I am trying to remove white-space from japanese word.
input "かいしゃ（会社）"
output "かいしゃ(会社)"
The space here is consumed by the parentheses. They are not your regular ASCII parentheses, they are of the "full width" flavor.
If you want to replace them with ASCII parentheses, you can do it like this:
compact_input = input.gsub("\uFF08", '(') # and a similar step for the closing parenthesis
Although this might make your string look weird in japanese (I don't know the language well enough, so can't say)
Related
I am new to ruby and I'm trying to work with regex.
I have a text which looks something like:
HEADING
Some text which is always non capitalized. Headings are always capitalized, followed by a space or nothing more.
YOU CAN HAVE MULTIPLE WORDS IN HEADING
I'm using this regular expression to choose all headings:
^[A-Z]{2,}\s?([A-Z]{2,}\s?)*$
However, it matches all headings which does not contain chars as Č, Š, Ž(slovenian characters).
So I'm guessing [A-Z] only matches ASCII characters? How could I get utf8?
You are right in that when you define the ASCII range A-Z, the match is made literally only for those characters. This is to do with the history of characters on computers, more and more characters have been added over time, and they are not always structured in an encoding in ways that are easy to use.
You could make a larger character class that matches the slovenian characters you need, by listing them.
But there is a shortcut. Someone else has already added necessary data to the Unicode data so that you can write shorter matches for "all uppercase characters": /[[:upper:]]/. See http://ruby-doc.org//core-2.1.4/Regexp.html for more.
Altering your regular expression with just this adjustment:
^[[:upper:]]{2,}\s?([[:upper:]]{2,}\s?)*$
You may need to adjust it further, for instance it would not match the heading "I AM A HEADING" due to the match insisting each word is at least two letters long.
Without seeing all your examples, I would probably simplify the group matching and just allow spaces anywhere:
^[[:upper:]\s]+$
You can use unicode upper case letter:
\p{Lu}
Your regex:
\b\p{Lu}{2,}(?:\s*\p{Lu}{2,})\b
RegEx Demo
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).
Hallo I would like to sanitize input from Ruby, but simultaneously not have it mess up my strings containing foreign characters.
string1=string.downcase.gsub( /<(.|\n)*?>/, '' ).gsub(" ", "").gsub(",","").gsub("'","").gsub("_", "").gsub(";", "").gsub("-", "").gsub(":","").gsub(".", "").gsub("?", "").gsub("!", "").gsub("^","").gsub("%", "").gsub("$","")
The string needs to be stripped of spaces, apostrophes, and everything but letters (not sure about numbers), in addition to be sanitized. I am not sure if I forgot something, and probably something is redundant.
My code works OK as long as string does not contain harmless non-english characters such as accented letters, which I'd like it to deal with, but they break my code. My guess is that they are converted to %25 and all that stuff and afterwards they break. In fact it breaks even if I don't sanitize at all. How can I tell Ruby to handle non-english characters correctly? Thank you a lot.
like this;
" ' ; te st".gsub(/\W+/, "") # "test"
I am using Ruby on Rails 3.0.9 and I would like to validate a string that can contain only characters (case insensitive characters), blank spaces and numbers.
More:
special characters are not allowed (eg: !"£$%&/()=?^) except - and _;
accented characters are allowed (eg: à, è, é, ò, ...);
The regex that I know from this question is ^[a-zA-Z\d\s]*$ but this do not validate special characters and accented characters.
So, how I should improve the regex?
I wrote the ^(?:[^\W_]|\s)*$ answer in the question you referred to (which actually would have been different if I'd known you wanted to allow _ and -). Not being a Ruby guy myself, I didn't realize that Ruby defaults to not using Unicode for regex matching.
Sorry for my lack of Ruby experience. What you want to do is use the u flag. That switches to Unicode (UTF-8), so accented characters are caught. Here's the pattern you want:
^[\w\s-]*$
And here it is in action at Rubular. This should do the trick, I think.
The u flag works on my original answer as well, though that one isn't meant to allow _ or - characters.
Something like ^[\w\s\-]*$ should validate characters, blank spaces, minus, and underscore.
Validation string only for not allowed characters. In this case |,<,>," and &.
^[^|<>\"&]*$
I am trying to come up with a regex to remove all special characters except some. For example, I have a string:
str = "subscripción gustaría♥"
I want the output to be "subscripción gustaría".
The way I tried to do is, match anything which is not an ascii character (00 - 7F) and not special character I want and replace it with blank.
str.gsub(/(=?[^\x00-\x7F])(=?^\xC3\xB3)(=?^\xC3\xA1)/,'')
This doesn't work. The last special character is not removed.
Can someone help? (This is ruby 1.8)
Update: I am trying to make the question a little more clear. The string is utf-8 encoded. And I am trying to whitelist the ascii characters plus ó and í and blacklist everything else.
Oniguruma has support for all the characters you care about without having to deal with codepoints. You can just add the unicode characters inside the character class you're whitelisting, followed by the 'u' option.
ruby-1.8.7-p248 > str = "subscripción gustaría♥"
=> "subscripci\303\263n gustar\303\255a\342\231\245"
ruby-1.8.7-p248 > puts str.gsub(/[^a-zA-Z\sáéíóúÁÉÍÓÚ]/u,'')
subscripción gustaría
=> nil
str.split('').find_all {|c| (0x00..0x7f).include? c.ord }.join('')
The question is a bit vague. There is not a word about encoding of the string. Also, you want to white-list characters or black list? Which ones?
But you get the idea, decide what you want, and then use proper ranges as colleagues here already proposed. Some examples:
if str = "subscripción gustaría♥" is utf-8
then you can blacklist all char above the range (excl. whitespaces):
str.gsub(/[^\x{0021}-\x{017E}\s]/,'')
if string is in ISO-8859-1 codepage you can try to match all quirky characters like the "heart" from the beginning of ASCII range:
str.gsub(/[\x01-\x1F]/,'')
The problem is here with regex, has nothing to do with Ruby. You probably will need to experiment more.
It is not completely clear which characters you want to keep and which you want to delete. The example string's character is some Unicode character that, in my browser, displays as a heart symbol. But it seems you are dealing with 8-bit ASCII characters (since you are using ruby 1.8 and your regular expressions point that way).
Nonetheless, you should be able to do it in one of two ways; either specify the characters you want to keep or, alternatively, specify the characters you want to delete. For example, the following specifies that all characters 0x00-0x7F and 0xC0-0xF6 should be kept (remove everything that is not in that group):
puts str.gsub(/[^\x00-\x7F\xC0-\xF6]/,'')
This next example specifies that characters 0xA1 and 0xC3 should be deleted.
puts str.gsub(/[\xA1\xC3]/,'')
I ended up doing this: str.gsub(/[^\x00-\x7FÁáÉéÍíÑñÓóÚúÜü]/,''). It doesn't work on my mac but works on linux.