I am trying to replace all occurrences of a whole word on emacs (say foo) using M-x replace-regexp.
The problem is that I don't want to replace occurrences of foo in underscored words such as word_foo_word
If I use \bfoo\b to match foo then it will match the underscored strings; because as I understand emacs considers underscores to be part of word boundaries, which is different to other RegEx systems such as Perl.
What would be the correct way to proceed?
The regexp \<foo\> or \bfoo\b matches foo only when it's not preceded or followed by a word constituent character (syntax code w, usually alphanumerics, so it matches in foo_bar but not in foo1).
Since Emacs 22, the regexp \_<foo_bar\_> matches foo_bar only when it's not preceded or followed by a symbol-constituent character. Symbol constituents include not only word constituents (alphanumerics) but also punctuation characters that are allowed in identifiers, meaning underscores in most programming languages.
You wrote:
as I understand emacs considers underscores to be part of word boundaries, which is different to other regex systems
The treatment of underscores, like everything else in emacs, is configurable. This question:
How to make forward-word, backward-word, treat underscore as part of a word?
...asks the converse.
I think you could solve your problem by changing the syntax of underscores in the syntax table so that they are not part of words, and then doing the search/replace.
To do that, you need to know the mode you are using, and the name of the syntax table for that mode. In C++, it would be like this:
(modify-syntax-entry ?_ "." c++-mode-syntax-table)
The dot signifies "punctuation", which implies not part of a word. For more on that, try M-x describe-function on modify-syntax-entry.
Related
I have this regular expression: ^[a-zA-Z]\s{3,16}$
What I want is to match any name with any spaces, for example, John Smith and that contains 3 to 16 characters long..
What am I doing wrong?
Background
There are a couple of things to note here. First, a quantifier (in this case, {3,16}) only applies to the last regex token. So what your current regex really is saying is to "Match any string that has a single alphabetical character (case-insensitive) followed by 3 to 16 whitespace characters (e.g. spaces, tabs, etc.)."
Second, a name can have more than 2 parts (a middle name, certain ethnic names like "De La Cruz") or include special characters such as accented vowels. You should consider if this is something you need to account for in your program. These things are important and should be considered for any real application.
Assumptions and Answer
Now, let's just assume you only want a certain format for names that consists of a first name, a last name, and a space. Let's also assume you only want simple ASCII characters (i.e. no special characters or accented characters). Furthermore, both the first and last names should start with a capital character followed by only lower-case characters. Other than that, there are no restrictions on the length of the individual parts of the name. In this case, the following regex would do the trick:
^(?=.{3,16}$)[A-Z][a-z]+ [A-Z][a-z]+$
Notes
The first token after the ^ character is what is called a positive lookahead. Basically a positive look ahead will match the regex between the opening (?= and closing ) without actually moving the position of the cursor that is matching the string.
Notice I removed the \s token, since you usually want only a (space). The space can be replaced with the \s token, if tabs and other whitespace is desired there.
I also added a restriction that a name must start with a capital letter followed by only lower-case letters.
Crude English Translation
To help your understanding, here is a simple English translation of what the regex is really doing. The part in italics is just copied from the first part of the English translation of the regex.
"Match any string that has 3-16 characters and starts with a capital alphabetical character followed by one or more (+) alphabetical characters followed by a single space followed by a capital alphabetical character followed by one or more (+) alphabetical characters and ends with any lowercase letter."
Tools
There are a couple of tools I like to use when I am trying to tackle a challenging regex. They are listed below in no particular order:
https://regex101.com/ - Allows you to test regex expressions in real time. It also has a nifty little library to help you along.
http://www.regular-expressions.info/ - Basically a repository of knowledge on regex.
Edit/Update
You mentioned in your comments that you are using your regex in JavaScript. JavaScript uses a forward slash surrounding the regex to determine what is a regex. For this simple case, there are 2 options for using a regex to match a string.
First, use String's match method as follows
"John Smith".match(/^(?=.{3,16}$)[A-Z][a-z]+ [A-Z][a-z]+$/);
Second, create a regex and use its test() method. For example,
/^(?=.{3,16}$)[A-Z][a-z]+ [A-Z][a-z]+$/.test("John Smith");
The latter is probably what you want as it simply returns true or false depending on whether the regex actually matches the string or not.
I know this is a elementary RegEx possibility, but I can't seem to determine the right expression to use.
What I am looking to do is find & replace "foo" and only "foo" within a set of different situations like; abc_foo, abc_foo[something], abc-foo-something, and all different combinations except when it becomes another word like "foobar". The basic 'whole word' search function was close but doesn't help when variables and underscores are factored in.
