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
Related
I have the following validation expresion on an asp.net web form that allows alphanumeric characters, spaces,at least one alpha character, and a minimum of 3 characters and a maximum of 20:
ValidationExpression="(?!^[0-9]$)(?!^[a-zA-Z]$)^([a-zA-Z0-9 _]{3,20})$"
Now I have been asked to allow hyphens and apostraphes but no other special characters.
How can I implement this in my current validation?
This (?!^[0-9'-]$)(?!^[a-zA-Z'-]$)^([a-zA-Z0-9 _'-]{3,20})$?
Well, the main trick here is that - sign should be placed at the end of the character group for it to be parsed as a literal hyphen.
Try this:
(?=.*?[A-Za-z]+)^[a-zA-Z0-9_\-' ]{3,20}$
I've a textbox in an ASP.NET application, for which I need to use a regular expression to validate the user input string. Requirements for regex are -
It should allow only one space between words. That is, total number of spaces between words or characters should only be one.
It should ignore leading and trailing spaces.
Matches:
Test
Test abc
Non Matches:
Test abc def
Test abc --> I wanted to include multiple spaces between the 2 words. However the editor ignores these extra spaces while posting a question.
Assuming there must be either one or two 'words' (i.e. sequences of non-space characters)
"\s*\S+(\s\S+)?\s*"
Change \S to [A-Za-z] if you want to allow only letters.
Pretty straightforward:
/^ *(\w+ ?)+ *$/
Fiddle: http://refiddle.com/gls
Maybe this one will do?
\s*\S+?\s?\S*\s*
Edit: Its a server-encoded regex, meaning that you might need to remove one of those escaping slashes.
How about:
^\s*(\w+\s)*\w+\s*$
I need a regular expression to validate whether text entered in an asp.net textbox has the following format
A-za-z123456789 /s = /s A-za-z123456789
Regular expression explained:
one or more alphanumeric characters
followed by any number of spaces
an equal sign
followed by any number of spaces
one or more alphanumeric characters
[a-zA-Z0-9]*\s*\=\s*[a-zA-Z0-9]*
Replace * with + if you want one or more rather than "any" (which includes zero)
Considering your answer to the comment about requiring one or more alphanumeric characters each side:
[a-zA-Z0-9]+\s*\=\s*[a-zA-Z0-9]+
This version will only match if there is at least one alphanumeric character each side of the "=".
If zero valid
"^[a-zA-Z\\d]+\\s*=\\s*[a-zA-Z\\d]+$"
If zero not valid
"^[a-zA-Z1-9]+\\s*=\\s*[a-zA-Z1-9]+$"
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.
I have a regular expression
^[a-zA-Z+#-.0-9]{1,5}$
which validates that the word contains alpha-numeric characters and few special characters and length should not be more than 5 characters.
How do I make this regular expression to accept a maximum of five words matching the above regular expression.
^[a-zA-Z+#\-.0-9]{1,5}(\s[a-zA-Z+#\-.0-9]{1,5}){0,4}$
Also, you could use for example [ ] instead of \s if you just want to accept space, not tab and newline. And you could write [ ]+ (or \s+) for any number of spaces (or whitespaces), not just one.
Edit: Removed the invalid solution and fixed the bug mentioned by unicornaddict.
I believe this may be what you're looking for. It forces at least one word of your desired pattern, then zero to four of the same, each preceded by one or more white-space characters:
^XX(\s+XX){0,4}$
where XX is your actual one-word regex.
It's separated into two distinct sections so that you're not required to have white-space at the end of the string. If you want to allow for such white-space, simply add \s* at that point. For example, allowing white-space both at start and end would be:
^\s*XX(\s+XX){0,4}\s*$
You regex has a small bug. It matches letters, digits, +, #, period but not hyphen and also all char between # and period. This is because hyphen in a char class when surrounded on both sides acts as a range meta char. To avoid this you'll have to escape the hyphen:
^[a-zA-Z+#\-.0-9]{1,5}$
Or put it at the beg/end of the char class, so that its treated literally:
^[-a-zA-Z+#-.0-9]{1,5}$
^[a-zA-Z+#.0-9-]{1,5}$
Now to match a max of 5 such words you can use:
^(?:[a-zA-Z+#\-.0-9]{1,5}\s+){1,5}$
EDIT: This solution has a severe limitation of matching only those input that end in white space!!! To overcome this limitation you can see the ans by Jakob.