How can I match a balanced pair of delimiters not escaped by backslash (that is itself not escaped by a backslash) (without the need to consider nesting)? For example with backticks, I tried this, but the escaped backtick is not working as escaped.
regex = /(?!<\\)`(.*?)(?!<\\)`/
"hello `how\` are` you"
# => $1: "how\\"
# expected "how\\` are"
And the regex above does not consider a backslash that is escaped by a backslash and is in front of a backtick, but I would like to.
How does StackOverflow do this?
The purpose of this is not much complicated. I have documentation texts, which include the backtick notation for inline code just like StackOverflow, and I want to display that in an HTML file with the inline code decorated with some span material. There would be no nesting, but escaped backticks or escaped backslashes may appear anywhere.
Lookbehind is the first thing everyone thinks of for this kind of problem, but it's the wrong tool, even in flavors like .NET that support unrestricted lookbehinds. You can hack something up, but it's going to be ugly, even in .NET. Here's a better way:
`[^`\\]*(\\.[^`\\]*)*`
The first part starts from the opening delimiter and gobbles up anything that's not the delimiter or a backslash. If the next character is a backslash, it consumes that and the character following it, whatever it may be. It could be the delimiter character, another backslash, or anything else, it doesn't matter.
It repeats those steps as many times as necessary, and when neither [^`\\] nor \\. can match, the next character must be the closing delimiter. Or the end of the string, but I'm assuming the input is well formed. But if it's not well formed, this regex will fail very quickly. I mention that because of this other approach I see a lot:
`(?:[^`\\]+|\\.)*`
This works fine on well-formed input, but what happens if you remove the last backtick from your sample input?
"hello `how\` are you"
According to RegexBuddy, after encountering the first backtick, this regex performed 9,252 distinct operations (or steps) before it could give up and report failure; mine failed in ten steps.
EDIT To extract just the par inside the delimiters, wrap that part in a capturing group. You'll still have to remove the backslashes manually.
`([^`\\]*(?:\\.[^`\\]*)*)`
I also changed the other group to non-capturing, which I should have done from the start. I don't avoid capturing religiously, but if you are using them to capture stuff, any other groups you use should be non-capturing.
EDIT I think I've been reading too much into the question. On StackOverflow, if you want to include literal backticks in an inline-code segment or a comment, you use three backticks as the the delimiter, not just one. Since there's no need to escape backticks, you can ignore backslashes as well. Your regex could turn out to be as simple as this:
```(.*?)```
Dealing with the possibility of false delimiters, you use the same basic technique:
```([^`]*(?:`(?!``)[^`]*)*)```
Is this what you're after?
By the way, this answer doesn't contradict #nneonneo's comment above. This answer doesn't consider the context in which the match is taking place. Is it in the source code of a program or web page? If it is, did the match occur inside a comment or a string literal? How do I even know the first backtick I found wasn't escaped? Regexes don't know anything about the context in which they operate; that's what parsers are for.
If you don't need nesting, regexes can indeed be a proper tool. Lexers of programming languages, for instance, use regexes to tokenize strings, and strings usually allow their own delimiters as an escaped content. Anything more complicated than that will probably need a full-blown parser though.
The "general formula" is to match an escaped character (\\.) or any character that's valid as content but don't need to be escaped ([^{list of invalid chars}]). A "naïve" solution would be joining them with or (|), but for a more efficient variant see #AlanMoore's answer.
The complete example is shown below, in two variants: the first assumes than backslashes should only be used for escaping inside the string, the second assumes that a backslash anywhere in the text escapes the next character.
`((?:\\.|[^`\\])*)`
(?:\\.|[^`\\])*`((?:\\.|[^`\\])*)`
Working examples here and here. However, as #nneonneo commented (and I endorsed), regexes are not meant to do a complete parse, so you'd better keep things simple if you want them to work out right (do you want to find a token in the text, or do you want to delimit it already knowing where it starts? The answer to that question is important to decide which strategy works best for your case).
Related
For a gradle script, I am composing strings that will be used as command line for a subsequent gradle Test-task. One of the strings is the user's password, which eventually will be passed to the called (exec'ed) "java ..." call using the JVM's -D option, e.g. -Dpassword=foobar.