It's actually not that elementary to match a string which does not contain word characters around itself:
If your language supports negative lookbehind, which is quite rare occasion, it would be simple:
(?<!\w)foo(?!\w)
However, there is a workaround to match the string with surrounding non-word characters (including _ which is a word character but you want to treat is as non-word) and use capturing groups to sort it all out:
(^|[\W_])foo([\W_]|$)
Debuggex Demo
e.g. in javascript syntax:
str.replace(/(^|[\W_])foo([\W_]|$)/g, "$1replacement$2");
You can use a negative lookahead assertion to do this. Using regex search, foo(?!bar) will match any instance of foo not followed by bar, and the following text is not part of the match, only foo is.
I have a very simple Vim syntax file for personal notes. I would like to highlight people's name and I chose a Twitter-like syntax #jonathan.
I tried:
syntax match notesPerson "\<#\S\+"
To mean: words beginning with # and having at least one non-whitespace character. The problem is that # seems to be a special character in Vim regular expressions.
I tried to escape \# and enclose in brackets [#], the usual tricks, but that didn't work. I could try something like (^|\s) (beginning of line or whitespace) but that's exactly the problem that word-boundary tries to solve.
Highlighting works on simplified regular expressions, so this is more a question of finding the right regex than anything else. What am I missing?
# is a special character only if you have enabled the “very magic”
mode by having \v somewhere in the pattern prior to that #.
You have another problem here: # does not start a new word. \< is
not just “word boundary” like perl/PCRE’s \b, but “left word
boundary” (in help: “beginning of the word”) meaning that \< must be
followed by some keyword character. As # is not normally a keyword
character, pattern \<# will never match. (And even if it was like
\b, it would match constructs like abc#def which is definitely not
what you want for the aforementioned reasons.)
You should use \k\#<!#\k\S* instead: \k\#<! ensures that # is not preceded by any keyword character, \k\S* makes sure that first character of the name is a keyword one (you could probably also use #\<\S\+).
There is another solution: include # into 'iskeyword' option and leave the regex as is:
:setlocal iskeyword+=#-#
See :help 'isfname' for the explanation why #-# is used here.
(The 'iskeyword' option has exactly the same syntax and will,
in fact, redirect you there for the explanation.)
how do you read this regex?
#(http|https|ftp)://([A-Z0-9][A-Z0-9_-]*(?:.[A-Z0-9][A-Z0-9_-]*)+):?(d+)?/?#i
this is a regex for links, but i'm having trouble to understand it
Thanks
Depending on what language you're in, regexes need a delimiter. Seems the # (pound sign or hash) is used here. So,
#...actual regex goes here...#
In javascript you need forward slashes (/..../).
Some regex engines allow you to pass flags that influence matching process. These appear after the closing delimiter:
#...actual regex goes here...#..flags go here..
In your example, there is one flag, the i and I am guessing that means: "case insensitive" (i for insensitive). Depending on the regex engine you can have flags that influence the syntax you can use for the actual regex (for example, the dot can match either any character or any character except newlines depending upon wheter a flag was passed), flags that influence how the matching is done (for example, in javascript a g indicates the global flag, and that means matching anywhere inside the string is done, and state is preserved), flags that determine whether whitespace is allowed as indentation inside the regex. And some have a m flag indicating whether the regex will be applied on a line by line basis, or on the entire text. There is AFAIK no standard set of flags, check your regex engine documentation.
If you have multiple flags, you just concatenate them together to a string of flags and put them after the closing delimiter.
Now for the actual regex. First, you start with a parenthesized expression:
(...group...)
This is also called a group. In many regex engines, these groups have special meaning, because when a match is found you can access the bits of text that matched the expression inside the group using a special variable (or sometimes, the match is returned as an array, where each element represents a group). If you can access the bits inside groups, it is called a "capturing group".
In this particular case the group uses "alternation" or "choice" and this is indicated by the | (pipe). The pipe is part of the regex syntax and means "or". So,
(http|https|ftp)
means: match "http", or if that doesn't match, "https", of if that doesn't match, "ftp". This also brings up another reason for using parenthesis: of all special regex syntax operators, the pipe has the lowest precedence, so the parenthesis would not have been there, it would have meant: match "http" or "https" or "ftp://...etc"
So far, we've seen these "special characters": | (pipe) and ( and ). After that we get
://
These are not special characters, and any non-special characters simply match themselves.
We then get another group, which makes up almost the rest of the regex:
([A-Z0-9][A-Z0-9_-]*(?:.[A-Z0-9][A-Z0-9_-]*)+)
Inside it, we see a bracketed expression:
[A-Z0-9]
The brackets [ and ] are special, and indicate a "character class". There are other ways to denote character classes, but in all cases a character class matches a single character. Which character depends on the nature of the class. In this case, the class is defined using two ranges:
A-Z
means characters A thru Z (and anything in between) and
0-9
means characters 0 thru 9 (and anything in between).
Basically, [A-Z0-9] matches any alpha-numeric character.
Note that the dash between the boundaries of the range is only a special character inside these bracketed expressions. Paradoxically, a dash inside the brackets can also simply mean a dash if it cannot be interpreted as a range.
