I'm trying to write a regular expression that will validate that user input is greater than X number of non-whitespace characters. I'm basically trying to filter out begining and ending whitespace while still ensuring that the input is greater than X characters; the characters can be anything, just not whitespace (space, tab, return, newline).
This the regex I've been using, but it doesn't work:
\s.{10}.*\s
I'm using C# 4.0 (Asp.net Regular Expression Validator) btw if that matters.
It may be easier to not use regex at all:
input.Where(c => !char.IsWhiteSpace(c)).Count() > 10
If whitespace shouldn't count in the middle, then this will work:
(\s*(\S)\s*){10,}
If you don't care about whitespace in between non-whitespace characters, the other answers have that scenario covered.
This regular expression looks for eight or more characters between the first and last non-whitespace characters, ignoring leading and trailing whitespace:
\s*\S.{8,}\S\s*
If your trying to check (like in my case a phone number that contains 8 digits) , you need to refer to a number below the one you need.
(\s*(\S)\s*){7,}
Related
I want to match strings that do not contain more than 3 of the same character repeated in a row. So:
abaaaa [no match]
abawdasd [match]
abbbbasda [no match]
bbabbabba [match]
Yes, it would be much easier and neater to do a regex match for containing the consecutive characters, and then negate that in the code afterwards. However, in this case that is not possible.
I would like to open out the question to x consecutive characters so that it can be extended to the general case to make the question and answer more useful.
Negative lookahead is supported in this case.
Use a negative lookahead with back references:
^(?:(.)(?!\1\1))*$
See live demo using your examples.
(.) captures each character in group 1 and the negative look ahead asserts that the next 2 chars are not repeats of the captured character.
To match strings not containing a character repeated more than 3 times consecutively:
^((.)\2?(?!\2\2))+$
How it works:
^ Start of string
(
(.) Match any character (not a new line) and store it for back reference.
\2? Optionally match one more exact copies of that character.
(?! Make sure the upcoming character(s) is/are not the same character.
\2\2 Repeat '\2' for as many times as you need
)
)+ Do ad nauseam
$ End of string
So, the number of /2 in your whole expression will be the number of times you allow a character to be repeated consecutively, any more and you won't get a match.
E.g.
^((.)\2?(?!\2\2\2))+$ will match all strings that don't repeat a character more than 4 times in a row.
^((.)\2?(?!\2\2\2\2))+$ will match all strings that don't repeat a character more than 5 times in a row.
Please be aware this solution uses negative lookahead, but not all not all regex flavors support it.
I'm answering this question :
Is there a regular expression for matching a string that has no more than 2 repeating characters?
which was marked as an exact duplicate of this question.
Its much quicker to negate the match instead
if (!Regex.Match("hello world", #"(.)\1{2}").Success) Console.WriteLine("No dups");
I am currently working on a regex which needs to match exactly 8 digits. But sometimes it occurs that there are spaces or dots between those numbers. This is the regex that i am currently using.
([0-9\ ?.?]{7,16})
It works fine most of the time, but the problem I am having is that it sometimes matches number with a lot of spaces tailing it so you will get something like 1234/s/s/s/s (/s stands for space) Or sometimes it is only matching spaces.
What i want is a regex that always matches at least 8 digits and also allows spaces and dots without detecting less then 8 digits. I know it may be stupid question, but I couldn't find anything I need elswhere.
Your ([0-9\ ?.?]{7,16}) expression matches 7 to 16 occurrences of any character that is either a digit, or a space, or a ?, or .. Yes, the ? inside [...] is a literal ?, not a quantifier.
You need to use an expression that will match a digit ([0-9]) and then exactly 7 sequences of a space or period ([ .]) followed with 1 digit, and to make sure you are not matching the digits in 123.156.78.146 you may use special boundaries:
(?<!\d[ .]?)\d(?:[. ]?\d){7}(?![ .]?\d)
if the space or . can only be 0 to 1 in between digits; or - if the space/dot can appear 0 or more times,
(?<!\d[ .]*)\d(?:[. ]*\d){7}(?![ .]*\d)
See the regex demo
The (?<!\d[ .]*) is a negative lookbehind that will fail any match if it starts with a digit that is followed with .(s) or space(s), and the (?![ .]*\d) negative lookahead will fail the match if the 7 digits you need are followed with .(s) or space(s) and a digit.
To solve this, describe the problem to yourself. You want to match one digit followed by seven repetitions of space-or-dot followed by a digit. This leads to a regular expression like \d([ .]?\d){7}. To avoid collecting the seven captures add a :? after the (. To capture the whole string, enclose it in brackets. Adding both changes gives the expression (\d(:?[ .]?\d){7}). If more than one space or dot is allowed between the digits then change the ? to a *.
To get just the eight digits out of the string I suggest using the string captured above and replacing any spaces or dots with nothing.
