I was just working on some validation and was stuck up on this though :( I want a text which contains only [a-z][A-Z][0-9][_] .
It should accept any of the above characters any number of times in any order. All other characters marks the text as invalid.
I tried this but it is not working !!
{
......
Regex strPattern = new Regex("[0-9]*[A-Z]*[a-z]*[_]*");
if (!strPattern.IsMatch(val))
{
return false;
}
return true
}
You want this:
Regex strPattern = new Regex("^[0-9A-Za-z_]*$");
Your expression does not work because:
It will accept any number of digits, followed by any number of uppercase letters, followed by any number of lowercase letters, followed by any number of underscores. For example, an underscore followed by a number would not match.
Your pattern is not anchored using the ^ and $ characters. This means that every string will match, because every string contains zero or more of the specified characters. (For example, the string "!##$" contains zero numbers, etc.!) Anchoring the expression to the start and end of the string means that the entire string much match the entire expression or the match will fail.
This pattern will still accept a zero-length string as valid. If you would like to enforce that the string be at least one character, change the * near the end of the expression to +. (* means "0 or more of the previous token" while + means "1 or more of the previous token.")
Try this:
new Regex("[0-9A-Za-z_]*");
Related
How can I check that the string is in correct format. I want the string to compare and pass only if matches exactly. Following are the correct formats :
0.#
0.##
0.###
0.####
0.#####
The hash (#) after the dot (.) can be upto 10 characters but it should only have 0.# nothing else is allowed.
Can someone please guide me how can I validate a string of this type ?
Im Regular Expression the carret (^) represent start-of-line and the ($) represents end-of-line (or before newline).
A regex with an exact match is just what you want enclosed by ^ and $. But you must ensure that special regular expression characters are quoted. For example the regex
^Hello World$
would match exactly on the String "Hello World" and nothing else.
You also can use numbers directly. You need to escape the dot "." as a dot in a regular expression means any character except newline. You escape a character by adding a backslash.
Next you should know about quantifiers. The usually ones are
-> 0 or many
-> 1 or many
{n} -> exactly n times
{n,} -> at least n times
{n,m} -> n to m times
So you can write:
^0\.#{1,10}$
If you use a normal string in C# with quotations (") you must use two backslashes
^0\\.#{1,10}$
I have a regex check:
Match matchLeft = Regex.Match(Name.Substring(subName.Length), #"\d*");
This basically checks for the first digits at the end of the subName. Now, I have noticed that with the use of * in the regex (* = 0 or more), if the next characters are not digits, it will return nothing. If they are however, it will return the string of digits.
But
If I use #"\d+" instead, it will look for 1 or more digits, and return the first instance of digits, regardless of there position after the substring.
So if I had a string ("abcdef123") and a substring ("abc"):
#"\d*" would match null
#"\d+" would match "123"
Alternatively, if the substring was "abcdef", both would match "123".
So my question is - why does the use of * return nothing if the directly following characters are not digits? Will this occur every time?
When you get the substring you end up with def123. The following are true:
\d+ tries to get at least one match in the string and will greedily match more. It must traverse the string to find the first match, arriving at the 123.
On the other hand, \d* will start at the beginning of the string and will successfully match the start of the string with zero digits. Even though it is greedy, it is completely satisfied with matching zero digits. It is a successful match and is zero-width.
You can change this behavior by making it \d*$ to anchor at the end of the matched string.
I think you answered your question yourself. This behavior is default and will occur every time.
See Quantifier Cheat Sheet
A+ One or more As, as many as possible (greedy), giving up characters if the engine needs to backtrack (docile)
A* Zero or more As, as many as possible (greedy), giving up characters if the engine needs to backtrack (docile)
Since \d* can match an empty string it will match an empty string as regex engine always tries to return a valid match, and can even match empty substrings at the beginning, end and between characters in a string.
I need a regular expression for my password format. It must ensure that password only contains letters a-z, digits 0-9 and special characters: .##$%&.
I am using .NET C# programming language.
This is my code:
Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]$");
if (!userAndPassPattern.IsMatch(username) || !userAndPassPattern.IsMatch(password))
return false;
The problem is that I always get back false.
