How to rewrite the [a-zA-Z0-9!$* \t\r\n] pattern to match hyphen along with the existing characters ?
The hyphen is usually a normal character in regular expressions. Only if it’s in a character class and between two other characters does it take a special meaning.
Thus:
[-] matches a hyphen.
[abc-] matches a, b, c or a hyphen.
[-abc] matches a, b, c or a hyphen.
[ab-d] matches a, b, c or d (only here the hyphen denotes a character range).
Escape the hyphen.
[a-zA-Z0-9!$* \t\r\n\-]
UPDATE:
Never mind this answer - you can add the hyphen to the group but you don't have to escape it. See Konrad Rudolph's answer instead which does a much better job of answering and explains why.
It’s less confusing to always use an escaped hyphen, so that it doesn't have to be positionally dependent. That’s a \- inside the bracketed character class.
But there’s something else to consider. Some of those enumerated characters should possibly be written differently. In some circumstances, they definitely should.
This comparison of regex flavors says that C♯ can use some of the simpler Unicode properties. If you’re dealing with Unicode, you should probably use the general category \p{L} for all possible letters, and maybe \p{Nd} for decimal numbers. Also, if you want to accomodate all that dash punctuation, not just HYPHEN-MINUS, you should use the \p{Pd} property. You might also want to write that sequence of whitespace characters simply as \s, assuming that’s not too general for you.
All together, that works out to apattern of [\p{L}\p{Nd}\p{Pd}!$*] to match any one character from that set.
I’d likely use that anyway, even if I didn’t plan on dealing with the full Unicode set, because it’s a good habit to get into, and because these things often grow beyond their original parameters. Now when you lift it to use in other code, it will still work correctly. If you hard‐code all the characters, it won’t.
[-a-z0-9]+,[a-z0-9-]+,[a-z-0-9]+ and also [a-z-0-9]+ all are same.The hyphen between two ranges considered as a symbol.And also [a-z0-9-+()]+ this regex allow hyphen.
use "\p{Pd}" without quotes to match any type of hyphen. The '-' character is just one type of hyphen which also happens to be a special character in Regex.
Is this what you are after?
MatchCollection matches = Regex.Matches(mystring, "-");
Related
I have a field in my application where users can enter a hashtag.
I want to validate their entry and make sure they enter what would be a proper HashTag.
It can be in any language and it should NOT precede with the # sign.
I am writing in JavaScript.
So the following are GOOD examples:
Abcde45454_fgfgfg (good because: only letters, numbers and _)
2014_is-the-year (good because: only letters, numbers, _ and -)
בר_רפאלי (good because: only letters and _)
арбуз (good because: only letters)
And the following are BAD examples:
Dan Brown (Bad because has a space)
OMG!!!!! (Bad because has !)
בר רפ#לי (Bad because has # and a space)
We had a regex that matched only a-zA-Z0-9, we needed to add language support so we changed it to ignore white spaces and forgot to ignore special characters, so here I am.
Some other StackOverflow examples I saw but didn't work for me:
Other languges don't work
Again, English only
[edit]
Added explanation why bad is bad and good is good
I don't want a preceding # character, but if I would to add a # in the beginning, it should be a valid hashtag
Basically I don't want to allow any special characters like !##$%^&*()=+./,[{]};:'"?><
If your disallowed characters list is thorough (!##$%^&*()=+./,[{]};:'"?><), then the regex is:
^#?[^\s!##$%^&*()=+./,\[{\]};:'"?><]+$
Demo
This allows an optional leading # sign: #?. It disallows the special characters using a negative character class. I just added \s to the list (spaces), and also I escaped [ and ].
Unfortunately, you can't use constructs like \p{P} (Unicode punctuation) in JavaScript's regexes, so you basically have to blacklist characters or take a different approach if the regex solution isn't good enough for your needs.
I don't understand why this question does not get more votes. Hashtag detection for multiple languages is a problem. The only working option I could find is posted by Lucas above (all other ones do not work so well).
It needs a modification though:
#[^\s!##$%^&*()=+.\/,\[{\]};:'"?><]+
DEMO
this detects all the hashtags, not only in the beginning of the string, fixes an unescaped character, and removes the unnecessary $ in the end.
First if we exclude all symbol it will not a handy solution. Because symbol depends on keyboard layout and there are hundreds of math symbols and so on. So use this..
