$.validator.addMethod('AZ09_', function (value) {
return /^[a-zA-Z0-9.-_]+$/.test(value);
}, 'Only letters, numbers, and _-. are allowed');
When I use somehting like test-123 it still triggers as if the hyphen is invalid. I tried \- and --
Escaping using \- should be fine, but you can also try putting it at the beginning or the end of the character class. This should work for you:
/^[a-zA-Z0-9._-]+$/
Escaping the hyphen using \- is the correct way.
I have verified that the expression /^[a-zA-Z0-9.\-_]+$/ does allow hyphens. You can also use the \w class to shorten it to /^[\w.\-]+$/.
(Putting the hyphen last in the expression actually causes it to not require escaping, as it then can't be part of a range, however you might still want to get into the habit of always escaping it.)
The \- maybe wasn't working because you passed the whole stuff from the server with a string. If that's the case, you should at first escape the \ so the server side program can handle it too.
In a server side string: \\-
On the client side: \-
In regex (covers): -
Or you can simply put at the and of the [] brackets.
Generally with hyphen (-) character in regex, its important to note the difference between escaping (\-) and not escaping (-) the hyphen because hyphen apart from being a character themselves are parsed to specify range in regex.
In the first case, with escaped hyphen (\-), regex will only match the hyphen as in example /^[+\-.]+$/
In the second case, not escaping for example /^[+-.]+$/ here since the hyphen is between plus and dot so it will match all characters with ASCII values between 43 (for plus) and 46 (for dot), so will include comma (ASCII value of 44) as a side-effect.
\- should work to escape the - in the character range. Can you quote what you tested when it didn't seem to? Because it seems to work: http://jsbin.com/odita3
A more generic way of matching hyphens is by using the character class for hyphens and dashes ("\p{Pd}" without quotes). If you are dealing with text from various cultures and sources, you might find that there are more types of hyphens out there, not just one character. You can add that inside the [] expression
Related
I am using this regex to parse URL from a semicolon separated string.
\b(?:https?:|http?:|www\.)\S+\b
It is working fine if my input text is in these formats:
"Google;\"https://google.com\""
//output - https://google.com
"Yahoo;\"www.yahoo.com\""
//output - www.yahoo.com
but in this case it gives incorrect string
"https://google.com;\"https://google.com\""
//output - https://google.com;\"https://google.com
how can I stop the parsing when I encounter the ';' ?
Looking at your examples, I would just match any URL between quotation marks. Something like this:
(?<=")(?:https?:|www\.)[^"]*
You can try it out here
Or as others have said, split the input string by the semicolon character using string.Split, and check each string sequentially for your desired match.
For your example data you might use a positive lookahead (?=) and a positive lookbehind (?<=)
(?<=")(?:https?:|www\.).+?(?=;?\\")
That would match
(?<=") Positive lookbehind to assert that what is on the left side is a double quote
(?:https?:|www\.) Match either http with an optional s or www.
.+? Match any character one or more times non greedy
(?=;?\\") Positive lookahead which asserts that what follows is an optional ; followed by\"
I would personally just modify the regex to look specifically for URLs and add some conditionals to the https:// protocols and www quantifier. Using \S+ can be kind of iffy because it will grab every non whitespace character, in which in a URL, it's limited on the characters you can use.
Something like this should work great for your particular needs.
(https?:\/{2})?([w]{3}.)?\w+\.[a-zA-Z]+
This sets up a conditional on the http (s also optional) protocol which would then be immediately be followed by the ://. Then, it will grab all letters, numbers, and underscores as many as possible until the ., followed by the last set of characters to end it. You can exchange the [a-zA-Z] character set for a explicit set of domains if you'd prefer.
I use the following regex, which is working, but I want to add a condition so as to accept spaces at the end of the value. Currently it is not working.
What am I missinghere?
^[a-zA-Z][a-zA-Z0-9_]+\s?$[\s]*$
Assumption: you added the two end of string anchors $ by mistake.
? quantifier, matching one or zero repetitions, makes the previous item optional
* quantifier, matching zero or more repetitions
So change your expression to
^[a-zA-Z][a-zA-Z0-9_]+\s*$
this is matching any amount of whitespace at the end of the string.
Be aware, whitespace is not just the space character, it is also tabs and newlines (and more)!
If you really want to match only space, just write a space or make a character class with all the characters you want to match.
^[a-zA-Z][a-zA-Z0-9_]+ *$
or
^[a-zA-Z][a-zA-Z0-9_]+[ \t]*$
Next thing is: Are you sure you only want plain ASCII letters? Today there is Unicode and you can use Unicode properties, scripts and blocks in your regular expressions.
Your expression in Unicode, allowing all letters and digits.
^\p{L}\w+\s*$
\p{L} Unicode property, any kind of letter from any language.
