I've a textbox for name field where I've used asp validation for proper name format. I want to validate multiple spaces between the strings. How can I do that? The leading and trail spaces are removed by trim() function but how can I validate multiple spaces between the strings? like
multiple spaces
no space
My validation code::
<label>
<span>Full name</span>
<input type="text" id="txt_name" runat="server" required="required"/>
<asp:RegularExpressionValidator ID="rev_txt_name" runat="server" ControlToValidate="txt_name" ForeColor="Red"
ErrorMessage="Invalid name!" SetFocusOnError="True" ValidationExpression="^[a-zA-Z'.\s]{2,50}"></asp:RegularExpressionValidator>
</label>
The pattern you are using allows matching whitespace anywhere inside the string and any occurrences, consecutive or not, since it is part of a rather generic character class. You need to use a grouping and quantify it accordingly:
^(?=.{2,50}$)[a-zA-Z'.]+(?:\s[a-zA-Z'.]+)*$
Note that the (?=.{2,50}$) lookahead requires the whole line to be of 2 to 50 chars long.
See the regex demo.
Details:
^ - start of string
(?=.{2,50}$) - a positive lookahead requiring any 2 to 50 chars other than a newline up to the end of the string
[a-zA-Z'.]+ - 1+ letters, single quote or dot chars
(?: - a non-capturing group start:
\s - 1 whitespace
[a-zA-Z'.]+ - 1+ letters, single quote or dot chars
)* - zero or more (*) occurrences
$ - end of string
Related
I've 2 regular expression:
string regex1 = "(?i)(^(?!^.*?admin)(?!^.*?admin[admin\d]).*$)"; this will check for 'admin' substring in the given string and case is insensitive.
string regex2 = "^[^<>?]{5,100}$"; this will check for special char(^<>?) and length between 5 to 100 only.
I want a regular expression where both the regex can be validated at once with the use of only single regex.
Ex-
<asp:RegularExpressionValidator ID="RegularExpressionValidator1" runat="server"
ControlToValidate="txtBox1" ErrorMessage="Validation Failed!"
ValidationExpression="(?i)(^(?!^.*?admin)(?!^.*?admin[admin\d]).*$)">
</asp:RegularExpressionValidator>
<asp:RegularExpressionValidator ID="RegularExpressionValidator2" runat="server"
ControlToValidate="txtBox2" ErrorMessage="Length Validation Failed!"
ValidationExpression="^[^<>?]{5,100}$">
</asp:RegularExpressionValidator>
Q. Can we have a single "RegularExpressionValidator" that serves both the above functionality?
The (?i)(^(?!^.*?admin)(?!^.*?admin[admin\d]).*$) regex is too redundant, it is equal to (?i)^(?!^.*?admin).*$. It basically matches any string that contains no admin substring.
The ^[^<>?]{5,100}$ regex disallows <, > and ? in the string and sets string length limit.
Combining the two is done by replacing the .* in the first pattern with the consuming part of the second regex ([^<>?]{5,100}):
(?i)^(?!^.*?admin)[^<>?]{5,100}$
Details
(?i) - case insensitive mode on
^ - start of string
(?!^.*?admin) - no admin substring allowed anywhere after 0 or more chars other than line break chars, as few as possible
[^<>?]{5,100} - five to a hundred chars other than <, > and ?
$ - end of string.
I have an ASP.NET MVC application containing a form field called 'First/last name'. I need to add some basic validation to ensure people enter at least two words. It doesn't need to be totally comprehensive in checking word length etc, we essentially just need to prevent people from entering just their first name which is what's happening currently. I don't want to limit to just alphabetic characters as some names include punctuation. I just want to ensure that people have entered at least two words separated by a space.
I have the following regex currently:
[RegularExpression(#"^((\b[a-zA-Z]{2,40}\b)\s*){2,}$", ErrorMessage = "Invalid first/last name")]
This works to an extent (it checks for 2 words) but it's invalid if punctuation is entered, which isn't what I'm looking for.
Could anyone suggest how to modify the above so that it doesn't matter if punctuation is used in the words? I'm not good with the regular expression syntax, hence asking here.
Thanks.
You want two words, so at least one space between them, and beyond that you want to allow everything else (e.g., punctuation). So keep it simple:
\w.*\s.*\w
Or if you must anchor it to start and end:
^.*\w.*\s.*\w.*$
These will match, for example, D' Addario (but not D'Artagnan by itself, since it counts as one word by the space criterion).
