Regex for allowing semi colon - c#

I have a regex for validating a string but it doesn't accept semicolons? Is it because I have to use some escape sequences? I tested my regex here and it passes i.e allows semi-colon but doesn't allow in my c# app.
EDITED I have following regex
^[A-Za-z0-9]{1}[A-Za-z.&0-9\s\\-]{0,21}$
And tried validating sar232 trading inc;

The & entity hints at the fact you have this regular expression inside some XML attribute, and that this & gets parsed as a single & symbol when the pattern is sent to the regex engine.
That means, your pattern lacks the semi-colon inside the second character class, and that is why your regex does not match the string you provided.
The solution is simple: add the semi-colon to the 2nd character class:
someattr="^[A-Za-z0-9][;A-Za-z.&0-9\s\\-]{0,21}$"
^
See the regex demo
Please also note that the {1} limiting quantifier is redundant since a [A-Za-z0-9] already matches only 1 symbol from the indicated ranges.

Related

RegEx expression works on regex101 but not in C# [duplicate]

https://regex101.com/r/sB9wW6/1
(?:(?<=\s)|^)#(\S+) <-- the problem in positive lookbehind
Working like this on prod: (?:\s|^)#(\S+), but I need a correct start index (without space).
Here is in JS:
var regex = new RegExp(/(?:(?<=\s)|^)#(\S+)/g);
Error parsing regular expression: Invalid regular expression:
/(?:(?<=\s)|^)#(\S+)/
What am I doing wrong?
UPDATE
Ok, no lookbehind in JS :(
But anyways, I need a regex to get the proper start and end index of my match. Without leading space.
Make sure you always select the right regex engine at regex101.com. See an issue that occurred due to using a JS-only compatible regex with [^] construct in Python.
JS regex - at the time of answering this question - did not support lookbehinds. Now, it becomes more and more adopted after its introduction in ECMAScript 2018. You do not really need it here since you can use capturing groups:
var re = /(?:\s|^)#(\S+)/g;
var str = 's #vln1\n#vln2\n';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
The (?:\s|^)#(\S+) matches a whitespace or the start of string with (?:\s|^), then matches #, and then matches and captures into Group 1 one or more non-whitespace chars with (\S+).
To get the start/end indices, use
var re = /(\s|^)#\S+/g;
var str = 's #vln1\n#vln2\n';
var pos = [];
while ((m = re.exec(str)) !== null) {
pos.push([m.index+m[1].length, m.index+m[0].length]);
}
console.log(pos);
BONUS
My regex works at regex101.com, but not in...
First of all, have you checked the Code Generator link in the Tools pane on the left?
All languages - "Literal string" vs. "String literal" alert - Make sure you test against the same text used in code, literal string, at the regex tester. A common scenario is copy/pasting a string literal value directly into the test string field, with all string escape sequences like \n (line feed char), \r (carriage return), \t (tab char). See Regex_search c++, for example. Mind that they must be replaced with their literal counterparts. So, if you have in Python text = "Text\n\n abc", you must use Text, two line breaks, abc in the regex tester text field. Text.*?abc will never match it although you might think it "works". Yes, . does not always match line break chars, see How do I match any character across multiple lines in a regular expression?
All languages - Backslash alert - Make sure you correctly use a backslash in your string literal, in most languages, in regular string literals, use double backslash, i.e. \d used at regex101.com must written as \\d. In raw string literals, use a single backslash, same as at regex101. Escaping word boundary is very important, since, in many languages (C#, Python, Java, JavaScript, Ruby, etc.), "\b" is used to define a BACKSPACE char, i.e. it is a valid string escape sequence. PHP does not support \b string escape sequence, so "/\b/" = '/\b/' there.
All languages - Default flags - Global and Multiline - Note that by default m and g flags are enabled at regex101.com. So, if you use ^ and $, they will match at the start and end of lines correspondingly. If you need the same behavior in your code check how multiline mode is implemented and either use a specific flag, or - if supported - use an inline (?m) embedded (inline) modifier. The g flag enables multiple occurrence matching, it is often implemented using specific functions/methods. Check your language reference to find the appropriate one.
