Hi guys I need a regex that only extracts punctuations/characters.
I have this so far :
[._^%$#!~#,-]+
but this works if there is at least 1 punctuations and still allows for any other char (digit or letter)
I need to to only allow punctuations/characters
Try this:
^[\p{S}\p{P}]+$
\p{S} matches any symbol character, and \p{P} matches any punctuation.
Note that your pattern will not match all the symbols and punctuations not present in the list.
Try anchoring the regex to the start and the end of the string (unless you're using Multiline matching) - i.e. ^ at the beginning and $ at the end:
^[._^%$#!~#,-]+$
Note - this does not endorse your actual pattern (I can't say whether this is matching all the 'special characters' you're talking about, but it will make it so that the entire string must be all 'special'.
[^a-zA-Z0-9]* u can try something like this. Should NOT accept those chars, cba to writte all beside chars beside one u typed.
\W
Matches any character that is not a word character (alphanumeric & underscore).
Related
I'm looking for a regular expression to extract a string from a file name
eg if filename format is "anythingatallanylength_123_TESTNAME.docx", I'm interested in extracting "TESTNAME" ... probably fixed length of 8. (btw, 123 can be any three digit number)
I think I can use regex match ...
".*_[0-9][0-9][0-9]_[A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z][A-Z].docx$"
However this matches the whole thing. How can I just get "TESTNAME"?
Thanks
Use parenthesis to match a specific piece of the whole regex.
You can also use the curly braces to specify counts of matching characters, and \d for [0-9].
In C#:
var myRegex = new Regex(#"*._\d{3}_([A-Za-z]{8})\.docx$");
Now "TESTNAME" or whatever your 8 letter piece is will be found in the captures collection of your regex after using it.
Also note, there will be a performance overhead for look-ahead and look-behind, as presented in some other solutions.
You can use a look-behind and a look-ahead to check parts without matching them:
(?<=_[0-9]{3}_)[A-Z]{8}(?=\.docx$)
Note that this is case-sensitive, you may want to use other character classes and/or quantifiers to fit your exact pattern.
In your file name format "anythingatallanylength_123_TESTNAME.docx", the pattern you are trying to match is a string before .docx and the underscore _. Keeping the thing in mind that any _ before doesn't get matched I came up with following solution.
Regex: (?<=_)[A-Za-z]*(?=\.docx$)
Flags used:
g global search
m multi-line search.
Explanation:
(?<=_) checks if there is an underscore before the file name.
(?=\.docx$) checks for extension at the end.
[A-Za-z]* checks the required match.
Regex101 Demo
Thanks to #Lucero #noob #JamesFaix I came up with ...
#"(?<=.*[0-9]{3})[A-Z]{8}(?=.docx$)"
So a look behind (in brackets, starting with ?<=) for anything (ie zero or more any char (denoted by "." ) followed by an underscore, followed by thee numerics, followed by underscore. Thats the end of the look behind. Now to match what I need (eight letters). Finally, the look ahead (in brackets, starting with ?=), which is the .docx
Nice work, fellas. Thunderbirds are go.
I am having issue with a reg ex expression and can't find the answer to my question.
I am trying to build a reg ex pattern that will pull in any matches that have # around them. for example #match# or #mt# would both come back.
This works fine for that. #.*?#
However I don't want matches on ## to show up. Basically if there is nothing between the pound signs don't match.
Hope this makes sense.
Thanks.
Please use + to match 1 or more symbols:
#+.+#+
UPDATE:
If you want to only match substrings that are enclosed with single hash symbols, use:
(?<!#)#(?!#)[^#]+#(?!#)
See regex demo
Explanation:
(?<!#)#(?!#) - a # symbol that is not preceded with a # (due to the negative lookbehind (?<!#)) and not followed by a # (due to the negative lookahead (?!#))
[^#]+ - one or more symbols other than # (due to the negated character class [^#])
#(?!#) - a # symbol not followed with another # symbol.
Instead of using * to match between zero and unlimited characters, replace it with +, which will only match if there is at least one character between the #'s. The edited regex should look like this: #.+?#. Hope this helps!
Edit
Sorry for the incorrect regex, I had not expected multiple hash signs. This should work for your sentence: #+.+?#+
Edit 2
I am pretty sure I got it. Try this: (?<!#)#[^#].*?#. It might not work as expected with triple hashes though.
Try:
[^#]?#.+#[^#]?
The [^ character_group] construction matches any single character not included in the character group. Using the ? after it will let you match at the beginning/end of a string (since it matches the preceeding character zero or more times. Check out the documentation here
I have a username validator IsValidUsername, and I am testing "baconman" but it is failing, could someone please help me out with this regex?
if(!Regex.IsMatch(str, #"^[a-zA-Z]\\w+|[0-9][0-9_]*[a-zA-Z]+\\w*$")) {
isValid = false;
}
I want the restrictions to be: (It's very close)
Be between 5 & 17 characters long
contain at least one letter
no spaces
no special characters
You're escaping unnecessarily: if you write your regex as starting with # outside the string, you don't need both \ - just one is fine.
Either:
#"\w"
or
"\\w"
Edit: I didn't make this clear: right now due to the double escaping, you're looking for a \ in your regex and a w. So your match would need [some character]\w to match (example: "a\w" or "a\wwwwww" would match.