Maybe just:
#"\w\s\w"
word white space word
Hi you can use this regex for validation
'^[a-zA-Z0-9]+ {1}[a-zA-Z0-9]+$`'
Demo http://rubular.com/r/YN8eFa1yFE
If you just want to allow a sequence of non-whitespace characters followed by 1 or more sequences of whitespace characters followed by non-whitespace characters, you can use
^\s*\S+(?:\s+\S+)+\s*$
See regex demo
It won't accept just First or First .
Regex breakdown:
^ - start of string
\s* - zero or more whitespace
\S+ - 1 or more non-whitespace symbols
(?:\s+\S+)+ - 1 or more sequences of ...
\s+ - 1 or more whitespace sequences (remove + to allow only 1 whitespace between words)
\S+ - 1 or more non-whitespace symbols
\s* - zero or more whitespace
$ - end of string
I'm new on Asp. I have a problem to using regex for checking password input. Here the regex
<asp:RegularExpressionValidator ID="Regex1" runat="server"
ErrorMessage="Password must contain: Minimum 8 characters atleast 1 UpperCase Alphabet, 1 LowerCase Alphabet, 1 Number and 1 Special Character"
Font-Italic="True" Font-Size="Small" ForeColor="Red"
ValidationExpression="^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&])[A-Za-z\d$#$!%*?&]{8,}"
ControlToValidate="TextBoxNewPassword" Display="Dynamic" />
When I input "Hamlida123#" regex did'nt allow it. How to solve this?
You need to include the '#` character specifically in the regex, like so:
ValidationExpression="^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&#])[A-Za-z\d$#$!%*?&#]{8,}"
Demo
Based on your current regex, I assume that you are only allowing certain non-word characters, and so you would need to list every allowable character in your regex as shown above.
I have the following input text:
#"This is some text #foo=bar #name=""John \""The Anonymous One\"" Doe"" #age=38"
I would like to parse the values with the #name=value syntax as name/value pairs. Parsing the previous string should result in the following named captures:
name:"foo"
value:"bar"
name:"name"
value:"John \""The Anonymous One\"" Doe"
name:"age"
value:"38"
I tried the following regex, which got me almost there:
#"(?:(?<=\s)|^)#(?<name>\w+[A-Za-z0-9_-]+?)\s*=\s*(?<value>[A-Za-z0-9_-]+|(?="").+?(?=(?<!\\)""))"
The primary issue is that it captures the opening quote in "John \""The Anonymous One\"" Doe". I feel like this should be a lookbehind instead of a lookahead, but that doesn't seem to work at all.
Here are some rules for the expression:
Name must start with a letter and can contain any letter, number, underscore, or hyphen.
Unquoted must have at least one character and can contain any letter, number, underscore, or hyphen.
Quoted value can contain any character including any whitespace and escaped quotes.
Edit:
Here's the result from regex101.com:
(?:(?<=\s)|^)#(?<name>\w+[A-Za-z0-9_-]+?)\s*=\s*(?<value>(?<!")[A-Za-z0-9_-]+|(?=").+?(?=(?<!\\)"))
(?:(?<=\s)|^) Non-capturing group
# matches the character # literally
(?<name>\w+[A-Za-z0-9_-]+?) Named capturing group name
\s* match any white space character [\r\n\t\f ]
= matches the character = literally
\s* match any white space character [\r\n\t\f ]
Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
(?<value>(?<!")[A-Za-z0-9_-]+|(?=").+?(?=(?<!\\)")) Named capturing group value
1st Alternative: [A-Za-z0-9_-]+
[A-Za-z0-9_-]+ match a single character present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
A-Z a single character in the range between A and Z (case sensitive)
a-z a single character in the range between a and z (case sensitive)
0-9 a single character in the range between 0 and 9
_- a single character in the list _- literally
2nd Alternative: (?=").+?(?=(?<!\\)")
(?=") Positive Lookahead - Assert that the regex below can be matched
" matches the characters " literally
.+? matches any character (except newline)
Quantifier: +? Between one and unlimited times, as few times as possible, expanding as needed [lazy]
(?=(?<!\\)") Positive Lookahead - Assert that the regex below can be matched
(?<!\\) Negative Lookbehind - Assert that it is impossible to match the regex below
\\ matches the character \ literally
" matches the characters " literally
You can use a very useful .NET regex feature where multiple same-named captures are allowed. Also, there is an issue with your (?<name>) capture group: it allows a digit in the first position, which does not meet your 1st requirement.