line-breaks - Line endings at regex101.com are LF only, you can't test strings with CRLF endings, see regex101.com VS myserver - different results. Solutions can be different for each regex library: either use \R (PCRE, Java, Ruby) or some kind of \v (Boost, PCRE), \r?\n, (?:\r\n?|\n)/(?>\r\n?|\n) (good for .NET) or [\r\n]+ in other libraries (see answers for C#, PHP). Another issue related to the fact that you test your regex against a multiline string (not a list of standalone strings/lines) is that your patterns may consume the end of line, \n, char with negated character classes, see an issue like that. \D matched the end of line char, and in order to avoid it, [^\d\n] could be used, or other alternatives.
php - You are dealing with Unicode strings, or want shorthand character classes to match Unicode characters, too (e.g. \w+ to match Стрибижев or Stribiżew, or \s+ to match hard spaces), then you need to use u modifier, see preg_match() returns 0 although regex testers work - To match all occurrences, use preg_match_all, not preg_match with /...pattern.../g, see PHP preg_match to find multiple occurrences and "Unknown modifier 'g' in..." when using preg_match in PHP?- Your regex with inline backreference like \1 refuses to work? Are you using a double quoted string literal? Use a single-quoted one, see Backreference does not work in PHP
phplaravel - Mind you need the regex delimiters around the pattern, see https://stackoverflow.com/questions/22430529
python - Note that re.search, re.match, re.fullmatch, re.findall and re.finditer accept the regex as the first argument, and the input string as the second argument. Not re.findall("test 200 300", r"\d+"), but re.findall(r"\d+", "test 200 300"). If you test at regex101.com, please check the "Code Generator" page. - You used re.match that only searches for a match at the start of the string, use re.search: Regex works fine on Pythex, but not in Python - If the regex contains capturing group(s), re.findall returns a list of captures/capture tuples. Either use non-capturing groups, or re.finditer, or remove redundant capturing groups, see re.findall behaves weird - If you used ^ in the pattern to denote start of a line, not start of the whole string, or used $ to denote the end of a line and not a string, pass re.M or re.MULTILINE flag to re method, see Using ^ to match beginning of line in Python regex
- If you try to match some text across multiple lines, and use re.DOTALL or re.S, or [\s\S]* / [\s\S]*?, and still nothing works, check if you read the file line by line, say, with for line in file:. You must pass the whole file contents as the input to the regex method, see Getting Everything Between Two Characters Across New Lines. - Having trouble adding flags to regex and trying something like pattern = r"/abc/gi"? See How to add modifers to regex in python?
c#, .net - .NET regex does not support possessive quantifiers like ++, *+, ??, {1,10}?, see .NET regex matching digits between optional text with possessive quantifer is not working - When you match against a multiline string and use RegexOptions.Multiline option (or inline (?m) modifier) with an $ anchor in the pattern to match entire lines, and get no match in code, you need to add \r? before $, see .Net regex matching $ with the end of the string and not of line, even with multiline enabled - To get multiple matches, use Regex.Matches, not Regex.Match, see RegEx Match multiple times in string - Similar case as above: splitting a string into paragraphs, by a double line break sequence - C# / Regex Pattern works in online testing, but not at runtime - You should remove regex delimiters, i.e. #"/\d+/" must actually look like #"\d+", see Simple and tested online regex containing regex delimiters does not work in C# code - If you unnecessarily used Regex.Escape to escape all characters in a regular expression (like Regex.Escape(#"\d+\.\d+")) you need to remove Regex.Escape, see Regular Expression working in regex tester, but not in c#
dartflutter - Use raw string literal, RegExp(r"\d"), or double backslashes (RegExp("\\d")) - https://stackoverflow.com/questions/59085824
javascript - Double escape backslashes in a RegExp("\\d"): Why do regex constructors need to be double escaped?
