As I put in the title I need a validator (regular expresion, customvalidator, etc) that allows the user to input everything (lettrs, special characters, spaces) but not numbers.
Valid string would be "D'Elia" or "Del Riego" Not Valid String would be " " or "p1" or "1" So I have to allow all letters, all special characters including space, no numbers, and not spaces at the beginning (well, al least that the field doesn't allow that only one space and nothing more can be written)
How can I do that?
Thank you,
Sabrina
You need to use a positive lookahead at the start to check for atleast one letter.
^(?=.*?[A-Za-z])[^0-9]+$
Related
Right now I need to duplicate a password expression validator for a website. The password is only required to be 8-25 characters (only alphabet characters) long. I thought this was weird and had been using this regex
(?!^[0-9]*$)(?!^[a-zA-Z]*$)^([a-zA-Z0-9]{8,25})
but it has to be optional to have a capital letter, special characters and/or numbers throughout the password. I'm not particularly apt at building regex's where there are optional characters. Any help would be be appreciated.
I am using asp.net's RegularExpressionValidator.
This pattern should work:
^[a-zA-Z]{8,25}$
It matches a string consisting of 8 to 25 Latin letters.
If you want to allow numbers as well, this pattern should work:
^[a-zA-Z0-9]{8,25}$
It matches a string consisting of 8 to 25 Latin letters or decimal digits.
If you want to allow special characters as well, this pattern should work:
^[a-zA-Z0-9$#!]{8,25}$
It matches a string consisting of 8 to 25 Latin letters, decimal digits, or symbols, $, # or ! (of course you can add to this set fairly easily).
Your current regex won't work because it will accept special characters as from 9th character (and anything after the 9th character in fact, even a 26th character because you don't have the end of string anchor) .
You probably want something like this:
^(?=.*[a-z])[A-Za-z0-9]{8,25}$
This first makes sure there are lowercase alphabets (you mentioned that uppercase and digits are optional, so this makes obligatory lowercase) and then allows only uppercase and digits.
EDIT: To allow any special characters, you can use this:
^(?=.*[a-z]).{8,25}$
My understanding of your problem is that the password's first requirement is that it has to contain lowercase alphabet characters. The option now is that it can also contain other characters. If this isn't right, let me know.
regex101 demo
The validator should allow only alphabetic characters (a-z and A-Z), dots(.),comma(,), slash(/) and hyphen (-). Please help to to find out one. Or tell me how to create one customized to my specifications.
I have tried [a-zA-Z,-/.] and it works but my requirements only allow for a maximum of 1 of each of the non-letter characters I specified (.,/-).
Try: ^[A-Za-z]*[-a-zA-Z,/.]{1}[A-Za-z]*$
Explanation
^ Anchor to start of string
[A-Za-z]* may be surrounded by multiple letters
[-a-zA-Z,/.]{1} Only one of the enclosed characters
[A-Za-z]* may be surrounded by multiple letters
$ Anchor to end of string
In my requirement a Textbox should allow Alphabets,Numeric s, Special Characters,Special Symbols With at least one Alphabet.
I will try like this but i am not getting.
^\d*[a-zA-Z][a-zA-Z0-9#*,$._&% -!><^#]*$
You may want to have 2 regular expression validators; one for validating the allowed characters, and one for validating that at least on alphabet has been provided. You may be able to get at least one, but this way, you can have two separate validation messages to show the user explaining why the input is wrong.
Just match for special characters until you encounter a letter, then match for everything until the end of the string:
^[0-9#*,$._&% -!><^#]*[a-zA-Z0-9#*,$._&% -!><^#]*$
Use lookaheads :
/^(?=.*[a-zA-Z])[\w#*,$.&%!><^#-]*$/
Edit :
I assume the - is meant as the actual - character and not a range of space to !.
I removed the space character. You can of course add it if you want.
[ -!]
Effectively means :
[ -!] # Match a single character in the range between “ ” and “!”
And I have no idea what that range entails!
Using ASP.NET syntax for the RegularExpressionValidator control, how do you specify restriction of two consecutive characters, say character 'x'?
You can provide a regex like the following:
(\\w)\\1+
(\\w) will match any word character, and \\1+ will match whatever character was matched with (\\w).
I do not have access to asp.net at the moment, but take this console app as an example:
Console.WriteLine(regex.IsMatch("hello") ? "Not valid" : "Valid"); // Hello contains to consecutive l:s, hence not valid
Console.WriteLine(regex.IsMatch("Bar") ? "Not valid" : "Valid"); // Bar does not contain any consecutive characters, so it's valid
Alexn is right, this is the way you match consecutive characters with a regex, i.e. (a)\1 matches aa.
However, I think this is a case of everything looking like a nail when you're holding a hammer. I would not use regex to validate this input. Rather, I suggest validating this in code (just looping through the string, comparing str[i] and str[i-1], checking for this condition).
This should work:
^((?<char>\w)(?!\k<char>))*$
It matches abc, but not abbc.
The key is to use so called "zero-width negative lookahead assertion" (syntax: (?! subexpression)).
Here we make sure that a group matched with (?<char>\w) is not followed by itself (expressed with (?!\k<char>)).
Note that \w can be replaced with any valid set of characters (\w does not match white-spaces characters).
You can also do it without named group (note that the referenced group has number 2):
^((\w)(?!\2))*$
And its important to start with ^ and end with $ to match the whole text.
If you want to only exclude text with consecutive x characters, you may use this
^((?<char>x)(?!\k<char>)|[^x\W])*$
or without backreferences
^(x(?!x)|[^x\W])*$
All syntax elements for .NET Framework Regular Expressions are explained here.
You can use a regex to validate what's wrong as well as what's right of course. The regex (.)\1 will match any two consecutive characters, so you can just reject any input that gives an IsValid result to that. If this is the only validation you need, I think this way is far easier than trying to come up with a regex to validate correct input instead.
So, I basically would like to test to see if a string contains a range of alphanumeric characters. It's to be used as a client-side validation and I don't want to prevent users from entering whatever they want. Best to give examples of what should/should not pass validation:
So to be specific, the expression I'm looking for is to test to make sure string contains anywhere from 3 to 10 alphanumeric characters. I'd like to plug into an ASP.NET client side validator.
NOTE: quotes not part of input (but could be!)
" f o o " should pass since there are 3 chars
"f_0_0" should pass
" fo " should not
"F......o......o......b.....a......r" should pass
thx
^([^a-zA-Z0-9]*[a-zA-Z0-9][^a-zA-Z0-9]*){3,10}$
Allows for exactly 3-10 alphanumeric characters, each surrounded by an arbitrary number of non-alphanumeric characters.
(Untested, but it should conform to the JScript subset of the standard .net Regex syntax, as required by the RegularExpressionValidator. Unfortunately, the shorthands \w and \W cannot be used since they include the underscore as an alphanumeric character.)
I'm not familiar with ASP.NET client-side validators, so I'm not sure if you need to do this in a regex, but potentially an easy solution is as follows:
Remove all non-alphanumeric characters (regex replace [^0-9A-Za-z] with nothing).
Check if string length is 3 or greater.