What complicates things here is, that this password can/should of course contain special characters, that may interfere with the use of the string as command line. In other words: I need to escape special characters (which is OS-specific). :-(
Now to my actual question:
I want to use the String.replaceAll method, i.e. replaceAll(list_of_special characters, EscapeCharacter + Ref_to_matched_character),
e.g. simplified something like replaceAll("[#$%^&]", "^$1")
'^' meaning the escape character and '$1' meaning the matched character here.
Is that possible, i.e. can one refer to the matched pattern in the second argument of replaceAll?
Is that possible, i.e. can one refer to the matched pattern in the second argument of replaceAll?
yes, it's possible
'a#b$c'.replaceAll('([#$%^&])', '^$1')
returns
a^#b^$c
Thanks for the responses and the reviews improving readability. Meanwhile I got my expression working. For those interested:
// handles gthe following: `~!##$%^&*()_+-={}|[]\:;"'<>?,./
escaped = original.replaceAll('[~!##\\$\\%\\^\\&\\*\\(\\)_\\+-={}\\|\\[\\]\\\\:;\"\\\'<>\\?,\\./]', '^$0') // for Windows - cmd.exe
I am trying to grasp the concept of Regular Expressions but seem to be missing something.
I want to ensure that someone enters a string that ends with .wav in a field. Should be a pretty simple Regular Expression.
I've tried this...
[RegularExpression(#"$.wav")]
but seem to be incorrect. Any help is appreciated. Thanks!
$ is the anchor for the end of the string, so $.wav doesn't make any sense. You can't have any characters after the end of the string. Also, . has a special meaning for regex (it just means 'any character') so you need to escape it.
Try writing
\.wav$
If that doesn't work, try
.*\.wav$
(It depends on if the RegularExpression attribute wants to match the whole string, or just a part of it. .* means 'any character, 0 or more times')
Another thing you should consider is what to do with extra whitespace in the field. Users have a terrible habit of adding extra white space in inputs - its why various .Trim() functions are so important. Here, RegularExpressionAttribute might be evaluated before you can trim the input, so you might want to write this:
.*\.wav[\s]*$
The [\s]* section means 'any whitespace character (tabs, space, linebreak, etc) 0 or more times'.
You should read a tutorial on regex. It's not so hard to understand for simple problems like this. When I was learning I found this site pretty handy: http://www.regular-expressions.info/
I need to find strings with * and / using reg-exes, I am writing in Ruby.The reason for this need to find lots of * and / is that I am building a tokenizer for an language and there are multi-line comments that use the C style of multi-line comments (/* */). I have the single line comments handled already.
Is there a way to use reg-ex without having to use the two foreword slashes to indicate some regular expression because I am finding it impossible to find my mistakes due to the insane amount of escaping. Or can someone give me advise on how to handle the escaping in a sane matter? I already tried writing the sequence first then escaping it.
Thank you for your time and advise.
One trick that might help is the %r literal:
%r{http://www\.google\.com}
I like to use pipes myself, when they're not in the regex.
%r|http://www\.google\.com|
You can also create new instances of Regexp via Regexp.new and pass a string.
Finally, you might also look at Regexp.quote:
Escapes any characters that would have special meaning in a regular expression. Returns a new escaped string, or self if no characters are escaped. For any string, Regexp.new(Regexp.escape(str))=~str will be true.
I know that you can use this to remove blank lines
sed /^$/d
and this to remove comments starting with #
sed /^#/d
but how to you do delete all the comments starting with // ?
You just need to "escape" the slashes with the backslash.
/\/\//
the ^ operator binds it to the front of the line, so your example will only affect comments starting in the first column. You could try adding spaces and tabs in there, too, and then use the alternation operator | to choose between two comment identifiers.
/^[ \t]*(\/\/|$)/
Edit:
If you simply want to remove comments from the file, then you can do something like:
/(\/\/|$).*/
I don't know what the 'd' operator at the end does, but the above expression should match for you modulo having to escape the parentheses or the alternation operator (the '|' character)
Edit 2:
I just realized that using a Mac you may be "shelling" that command and using the system sed. In that case, you could try putting quotation marks around the search pattern so that the shell doesn't do anything crazy to all of your magic characters. :) In this case, 'd' means "delete the pattern space," so just stick a 'd' after the last example I gave and you should be set.
