The vim help example gives this line for a syntax statement:
:syntax region String start=+"+ skip=+\\"+ end=+"+
Is the + sign here an alternative to quoting? I couldn't find a reference to this in the help pages on patterns.
An introduction to patterns, see :h pattern.txt. As in the example of chapter 10 (:h :match), :match MyGroup /TODO/, instead of // any character can be used to mark the start and end of the pattern.
Like your question (:h syntax.txt), this is explained in Chapter 8 (:h :syn-pattern) :
In the syntax commands, a pattern must be surrounded by two identical characters. This is like it works for the ":s" command. The most common to use is the double quote. But if the pattern contains a double quote, you can use another character that is not used in the pattern. Examples: :syntax region String start=+"+ end=+"+ skip=+\\"+
Just like the :s command, see :h :s and :h pattern-delimiter, the benefit and convenience is the handling of escape characters. Using the :s command as an example is more convenient for you to verify quickly.
If you want to replace a with b, you can use :s/a/b/ or :s+a+b+. But if you want to replace / with //, using delimiter / require :s/\//\/\//, we can change the delimiter / to + (:s+\/+\/\/+). Actually, here we no longer need to escape /, so in the end there is only a concise :s+/+//+.
Back to your question, if the pattern contains a lot of double quotes, we can use another character that is not used in the pattern as delimiter (e.g. +), otherwise each double quote in the pattern needs to be escaped.
That sample command in your question isn't provided without context. It is provided as an illustration for the paragraph right above it:
In the syntax commands, a pattern must be surrounded by two identical
characters. This is like it works for the ":s" command. The most common to
use is the double quote. But if the pattern contains a double quote, you can
use another character that is not used in the pattern.
Related
/^[a-zA-Z0-9]{1,5}$
For example, I want to match letters or digits less than 5 in length, but the above doesn't work. In particular, it is the quantifier part is wrong.
How to correct it?
This very similar StackOverflow question has a good answer on the same topic.
Your regex isn't working because the default search pattern interpretation mode in Vim requires curly braces beginning a quantifier to be escaped. When the beginning brace isn't escaped using a slash \{, Vim searches for a literal curly brace { instead.
For example, assume we're searching the literal text: abc{1,5}
Searching with /[a-z]{1,5} matches c{1,5}
while searching with /[a-z]\{1,5} matches abc
That being said, your search pattern works when written as /^[a-zA-Z0-9]\{1,5}$ (notice the added leading slash before the quantifier opening brace). It seems like this works fine without escaping the closing brace, however I'm not sure if that will always be the case.
/\v^[a-zA-Z0-9]{1,5}$ will also work. Notice how the leading \v is used to enable "very magic" mode (see :help magic for a full explanation), and that escaping the curly brace is no longer required as a result.
Additional Info
Vim has a few different ways in which the search pattern can be interpreted. See the built-in :help magic page for a full explanation including a table of characters which need to be escaped at each "magic" level.
For instance, if I wanted to a find and replace with strings containing backward or forward slashes, how would this be accomplished in vim?
Examples
Find & Replace is: :%s/foo/bar/g
what if I wanted to find all occurrences of <dog/> and replace it with <cat\>
Same way you escape characters most anywhere else in linuxy programs, with a backslash:
:%s/<dog\/>/<cat\\>
But note that you can select a different delimiter instead:
:%s#<doc/>#<cat\\>#
This saves you all typing all those time-consuming, confusing backslashes in patterns with a ton of slashes.
From the documentation:
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.
%s:<dog/>:<cat>
You can replace the / delimiters if they become annoying for certain patterns.
Quote them with a backslash. Also, it often helps to use another delimiter besides slash.
:%s#<dog/>#<cat\\>#
or if you have to use slash as the substitute command delimiter
:%s/<dog\/>/<cat\\>/
I was looking for something similar, to search for register values containing the / character (to record a macro). The solution was to search using the ? token instead of the /.
The syntax is:
:%s/<dog\/>/<cat\\>/g
backslash slash backslash star
/(<- the prompt)\/\*
so after you type it looks like
/\/\*
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 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.