[\p{sc=Bengali}|\p{L}_\p{N}]+
1. If you think if language need extra care include like \p{sc=Bengali}|\p{sc=Spanish} etc. Suppose bangla has surrogate alphabet like া, ে ৌ etc so codepoint need to recognize Bangla separately first by \p{sc=Bengali}
2. Than use \p{L} that matches anything that is a Unicode letter a-z and letters like é,ü,ğ,i,ç too or normal any alphabet without complex...matches a single code point in the category "letter"
3. _ underscore allowed
4. \p{N} matches any kind of numeric character in any script. (\d matches only a digit (equal to [0-9]) but for allowed Unicode digit \p{N} only option, because its works with any digit codepoint)
I want to replace invalid characters when entering a phone number
The rules i want are below.
The first character can be "+"
The remaining characters have to be digits 0-9
This is what I have already
phoneNumber.getValue().replace(/[^0-9,+]+/g, "");
this works, kind of but not fully
however I can have a "+" anywhere in the string
I want to remove this if it is not the first character does anyone now how to-do this
Thanks
AJ
Assuming that you don't want to include commas, (i.e. the result should be only decimal digits preceded by an optional +), then this will do it:
phoneNumber.getValue().replace(/(^\+)|\D+/g, '$1');
Notes:
Makes use of the \D non-decimal digit character class shorthand. (i.e. \D is the same as: [^0-9].)
Makes use of the fact that when a capturing group does not participate in the match, it can still be referenced in the replacement string - (it is replaced with the empty string).
Make a slightly smarter regex.
phoneNumber.getValue().replace(/.*(?=\+)|[^0-9]/g,'')
This regex sort of abuses the fact that alternation in replacements work from left to right. Because the + has sort of been "covered" by the lookahead in the first alternation, it is effectively immune to being deleted by the [^0-9] part.
Another one would be
phoneNumber.getValue().replace(/[^\+0-9]/g, '')
When using "+1234abc+5678" as value this results in "+1234+5678"
Whereas the regex of Niet the Dark Absol results in "+5678"
Probably you have to combine that using substring to remove the second plus.
That would be pretty easy.
I have a js code：
/^([a-zA-Z0-9]+[_|\_|\.]?)*[a-zA-Z0-9]+#([a-zA-Z0-9]+[_|\_|\.]?)*[a-zA-Z0-9]+\.[a-zA-Z]{2,3}$/
But what's meaning of [_|\_|\.]?（js regexp）
If we use a resource like Regexper, we can visualise this regular expression:
From this we can conclude that [_|\_|\.] requires one of either "_", "|" or ".". We can also see that the double declaration of "_" and "|" is unnecessary. As HamZa commented, this segment can be shortened to [_|.] to achieve the same result.
In fact, we can even use resources like Regexper to visualise the entire expression.
REGEX101 is a very good tool for
understanding regular expression
Char class [_|\_|\.] 0 to 1 times [greedy] matches:
[_|\_|\. One of the following characters _|_|.
[_|\_|\.] requires one of either "_", "|" or "."
See This Link of RegEx101 here
Your Expression explanation
It matches a pipe character, an underscore, or a period.
It is unnecessarily convoluted, however. It could be simpler.
It could be shortened to this
[|_.]
[_|\_|\.] is probably meant to match an underscore (_) or a period (.), and should have been written as [_.].
I'm reasonably sure the author is using the pipe (|) to mean "or" (i.e., alternation), which isn't necessary inside a character class. As the other responders said, the pipe actually matches a literal pipe, but I don't believe that was the author's intent. It's a very common beginner's mistake.
The dot (.) is another special character that loses its special meaning when it appears in a character class. There's no need to escape it with a backslash as the author did, though it does no harm. And the underscore never has any special meaning; I won't even try to guess why the author listed it twice, once with a backslash and once without.
You didn't ask about it, but the ? doesn't belong there either. That's what makes the regex so horribly inefficient, as Kobi remarked. The idea was to match one or more alphanumerics, then optionally match a separator character (dot or underscore), which must be followed by some more alphanumerics, repeating as needed. Here's how I would write that:
[a-zA-Z0-9]+([_.][a-zA-Z0-9]+)*
If it runs out of alphanumerics and the next character is not _ or ., it skips that whole section and tries to match the next part. And if it can't do that, it can bail out immediately because no match is possible. But the way your regex is written, the separator is optional independently of the things it's supposed to separate, which makes it useless. The regex engine has to keep backing up, trying to match characters that it has already consumed in endless, pointless combinations before it can give up. And that, unfortunately, is another common mistake.
I'm looking to match /(?=\W)(gimme)(?=\W)/gi or alike. The \W are supposed to be zero-width characters to surround my actual match.