I want regex which will allow following format
1234567-8
123456B
Now here if user enter second pattern then he should be lock to enter maximum 7 characters so
1234568B
123456V1
this becomes invalid
I have tried
[0-9]{7}-[0-9]|[[0-9]{6}[A-z]{1}]{7,7}
but this fails
For the sample input you provided, you may use ^([0-9]{7}-[0-9]|[0-9]{6}[A-Za-z])$.
A bit more contracted version: ^[0-9]{6}(?:[0-9]-[0-9]|[A-Za-z])$.
Note that 1234567-8 has 7 digits and a hyphen followed with a digit, so the whole string length cannot be limited to just 7 characters all in all.
In .NET and almost all other regex flavors [A-z] is a mistake, as it can match more than just letters.
Placing a quantifier {1} into a character class makes it a simple symbol combination, so [{1}] matches either { or 1 or }.
The {7,7} (={7}) will not limit the whole string length to 7, as you do not have anchors (^ and $) around the expression and you "ruined" the preceding quantifiers by putting them into a character class.
I think this is what you need:
^(\d{7}-\d|\d{6}[A-Z])$
7 digits, dash, digit OR 6 digits, 1 large latin letter.
^\d{6}(?:\d-\d|[A-Z])$
It can satisfy well with 2 your above formats
1234567-8
123456B
I have to check whether the user entered string is in a particular format as below eg:
123-1234-1234567-1
ie after first 3 digit a hyphen, then after 4 digit another hyphen, after seven digit an hyphen then a single digit.
I used the below regular expression
#"^\(?([0-9]{3})\)?[-. ]?([0-9]{4})[-. ]?([0-9]{7})[-. ]?([0-9]{1})$"
It is working fine for above expression but it will also pass the expression without - also
eg:- 123-1234-1234567-1 //pass
123123412345671 //also getting pass.
The second string should fail. What change i should do in the regular expression to achieve the same?
You can simply use:
^\d{3}-\d{4}-\d{7}-\d$
If you want to allow dot and space also as delimiter then use:
^\d{3}[-. ]\d{4}[-. ]\d{7}[-. ]\d$
The problems is that you are having optional quantifier ? after [. ].
Remove them, and it should work fine
#"^\(?([0-9]{3})\)?[-. ]([0-9]{4})[-. ]([0-9]{7})[-. ]([0-9]{1})$"
Regex demo
The ? makes the preceding pattern optional as it matches 0 or 1 character. So in the second example the regex engine safely matches zero - to get to match the entire string
I need Regexp to validate string has minimum length 6 and it is contains at least one non-alphanumeric character e.g: "eN%{S$u)", "h9YI!>4j", "{9YI!;4j", "eN%{S$usdf)", "dfh9YI!>4j", "ghffg{9YI!;4j".
This one is working well ^.*(?=.{6,})(?=.*\\d).*$" but in cases when string does not contain any numbers(e.g "eN%{S$u)") it is not working.
^(?=.{6})(.*[^0-9a-zA-Z].*)$
We use positive lookahead to assure there are at least 6 characters. Then we match the pattern that looks for at least one non-alphanumeric character ([^0-9a-zA-Z]). The .*'s match any number of any characters around this one non-alphanumeric character, but by the time we've reached here we've already checked that we're matching at least 6.
^.*(?=.{6,})(?=.*\\d).*$"
is the regex you tried. Here are some suggestions:
You don't need to match more than 6 characters in the lookahead. Matching only 6 here does no restrict the rest of the regular expression from matching more than 6.
\d matches a digit, and (?=.*\\d) is a lookahead for one of them. This is why you are experiencing the problems you mentioned with strings like eN%{S$u).
Even if the point above wasn't incorrect and the regular expression here was correct, you can combine the second lookahead with the .* that follows by just using .*\\d.*.
marcog's answer is pretty good, but I'd do it the other way around so that it's easier to add even more conditions (such as having at least one digit or whatever), and I'd use lazy quantifiers because they are cheaper for certain patterns:
^(?=.*?[^0-9a-zA-Z]).{6}
So if you were to add the mentioned additional condition, it would be like this:
^(?=.*?[^0-9a-zA-Z])(?=.*?[0-9]).{6}
As you can see, this pattern is easily extensible. Note that is is designed to be used for checking matches only, its capture is not useful.
Keep it easy.
// long enough and contains something not digit or a-z
x.Length >= 6 && Regex.IsMatch(x, #"[^\da-zA-Z]")
Happy coding.
Edit, pure "regular expression":
This first asserts there are 6 letters of anything in the look-ahead, and then ensures that within the look-ahead there is something that is not alpha-numeric (it will "throw away" up to the first 5 characters trying to match).
(?=.{6}).{0,5}[^\da-zA-Z]
What about that(fixed): ^(?=.{6})(.*[^\w].*)$
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