Your requirements are best taken care of in normal C#. They don't map well to a regular expression. Just code them up using LINQ which works on strings like it would on an IEnumerable<char>.
Also, understanding a query of a string is much easier than understanding a Regex with the requirements that you have.
It is possible to do everything as part of a Regex, however it is not pretty :-)
^(\w(?=\w*[a-zA-Z])|[a-zA-Z]|\w(?<=[a-zA-Z]\w*)){5,17}$
It does 3 checks that always results in 1 character being matched (so we can perform the length check in the end)
Either the character is any word character \w which is before [a-zA-Z]
Or it is [a-zA-Z]
Or it is any word character \w which is after [a-zA-Z]
I have this code that replaces all non alphanumeric characters with "-" char.
return Regex.Replace(strIn, #"[\W|_]+", "-", RegexOptions.IgnorePatternWhitespace | RegexOptions.IgnoreCase);
but I need to change it to allow pass some special characters (one or more) for example: #,*,%
how to change this regular expression?
Use
[^\p{L}\p{N}#*%]+
This matches one or more characters that are neither letters nor digits nor any of #, * or %.
Another option, you can use charcter class subtractioninfo, for example to remove # from the character class:
[\W_-[#]]+
Just add other accepted special chars after the #. Live example here: http://rextester.com/rundotnet?code=YFQ40277
How about this one:
[^a-zA-Z0-9#*%]+
If you are using unicode you can do (as Tim's answer):
[^\p{L}\p{N}#*%]+
Use this.
([^\w#*%]|_)
Add any other special characters after the %.
It is basically saying, match any character that is not (^) a word character(\w), #, * or % OR match _.
It seems this way is the best solution for you
#"(?!.*[^\w#*%])"
You can use set subtraction for that:
#"[\W_-[#*%]]+"
This matches the set of all non-word characters and the underscore, minus the set of #, * and %.
Note that you don't have to use | for "or" in a character class, since that's implied. In fact, the | in your regex just matches |.
Note also that in .NET, \w matches a few other "connector punctuation" characters besides the underscore. If you want to match the other characters too, you can use
#"[\W\p{Pc}-[#*%]]+"
I want to check in a C# program, if a user input is a single word. The word my only have characters A-Z and a-z. No spaces or other characters.
I try [A-Za-z]* , but this doesn't work. What is wrong with this expression?
Regex regex = new Regex("[A-Za-z]*");
if (!regex.IsMatch(userinput);)
{
...
}
Can you recomend website with a comprensiv list of regex examples?!
It probably works, but you aren't anchoring the regular expression. You need to use ^ and $ to anchor the expression to the beginning and end of the string, respectively:
Regex regex = new Regex("^[A-Za-z]+$");
I've also changed * to + because * will match 0 or more times while + will match 1 or more times.
You should add anchors for start and end of string: ^[A-Za-z]+$
Regarding the question of regex examples have a look at http://regexlib.com/.
For the regex, have a look at the special characters ^ and $, which represent starting and ending of string. This site can come in handy when constructing regexes in the future.
The asterisk character in regex specifies "zero or more of the preceding character class".
This explains why your expression is failing, because it will succeed if the string contains zero or more letters.
What you probably intended was to have one or more letters, in which case you should use the plus sign instead of the asterisk.
Having made that change, now it will fail if you enter a string that doesn't contain any letters, as you intended.
However, this still won't work for you entirely, because it will allow other characters in the string. If you want to restrict it to only letters, and nothing else, then you need to provide the start and end anchors (^ and $) in your regex to make the expression check that the 'one or more letters' is attached to the start and end of the string.
^[a-zA-Z]+$
This should work as intended.
Hope that helps.
For more information on regex, I recommend http://www.regular-expressions.info/reference.html as a good reference site.
I don't know what the C#'s regex syntax is, but try [A-Za-z]+.
Try ^[A-Za-z]+$ If you don't include the ^$ it will match on any part of the string that has a alpha characters in it.
I know the question is only about strictly alphabetic input, but here's an interesting way of solving this which does not break on accented letters and other such special characters.
The regex "^\b.+?\b" will match the first word on the start of a string, but only if the string actually starts with a valid word character. Using that, you can simply check if A) the string matches, and B) the length of the matched string equals your full string's length:
public Boolean IsSingleWord(String userInput)
{
Regex firstWordRegex = new Regex("^\\b.+?\\b");
Match firstWordMatch = firstWordRegex.Match(userInput);
return firstWordMatch.Success && firstWordMatch.Length == userInput.Length;
}
The other persons have wrote how to resolve the problem you know. Now I'll speak about the problem you perhaps don't know: diacritics :-) Your solution doesn't support àèéìòù and many other letters. A correct solution would be:
^(\p{L}\p{M}*)+$
where \p{L} is any letter plus \p{M}* that is 0 or more diacritic marks (in unicode diacritics can be "separated" from base letters, so you can have something like a + ` = à or you can have precomposed characters like the standard à)
if you just need the characters a-zA-Z you could simply iterate over the characters and compare the single characters if they are inside your range
for example:
for each character c: ('a' <= c && c <= 'z') || ('A' <= c && c <= 'Z')
This could increase your performance