Maybe some background. I want te replace certain words (always \w+) with some literal padding added, but only if it's not surrounded by a \w. (That does sound like a negative lookaround, but I hear JS doesn't do those!?)
(Btw: the above "gimme" is the word literal I want to replace. If that wasn't obvious.)
It has to be (?) a lookaround, because the \W have to be zero-width, because the intention is a .replace(...) and I cannot replace/copy the surrounding characters.
So this won't work:
text.replace(/(?=\W)(gimme)(?=\W)/gi, function(l, match, r) {
return l + doMagic(match) + r;
})
The zero-width chars have to be ignored, so the function can return (and replace) only doMagic(match).
I have only very limited lookaround experience and non of it in JS. Grazie.
PS. Or maybe I need a lookbehind and those aren't supported in JS..? I'm confused?
PS. A little bit of context: http://jsfiddle.net/rudiedirkx/kMs2N/show/ (ooh a link!)
you can use word boundary shortcut \b to assert that it's the whole word that you are matching.
The easiest way to achieve what you want to do is probably to match:
/(\s+gimme)(?=\W)/gi
and replace with [yourReplacement] - i.e. capture the whitespaces before 'gimme' and then include one in the replacement.
Another way to approach this would be capturing more characters before and after the gimme literal and then using the groups with backreference:
(\W+?)gimme(\W+?) - your match - note that this time the before and after characters are in the capturing groups 1 and 2
And you'd want to use \1[yourReplacement]\2 as replacement string - not sure how you use backreference in JS, but the idea is to tell the engine that with \1 you mean whatever was matched by the first captuing parenthesis. In some languages these are accessed with $1.
What you currently have will not work, for the following reason, (?=\W) means "the next character is not a word character", and the next thing you try to match is a literal g, so you have a contradiction ("next character is a g, but isn't a letter").
You do in fact need a lookbehind, but they are not supported by JavaScript.
Check out this article on Mimicking Lookbehind in JavaScript for a possible approach.
Have you considered using a lexer/parser combo?
This one is javascript based, and comes with a spiffy demonstration.
I'm learning Javascript via an online tutorial, but nowhere on that website or any other I googled for was the jumble of symbols explained that makes up a regular expression.
Check if all numbers: /^[0-9]+$/
Check if all letters: /^[a-zA-Z]+$/
And the hardest one:
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
What do all the slashes and dollar signs and brackets mean? Please explain.
(By the way, what languages are required to create a flexible website? I know a bit of Javascript and wanna learn jQuery and PHP. Anything else needed?)
Thanks.
There are already a number of good sites that explain regular expressions so I'll just dive a bit into how each of the specific examples you gave translate.
Check if all numbers: ^ anchors the start of the expression (e.g. start at the beginning of the text). Without it a match could be found anywhere. [0-9] finds the characters in that character class (e.g. the numbers 0-9). The + after the character class just means "one or more". The ending $ anchors the end of the text (e.g. the match should run to the end of the input). So if you put that together, that regular expression would allow for only 1 or more numbers in a string. Note that the anchors are important as without them it might match something like "foo123bar".
Check if all letters: Pretty much the same as above but the character classes are different. In this example the character class [a-zA-Z] represents all lowercase and uppercase characters.
The last one actually isn't any more difficult than the other two it's just longer. This answer is getting quite long so I'll just explain the new symbols. A \w in a character class will match word characters (which are defined per regex implementation but are generally 0-9a-zA-Z_ at least). The backslash before the # escapes the # so that it isn't seen as a token in the regex. A period will match any character so .+ will match one or more of any character (e.g. a, 1, Z, 1a, etc). The last part of the regex ({2,4}) defines an interval expression. This means that it can match a minimum of 2 of the thing that precedes it, and a maximum of 4.
Hope you got something out of the above.
There is an awesome explanation of regular expressions at http://www.regular-expressions.info/ including notes on language and implementation specifics.
Let me explain:
Check if all numbers: /^[0-9]+$/
So, first thing we see is the "/" at the beginning and the end. This is a deliminator, and only serves to show the beginning and end of the regular expression.
Next, we have a "^", this means the beginning of the string. [0-9] means a number from 0-9. + is a modifier, which modifies the term in front of it, in this case, it means you can have one or more of something, so you can have one or more numbers from 0-9.
Finally, we end with "$", which is the opposite of "^", and means the end of the string. So put that all together and it basically makes sure that inbetween the start and end of the string, there can be any number of digits from 0-9.
Check if all letters: /^[a-zA-Z]+$/
We notice this is very similar, but instead of checking for numbers 0-9, it checks for letters a-z (lowercase) and A-Z (uppercase).