\w shorthand character class for word characters (letters, digits and connector characters like "_") [\p{L}\p{Nd}\p{Pc}] as character class with Unicode properties. Definition on msdn
why two dollars?
^[a-zA-Z][a-zA-Z0-9_]+\s*$
or make it this :
"^[a-zA-Z][a-zA-Z0-9_]+\s?\$\s*$"
if you want to literally match the dollar.
Try this -
"^[a-zA-Z][a-zA-Z0-9_]+(\s)?$"
or this -
"^[a-zA-Z][a-zA-Z0-9_]+((\s){,})$"
$ indicates end of expression, if you are looking $ as character, then escape it with \
Brand new to using Regular Expressions. I have one that currently accepts alphanumeric characters only. I need to add the following special characters to the regex:
# #$%*():;"',/? !+=-_
Here is the regular expression:
RegularExpression(#"^[a-zA-Z\s.,0-9-]{1,30}$",
When I try to add the special characters, I alter the Regex like so:
RegularExpression(#"^[a-zA-Z\s.,0-9-# #$%*():;"',/? !+=-_]{1,30}$"
However this throws an error starting with the ' character that says Newline in constant.
I've tied to escape both the " and the ' characters, however without any luck.
the problem comes from the double quote that need to be escaped (""), not from the single quote.
#"^[a-zA-Z\s.,0-9##$%*():;""'/?!+=_-]{1,30}$"
note that the - must be at the last (or first) position in a character class, since it has a special meaning (define ranges)
These regexs' are equivalent to yours.
Both use tilde ~ as the delimeter.
Both use double quotes on the regex strings.
Note that in order for the the dash - in class to be interpreted literally and not as a range operator, it must exist somewhere disambiguous, or be escaped.
A good place to put it is between valid ranges (or at the beginning or end of a class).
For example [a-z-0-9] is a good place.
Edit - '-' Literal may have to be escaped or beginning/end of class. (This case was for Perl/PCRE engines)
This one ^[a-z-A-Z0-9_\s.,##$%*():;"',/?!+=]{1,30}$ is your regex without duplicate chars.
To make it more readable noting that the word class is contained, it can be reduced to
^[\w-\s.,##$%*():;"',/?!+=]{1,30}$
Edit - Php test cases removed.
I am trying to define a regular to use with a regular expression validator that limits the content of a textbox to only alphanumeric characters, slash (/), hash (#), left and right parentheses (()), period (.), apostrophe ('), quote ("), hyphen (-) and spaces.
I am having troubles with the hash and quote, the other restrictions are working, but when I insert one of these chars the evaluation fails and I get the error message. I have tried to escape these characters without and also using verbatim which was my last attempt.
#"[ a-zA-ZÀ-ÿ/().\'-""#]"
Any thoughts on these? Thank you
The regex language is smart enough to understand that periods and parentheses within a character class actually refer to the characters and not to the patterns they usually do when they appear outside of character classes.
Within your character class, you need to escape the slash (\) and the hyphen(-), but that's it:
#"[ a-zA-ZÀ-ÿ/().\\'\-""#]"
If you move your hyphen to the end of the character class, you won't even need to escape that:
#"[ a-zA-ZÀ-ÿ/().\\'""#-]"
And of course this still only matches one a single character. If you want to ensure that the entire string consists only of these characters, you'll need to use start (^) and end ($) anchors and a quantifier (* or +) after your character class.
I believe your final pattern should look like this:
#"^[ a-zA-ZÀ-ÿ/().\\'""#-]*$"
I have a username validator IsValidUsername, and I am testing "baconman" but it is failing, could someone please help me out with this regex?
if(!Regex.IsMatch(str, #"^[a-zA-Z]\\w+|[0-9][0-9_]*[a-zA-Z]+\\w*$")) {
isValid = false;
}
I want the restrictions to be: (It's very close)
Be between 5 & 17 characters long
contain at least one letter
no spaces
no special characters
You're escaping unnecessarily: if you write your regex as starting with # outside the string, you don't need both \ - just one is fine.
Either:
#"\w"
or
"\\w"
Edit: I didn't make this clear: right now due to the double escaping, you're looking for a \ in your regex and a w. So your match would need [some character]\w to match (example: "a\w" or "a\wwwwww" would match.
Your requirements are best taken care of in normal C#. They don't map well to a regular expression. Just code them up using LINQ which works on strings like it would on an IEnumerable<char>.
Also, understanding a query of a string is much easier than understanding a Regex with the requirements that you have.
It is possible to do everything as part of a Regex, however it is not pretty :-)
^(\w(?=\w*[a-zA-Z])|[a-zA-Z]|\w(?<=[a-zA-Z]\w*)){5,17}$
It does 3 checks that always results in 1 character being matched (so we can perform the length check in the end)
Either the character is any word character \w which is before [a-zA-Z]
Or it is [a-zA-Z]
Or it is any word character \w which is after [a-zA-Z]