So, I suggest:
(?si)(?:(?<=\s)|^)#(?<name>\w+[a-z0-9_-]+?)\s*=\s*(?:(?<value>[a-z0-9_-]+)|(?:"")?(?<value>.+?)(?=(?<!\\)""))
See demo
Note that you cannot debug .NET-specific regexes at regex101.com, you need to test them in .NET-compliant environment.
Use string methods.
Split
string myLongString = ""#"This is some text #foo=bar #name=""John \""The Anonymous One\"" Doe"" #age=38"
string[] nameValues = myLongString.Split('#');
From there either use Split function with "=" or use IndexOf("=").
I am new to regular expressions and need a regular expression for address, in which user cannot enter repeating special characters such as: ..... or ,,,.../// etc and none of the special characters could be entered more than 5 times in the string.
...,,,....// =>No Match
Street no. 40. hello. =>Match
Thanks in advance!
I have tried this:
([a-zA-Z]+|[\s\,\.\/\-]+|[\d]+)|(\(([\da-zA-Z]|[^)^(]+){1,}\))
It selects all alphanumeric n some special character with no empty brackets.
You can use Negative lookahead construction that asserts what is invalid to match. Its format is (?! ... )
For your case you can try something like this:
This will not match the input string if it has 2 or more consecutive dots, commas or slashes (or any combination of them)
(?!.*[.,\/]{2}) ... rest of the regex
This will not match the input string if it has more than 5 characters 'A'.
(?!(.*A.*){5}) ... rest of the regex
This will match everything except your restrictions. Repplace last part (.*) with your regex.
^(?!.*[.,\/]{2})(?!(.*\..*){5})(?!(.*,.*){5})(?!(.*\/.*){5}).*$
Note: This regex may no be optimized. It may be faster if you use loop to iterate over string characters and count their occurences.
You can use this regex:
^(?![^,./-]*([,./-])\1)(?![^,./-]*([,./-])(?:[^,./-]*\2){4})[ \da-z,./-]+$
In C#:
foundMatch = Regex.IsMatch(yourString, #"^(?![^,./-]*([,./-])\1)(?![^,./-]*([,./-])(?:[^,./-]*\2){4})[ \da-z,./-]+$", RegexOptions.IgnoreCase);
Explanation
The ^ anchor asserts that we are at the beginning of the string
The negative lookahead (?![^,./-]*([,./-])\1) asserts that it is not possible to match any number of special chars, followed by one special char (captured to Group 1) followed by the same special char (the \1 backreference)
The negative lookahead (?![^,./-]*([,./-])(?:[^,./-]*\2){4}) ` asserts that it is not possible to match any number of special chars, followed by one special char (captured to Group 2), then any non-special char and that same char from Group 2, four times (five times total)
The $ anchor asserts that we are at the end of the string
A regular expression string to detect invalid strings is:
[^\w \-\r\n]{2}|(?:[\w \-]+[^\w \-\r\n]){5}
As C# string literal (regular and verbatim):
"[^\\w \\-\\r\\n]{2}|(?:[\\w \\-]+[^\\w \\-\\r\\n]){5}"
#"[^\w \-\r\n]{2}|(?:[\w \-]+[^\w \-\r\n]){5}"
It is much easier to find a string than to validate if a string does not contain ...
It can be checked with this expression if the string entered by the user is invalid because of a match of 2 special characters in sequence OR 5 special characters used in the string.
Explanation:
[^...] ... a negative character class definition which matches any character NOT being one of the characters listed within the square brackets.
\w ... a word character which is either a letter, a digit or an underscore.
The next character is simply a space character.
\- ... the hyphen character which must be escaped with a backslash within square brackets as otherwise the hyphen character would be interpreted as "FROM x TO z" (except when being the first or the last character within the square brackets).
\r ... carriage return
\n ... line-feed
Therefore [^\w \-\r\n] finds a character which is NOT a letter, NOT a digit, NOT an underscore, NOT a space, NOT a hyphen, NOT a carriage return and also NOT a line-feed.
{2} ... the preceding expression must match 2 such characters.
So with the expression [^\w \-\r\n]{2} it can be checked if the string contains 2 special characters in a sequence which makes the string invalid.
| ... OR
(?:...) ... none marking group needed here for applying the expression inside with the multiplier {5} at least 5 times.
[...] ... a positive character class definition which matches any character being one of the characters listed within the square brackets.
[\w \-]+ ... find a word character, or a space, or a hyphen 1 or more times.
[^\w \-\r\n] ... and next character being NOT a word character, space, hyphen, carriage return or line-feed.
Therefore (?:[\w \-]+[^\w \-\r\n]){5} finds a string with 5 "special" characters between "standard" characters.