And the hardest one:
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
"\w" means that it is a word, in this case we can have any number of letters or numbers, as well as the period means that it can be pretty much any character.
The new thing here is escape characters. Many symbols cannot be used without escaping them by placing a slash in front, as is the case with "\#". This means it is looking directly for the symbol "#".
Now it looks for letters and symbols, a period (this one seems incorrect, it should be escaping the period too, though it will still work, since an unescaped period will make any symbol). Numbers inside {} mean that there is inbetween this many terms in the previous term, so of the [a-zA-Z0-9], there should be 2-4 characters (this part here is the website domain, such as .com, .ca, or .info). Note there's another error in this one here, the [a-zA-z0-9] should be [a-zA-Z0-9] (capital Z).
Oh, and check out that site listed above, it is a great set of tutorials too.
Regular Expressions is a complex beast and, as already pointed out, there are quite a few guides off of google you can go read.
To answer the OP questions:
Check if all numbers: /^[0-9]+$/
regexps here are all delimated with //, much like strings are quoted with '' or "".
^ means start of string or line (depending on what options you have about multiline matching)
[...] are called character classes. Anything in [] is a list of single matching characters at that position in this case 0-9. The minus sign has a special meaning of "sequence of characters between". So [0-9] means "one of 0123456789".
+ means "1 or more" of the preceeding match (in this case [0-9]) so one or more numbers
$ means end of string/line match.
So in summary find any string that contains only numbers, i.e '0123a' will not match as [0-9]+ fails to match a before $).
Check if all letters: /^[a-zA-Z]+$/
Hopefully [A-Za-z] makes sense now (A-Z = ABCDEF...XYZ and a-z abcdef...xyz)
Validate Email: /^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
Not all regexp parses know the \w sequence. Javascript, java and perl I know do support it.
I have already have covered '/^ at the beginning, for this [] match we are looking for
\w - . and +. I think that regexp is incorrect. Either the minus sign should be escaped with \ or it should be at the end of the [] (i.e [\w+.-]). But that is an aside they are basically attempting to allow anything of abcdefghijklmnopqrstuvwxyz01234567890-.+
so fred.smith-foo+wee#mymail.com will match but fred.smith%foo+wee#mymail.com wont (the % is not matched by [\w.+-]).
\# is the litteral atsil sign (it is escaped as perl expands # an array variable reference)
[a-zA-Z0-9.-]+ is the same as [\w.-]+. Very much like the user part of the match, but does not match +. So this matches foo.com. and google.co. but not my+foo.com or my***domain.co.
. means match any one character. This again is incorrect as fred#foo%com will match as . matches %*^%$£! etc. This should of been written as \.
The last character class [a-zA-z0-9]{2,4} looks for between 2 3 or 4 of the a-zA-Z0-9 specified in the character class (much like + looks for "1 more more" {2,4} means at least 2 with a maximum of 4 of the preceeding match. So 'foo' matches, '11' matches, '11111' does not match and 'information' does not.
The "tweaked" regexp should be:
/^[\w.+-]+\#[a-zA-Z0-9.-]+\.[a-zA-z0-9]{2,4}$/
I'm not doing a tutorial on RegEx's, that's been done really well already, but here are what your expressions mean.
/^<something>$/ String begins, has something in the middle, and then immediately ends.
/^foo$/.test('foo'); // true
/^foo$/.test('fool'); // false
/^foo$/.test('afoo'); // false
+ One or more of something:
/a+/.test('cot');//false
/a+/.test('cat');//true
/a+/.test('caaaaaaaaaaaat');//true
[<something>] Include any characters found between the brackets. (includes ranges like 0-9, a-z, and A-Z, as well as special codes like \w for 0-9a-zA-Z_-
/^[0-9]+/.test('f00')//false
/^[0-9]+/.test('000')//true
{x,y} between X and Y occurrences
/^[0-9]{1,2}$/.test('12');// true
/^[0-9]{1,2}$/.test('1');// true
/^[0-9]{1,2}$/.test('d');// false
/^[0-9]{1,2}$/.test('124');// false
So, that should cover everything, but for good measure:
/^[\w-.+]+\#[a-zA-Z0-9.-]+.[a-zA-z0-9]{2,4}$/
Begins with at least character from \w, -, +, or .. Followed by an #, followed by at least one in the set a-zA-Z0-9.- followed by one character of anything (. means anything, they meant \.), followed by 2-4 characters of a-zA-z0-9
As a side note, this regular expression to check emails is not only dated, but it is very, very, very incorrect.