So I want to split a string in java on any non-alphanumeric characters.
Currently I have been doing it like this
words= Str.split("\\W+");
However I want to keep apostrophes("'") in there. Is there any regular expression to preserve apostrophes but kick the rest of the junk? Thanks.
words = Str.split("[^\\w']+");
Just add it to the character class. \W is equivalent to [^\w], which you can then add ' to.
Do note, however, that \w also actually includes underscores. If you want to split on underscores as well, you should be using [^a-zA-Z0-9'] instead.
For basic English characters, use
words = Str.split("[^a-zA-Z0-9']+");
If you want to include English words with special characters (such as fiancé) or for languages that use non-English characters, go with
words = Str.split("[^\\p{L}0-9']+");
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
I have regular expression like this:
regularExp = "^[-]{0,1}([0-9]|[a-z]|[A-Z]|[\s]){0," & decNum & "}\.$"
Here I need to add all Special Character's, like ~!##$%^&*()_+{}|:"<>?[]\;',./ in VB6.0
I guess you are looking for something like POSIX bracket extensions and a special character class which matches all punctuation characters without listing them explicitly.
Unfortunately you are out of luck, since the Regular Expressions available in Visual Basic 6 are provided by the same VBScript RegExp engine which was available in IE 5.5. That engine was not updated in 15 years, so many features are missing.
Having said that, your only option is to "handpick" each and every character you want to match and put them in a character class, like this
[~!##$%^&*()_+{}|:"<>?[\]\\;',./]
Fortunately you don't have to escape all special characters within character classes, only the ones which confuse the parser. (Namely \, ^, - and ])
you can use
^[a-zA-Z._^%$#!~#,-] as referance and add more special characters which you want to allow.
You can use add special characters as below
[^%$#!~#()*\s]
I'm looking for words starting with a hashtag: "#yolo"
My regex for this was very simple: /#\w+/
This worked fine until I hit words that ended with a question mark: "#yolo?".
I updated my regex to allow for words and any non whitespace character as well: /#[\w\S]*/.
The problem is I sometimes need to pull a match from a word starting with two '#' characters, up until whitespace, that may contain a special character in it or at the end of the word (which I need to capture).
Example:
"##yolo?"
And I would like to end up with:
"#yolo?"
Note: the regular expressions are for Ruby.
P.S. I'm testing these out here: http://rubular.com/
Maybe this would work
#(#?[\S]+)
What about
#[^#\s]+
\w is a subset of ^\s (i.e. \S) so you don't need both. Also, I assume you don't want any more #s in the match, so we use [^#\s] which negates both whitespace and # characters.
I want to extract #hashtags from a string, also those that have special characters such as #1+1.
Currently I'm using:
#hashtags ||= string.scan(/#\w+/)
But it doesn't work with those special characters. Also, I want it to be UTF-8 compatible.
How do I do this?
EDIT:
If the last character is a special character it should be removed, such as #hashtag, #hashtag. #hashtag! #hashtag? etc...
Also, the hash sign at the beginning should be removed.
The Solution
You probably want something like:
'#hash+tag'.encode('UTF-8').scan /\b(?<=#)[^#[:punct:]]+\b/
=> ["hash+tag"]
Note that the zero-width assertion at the beginning is required to avoid capturing the pound sign as part of the match.
References
String#encode
Ruby's POSIX Character Classes
This should work:
#hashtags = str.scan(/#([[:graph:]]*[[:alnum:]])/).flatten
Or if you don't want your hashtag to start with a special character:
#hashtags = str.scan(/#((?:[[:alnum:]][[:graph:]]*)?[[:alnum:]])/).flatten
How about this:
#hashtags ||=string.match(/(#[[:alpha:]]+)|#[\d\+-]+\d+/).to_s[1..-1]
Takes cares of #alphabets or #2323+2323 #2323-2323 #2323+65656-67676
Also removes # at beginning
Or if you want it in array form:
#hashtags ||=string.scan(/#[[:alpha:]]+|#[\d\+-]+\d+/).collect{|x| x[1..-1]}
Wow, this took so long but I still don't understand why scan(/#[[:alpha:]]+|#[\d\+-]+\d+/) works but not scan(/(#[[:alpha:]]+)|#[\d\+-]+\d+/) in my computer. The difference being the () on the 2nd scan statement. This has no effect as it should be when I use with match method.
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 &.
^[^|<>\"&]*$