['sessionUser'].
However, no matter what search string I use, I always get a "pattern not found" as the result of it.
So, what would be the correct search string? Do I need to escape any of these characters?
:%s/$data['user']/$data['sessionUser']/g
:%s/\$data\[\'user\'\]/$data['sessionUser']/g
I did not test this, but I guess it should work.
Here's a list of all special search characters you need to escape in Vim: `^$.*[~)+/
There's nothing wrong with with the answers given, but you can do this:
:%s/$data\['\zsuser\ze']/sessionUser/g
\zs and \ze can be used to delimit the part of the match that is affected by the replacement.
You don't need to escape the $ since it's the at the start of the pattern and can't match an EOL here. And you don't need to escape the ] since it doesn't have a matching starting [. However there's certainly no harm in escaping these characters if you can't remember all the rules. See :help pattern.txt for the full details, but don't try to digest it all in one go!
If you want to get fancy, you can do:
:%s/$data\['\zsuser\ze']/session\u&/g
& refers to the entire matched text (delimited by \zs and \ze if present), so it becomes 'user' in this case. The \u when used in a replacement string makes the next character upper-case. I hope this helps.
Search and replace in vim is almost identical to sed, so use the same escapes as you would with that:
:%s/\$data\['user'\]/$data['session']/g
Note that you only really need to escape special characters in the search part (the part between the first set of //s). The only character you need to escape in the replace part is the escape character \ itself (which you're not using here).
The [ char has a meaning in regex. It stands for character ranges. The $ char has a meaning too. It stands for end-line anchor. So you have to escape a lot of things. I suggest you to try a little plugin like this or this one and use a visual search.
I have a file that was converted from EBCDIC to ASCII. Where there used to be new lines there are now characters that show up as <85> (a symbol representing a single character, not the four characters it appears to be) and the whole file is on one line. I want to search for them and replace them all with new lines again, but I don't know how.
I tried putting the cursor over one and using * to search for the next occurrence, hoping that it might show up in my / search history. That didn't work, it just searched for the word that followed the <85> character.
I searched Google, but didn't see anything obvious.
My goal is to build a search and replace string like:
:%s/<85>/\n/g
Which currently just gives me:
E486: Pattern not found: <85>
I found "Find & Replace non-printable characters in vim" searching Google. It seems like you should be able to do:
:%s/\%x85/\r/gc
Omit the c to do the replacement without prompting, try with c first to make sure it is doing what you want it to do.
In Vim, typing :h \%x gives more details. In addition to \%x, you can use \%d, \%o, \%u and \%U for decimal, octal, up to four and up to eight hexadecimal characters.
For special character searching, win1252 for example, for the case of <80>,<90>,<9d>...
type:
/\%u80, \/%u90, /\%u9d ...
from the editor.
Similarly for octal, decimal, hex, type: /\%oYourCode, /\%dYourCode, /\%xYourCode.
try this: :%s/<85>/^M/g
note: press Ctrl-V together then M
or if you don't mind using another tool,
awk '{gsub("<85>","\n")}1' file