This is folllowed by yet another character class:
[A-Z0-9_-]
Almost the same as the previous on, it just adds the underscore and the dash. This last dash cannot be interpreted as a range separator, so it simply means a dash. This character class will match any alpha-numeric character as well as underscore and dash.
This class is followed by a * (asterisk) and this is a special character indicating a cardinality. Cardinalities specify how often the immediately preceding element may occur. These are the common cardinalities:
* (asterisk) means zero or more times.
? (question mask) means zero or once.
+ (plus) means one or more times.
Now the entire bit starts to make sense:
[A-Z0-9][A-Z0-9_-]*
means: a sequence starting with one alphanumeric caracter, optionally followed by a string of "word" characters (that is, alphanumeric, dash and underscore).
The following bit of the regex is this:
(?:.[A-Z0-9][A-Z0-9_-]*)+
I think this is trying to match the domain parts. So that if you have say:
https://mail.google.com
The .google and .com bits would be matched by this part. The initial (?: bit is meant to tell the regex engine to not create a "backreference". This is not really my stronghold, maybe someone else can explain. But the rest of that group is quite clear and resembles what we saw before. I think there is a mistake though: the dot (.) that appears immediately before the bracketed character class usually means "match any character" or "match any non-newline character", not "match a literal dot". Typically if you want a literal dot, you need to escape it. This would be the syntax in javascript and I think perl:
(\.[A-Z0-9][A-Z0-9_-]*)+
(note the backslash immediately before the dot to indicate a literal dot)
The final bits of the regex seem an attempt to match a port number:
:?(d+)?
However, the d+ bit is probably wrong: right now it matches "one or more d's". It should probably be:
:?(\d+)?
meaning: optionally match a colon (:), optionally followed by a bunch of digits. The \d is also a character class, but a predefined one. I think most regex engines use \d to denote a digit, but you should check the documentation of your engine to see the exact convention. So in say:
http://domain.server.extension:8080/
this part of the regex would match :8080 (provided you fix the d+ thing).
Finally, we see
/?
Meaning the entire thing can be followed optionally by a forward slash.
So, all in all, I don;'t think this matches a "link", rather it matches the inital part of a URL. To match an entire url, you would need a bit more, at least I don't see any expression that could match the path, resource, hash and query bits that may occur in a proper URL.
When you say you have trouble understanding it, it means you tried something and are stuck somewhere?
Please ask more specific questions.
I can give you some keywords that you can lookup them more easy, a good place for that is regular-expressions.info
(http|https|ftp) is an alternation
[A-Z0-9] is a character class
*, + and ? are quantifiers
(...) is a (capturing) group, (?:...) is a non capturing group
The # at the start and end are regex delimiters, the i at the very end is a modifier/option (match case independent).
The (d+)? at the end would match one or more (optional) letters "d". This is quite strange. I assume it should be (\d+)? that would be one or more (optional) digits.
I'm new to this site and don't know if this is the place to ask this question here?
I was wondering if someone can explain the 3 regex code examples below in detail?
Thanks.
Example 1
`&([a-z]{1,2})(acute|uml|circ|grave|ring|cedil|slash|tilde|caron|lig);`i
Example 2
\\1
Example 3
`[^a-z0-9]`i','`[-]+`
The first regex looks like it'll match the HTML entities for accented characters (e.g., é is é; ø is ø; æ is æ; and  is Â).
To break it down, & will match an ampersand (the start of the entity), ([a-z]{1,2}) will match any lowercase letter one or two times, (acute|uml|circ|grave|ring|cedil|slash|tilde|caron|lig) will match one of the terms in the pipe-delimited list (e.g., circ, grave, cedil, etc.), and ; will match a semicolon (the end of the entity). I'm not sure what the i character means at the end of the line; it's not part of the regex.
All told, it will match the HTML entities for accented/diacritic/ligatures. Compared, though, to this page, it doesn't seem that it matches all of the valid entities (although ti does catch many of them). Unless you run in case-insensitive mode, the [a-z] will only match lowercase letters. It will also never match the entities ð or þ (ð, þ, respectively) or their capital versions (Ð, Þ, also respectively).
The second regex is simpler. \1 in a regex (or in regex find-replace) simply looks for the contents of the first capturing group (denoted by parentheses ()) and (in a regex) matches them or (in the replace of a find) inserts them. What you have there \\1 is the \1, but it's probably written in a string in some other programming language, so the coder had to escape the backslash with another backslash.
For your third example, I'm less certain what it does, but I can explain the regexes. [^a-z0-9] will match any character that's not a lowercase letter or number (or, if running in case-insensitive mode, anything that's not a letter or a number). The caret (^) at the beginning of the character class (that's anything inside square brackets []) means to negate the class (i.e., find anything that is not specified, instead of the usual find anything that is specified). [-]+ will match one or more hyphens (-). I don't know what the i',' between the regexes means, but, then, you didn't say what language this is written in, and I'm not familiar with it.