!A || !B is logically equivalent to !(A && B)
So you could write better
!(userAndPassPattern.IsMatch(username) && userAndPassPattern.IsMatch(password))
Then you have a special character $ in you character class, maybe you need to mask it \$
I'm not quite sure about this, because in a character class it is not a special character. Maybe it depends on the RegEx engine in use. If you mask the $ it should do no harm ([a-z0-9.##\$%&])
Then you have just a single character to match. You need a quantifier
[a-z0-9.##$%&] means one single character out of the given, will match aor b or 0 but not ab
[a-z0-9.##$%&]+ many characters out of the given, from 1 to endless appearances, will match a, b, and ab and ba etc.
edit
This is what you want
Regex userAndPassPattern = new Regex("^[a-z0-9\.##\$%&]+$");
if (!(userAndPassPattern.IsMatch(username) && userAndPassPattern.IsMatch(password))) {
return false;
}
You forgot to add the '+' for matching one or more times:
Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]+$");
This code Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]$"); will only match a username or password that is a single character long.
You are looking for something like this Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]+$"); which will match one or more of the characters in your class. The + symbol tells it to match one or more of the previous atom (which in this case is the character class you specified in the square brackets)
Also, if you did not mean to constrain the match to lowercase characters, you should add 'A-Z' to the character class Regex userAndPassPattern = new Regex("^[A-Za-z0-9.##$%&]$");
You might also want to implement a minimum length restriction which can be accomplished by replacing the + with the {n,} construct, where n is the minimum length you want to match. For example:
this would match a minimum of 6 characters
Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]{6,}$");
this would match a minimum of 6 and a maximum of 12
Regex userAndPassPattern = new Regex("^[a-z0-9.##$%&]{6,12}$");
You have two problems. First . and $ need to be escaped. Second you are matching only 1 character. Add a + before the last $:
^[a-z0-9\.##\$%&]+$
Edit: Another suggestion, if you have a minimum/maximum length you can replace the + with, for example, {6,16} or whatever you think is appropriate. This will match strings that are 6 to 16 character inclusive and reject any shorter or longer strings. If you don't care about an upper limit, you could use {6,}.
Have you tried using a verbatim string literal when you're using regex escape sequences?
Regex userAndPassPattern = new Regex(#"^[a-z0-9##\$%&]+$");
if (!userAndPassPattern.IsMatch(username) || !userAndPassPattern.IsMatch(password))
return false;
Your pattern only allows a single character set you probably want a repetition operator like * + or {10,}.
Your character set includes . which matches any character, defeating the object of the character class. If you wanted to match "." then you need to escape it with \.
I want to check whether there is string starting from number and then optional character with the help of the regex.So what should be the regex for matching the string which must be started with number and then character might be there or not.Like there is string "30a" or "30" it should be matched.But if there is "a" or some else character or sereis of characters, string should not be matched.
Sounds like there should be able to be any number of numeric characters at the beginning followed by optional other characters. To match any other character after a series of numbers at the beginning I would use:
\d+.*
To match only alpha numeric characters after the mandatory numeric beginning I would use:
\d+\w*
Note: as pointed out by Dav, if you add a ^ to the start of the expression and a $ to the end of the expression like this ^\d+\w*$ you will ensure the whole string matches. However if you leave those off, you will be able to search the input string for what you need. It just depends on what your needs are.
^\d.*
The ^ matches the start of the string, \d matches a single digit, and then the .* matches any number of additional characters.
Thus, the net result is that it will only match if the string begins with a digit.
NET. I have created a regex validator to check for special characters means I donot want any special characters in username. The following is the code
Regex objAlphaPattern = new Regex(#"[a-zA-Z0-9_#.-]");
bool sts = objAlphaPattern.IsMatch(username);
If I provide username as $%^&asghf then the validator gives as invalid data format which is the result I want but If I provide a data s23_#.-^&()%^$# then my validator should block the data but my validator allows the data which is wrong
So how to not allow any special characters except a-z A-A 0-9 _ # .-
Thanks
Sunil Kumar Sahoo
There's a few things wrong with your expression. First you don't have the start string character ^ and end string character $ at the beginning and end of your expression meaning that it only has to find a match somewhere within your string.
Second, you're only looking for one character currently. To force a match of all the characters you'll need to use * Here's what it should be:
Regex objAlphaPattern = new Regex(#"^[a-zA-Z0-9_#.-]*$");
bool sts = objAlphaPattern.IsMatch(username);
Your pattern checks only if the given string contains any "non-special" character; it does not exclude the unwanted characters. You want to change two things; make it check that the whole string contains only allowed characters, and also make it check for more than one character:
^[a-zA-Z0-9_#.-]+$
Added ^ before the pattern to make it start matching at the beginning of the string. Also added +$ after, + to ensure that there is at least one character in the string, and $ to make sure that the string is matched to the end.
Change your regex to ^[a-zA-Z0-9_#.-]+$. Here ^ denotes the beginning of a string, $ is the end of the string.