Edit 3:
Oh I just realized, you'll want to beware that if you don't catch things inside of quotes (i.e. you don't want to delete from # to end of line if it's in a string!). The regexp becomes quite a bit more complicated in that case, unfortunately, unless you just forgo checking lines with strings for comments. ...but then you'd need to use the substitution operation to sed rather than search-and-delete-match. ...and you'd need to put in more escapes, and it becomes madness. I suggest searching for an online sed helper (there are good regex testers out there, maybe there's one for sed?).
Sorry to sort of abandon the project at this point. This "problem" is one that sed can do but it becomes substantially more complex at every stage, as opposed to just whipping up a bit of Python to do it.
Which style of Ruby string quoting do you favour? Up until now I've always used 'single quotes' unless the string contains certain escape sequences or interpolation, in which case I obviously have to use "double quotes".
However, is there really any reason not to just use double quoted strings everywhere?
Don't use double quotes if you have to escape them. And don't fall in "single vs double quotes" trap. Ruby has excellent support for arbitrary delimiters for string literals:
Mirror of Site - https://web.archive.org/web/20160310224440/http://rors.org/2008/10/26/dont-escape-in-strings
Original Site -
http://rors.org/2008/10/26/dont-escape-in-strings
I always use single quotes unless I need interpolation.
Why? It looks nicer. When you have a ton of stuff on the screen, lots of single quotes give you less "visual clutter" than lots of double quotes.
I'd like to note that this isn't something I deliberately decided to do, just something that I've 'evolved' over time in trying to achieve nicer looking code.
Occasionally I'll use %q or %Q if I need in-line quotes. I've only ever used heredocs maybe once or twice.
Like many programmers, I try to be as specific as is practical. This means that I try to make the compiler do as little work as possible by having my code as simple as possible. So for strings, I use the simplest method that suffices for my needs for that string.
<<END
For strings containing multiple newlines,
particularly when the string is going to
be output to the screen (and thus formatting
matters), I use heredocs.
END
%q[Because I strongly dislike backslash quoting when unnecessary, I use %Q or %q
for strings containing ' or " characters (usually with square braces, because they
happen to be the easiest to type and least likely to appear in the text inside).]
"For strings needing interpretation, I use %s."%['double quotes']
'For the most common case, needing none of the above, I use single quotes.'
My first simple test of the quality of syntax highlighting provided by a program is to see how well it handles all methods of quoting.
I use single quotes unless I need interpolation. The argument about it being troublesome to change later when you need interpolation swings in the other direction, too: You have to change from double to single when you found that there was a # or a \ in your string that caused an escape you didn't intend.
The advantage of defaulting to single quotes is that, in a codebase which adopts this convention, the quote type acts as a visual cue as to whether to expect interpolated expressions or not. This is even more pronounced when your editor or IDE highlights the two string types differently.
I use %{.....} syntax for multi-line strings.
I usually use double quotes unless I specifically need to disable escaping/interpolation.
I see arguments for both:
For using mostly double quotes:
The github ruby style guideline advocates always using double quotes:
It's easier to search for a string foobar by searching for "foobar" if you were consistent with quoting. However, I'm not. So I search for ['"]foobar['"] turning on regexps.
For using some combination of single double quotes:
Know if you need to look for string interpolation.
Might be slightly faster (although so slight it wasn't enough to affect the github style guide).
I used to use single quotes until I knew I needed interpolation. Then I found that I was wasting a lot of time when I'd go back and have to change some single-quotes to double-quotes. Performance testing showed no measurable speed impact of using double-quotes, so I advocate always using double-quotes.
The only exception is when using sub/gsub with back-references in the replacement string. Then you should use single quotes, since it's simpler.
mystring.gsub( /(fo+)bar/, '\1baz' )
mystring.gsub( /(fo+)bar/, "\\1baz" )
I use single quotes unless I need interpolation, or the string contains single quotes.
However, I just learned the arbitrary delimiter trick from Dejan's answer, and I think it's great. =)
Single quote preserve the characters inside them. But double quotes evaluate and parse them. See the following example:
"Welcome #{#user.name} to App!"
Results:
Welcome Bhojendra to App!
But,
'Welcome #{#user.name} to App!'
Results:
Welcome #{#user.name} to App!