- (Negative) lookbehinds unsupported by most browsers: Regex works on browser but not in Node.js - Strings are immutable, assign the .replace result to a var - The .replace() method does change the string in place - Retrieve all matches with str.match(/pat/g) - Regex101 and Js regex search showing different results or, with RegExp#exec, RegEx to extract all matches from string using RegExp.exec- Replace all pattern matches in string: Why does javascript replace only first instance when using replace?
javascriptangular - Double the backslashes if you define a regex with a string literal, or just use a regex literal notation, see https://stackoverflow.com/questions/56097782
java - Word boundary not working? Make sure you use double backslashes, "\\b", see Regex \b word boundary not works - Getting invalid escape sequence exception? Same thing, double backslashes - Java doesn't work with regex \s, says: invalid escape sequence - No match found is bugging you? Run Matcher.find() / Matcher.matches() - Why does my regex work on RegexPlanet and regex101 but not in my code? - .matches() requires a full string match, use .find(): Java Regex pattern that matches in any online tester but doesn't in Eclipse - Access groups using matcher.group(x): Regex not working in Java while working otherwise - Inside a character class, both [ and ] must be escaped - Using square brackets inside character class in Java regex - You should not run matcher.matches() and matcher.find() consecutively, use only if (matcher.matches()) {...} to check if the pattern matches the whole string and then act accordingly, or use if (matcher.find()) to check if there is a single match or while (matcher.find()) to find multiple matches (or Matcher#results()). See Why does my regex work on RegexPlanet and regex101 but not in my code?
scala - Your regex attempts to match several lines, but you read the file line by line (e.g. use for (line <- fSource.getLines))? Read it into a single variable (see matching new line in Scala regex, when reading from file)
kotlin - You have Regex("/^\\d+$/")? Remove the outer slashes, they are regex delimiter chars that are not part of a pattern. See Find one or more word in string using Regex in Kotlin - You expect a partial string match, but .matchEntire requires a full string match? Use .find, see Regex doesn't match in Kotlin
mongodb - Do not enclose /.../ with single/double quotation marks, see mongodb regex doesn't work
c++ - regex_match requires a full string match, use regex_search to find a partial match - Regex not working as expected with C++ regex_match - regex_search finds the first match only. Use sregex_token_iterator or sregex_iterator to get all matches: see What does std::match_results::size return? - When you read a user-defined string using std::string input; std::cin >> input;, note that cin will only get to the first whitespace, to read the whole line properly, use std::getline(std::cin, input); - C++ Regex to match '+' quantifier - "\d" does not work, you need to use "\\d" or R"(\d)" (a raw string literal) - This regex doesn't work in c++ - Make sure the regex is tested against a literal text, not a string literal, see Regex_search c++
go - Double backslashes or use a raw string literal: Regular expression doesn't work in Go - Go regex does not support lookarounds, select the right option (Go) at regex101.com before testing! Regex expression negated set not working golang
groovy - Return all matches: Regex that works on regex101 does not work in Groovy
r - Double escape backslashes in the string literal: "'\w' is an unrecognized escape" in grep - Use perl=TRUE to PCRE engine ((g)sub/(g)regexpr): Why is this regex using lookbehinds invalid in R?
oracle - Greediness of all quantifiers is set by the first quantifier in the regex, see Regex101 vs Oracle Regex (then, you need to make all the quantifiers as greedy as the first one)] - \b does not work? Oracle regex does not support word boundaries at all, use workarounds as shown in Regex matching works on regex tester but not in oracle
firebase - Double escape backslashes, make sure ^ only appears at the start of the pattern and $ is located only at the end (if any), and note you cannot use more than 9 inline backreferences: Firebase Rules Regex Birthday
firebasegoogle-cloud-firestore - In Firestore security rules, the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols, i.e. use allow create: if docId.matches("^\\d+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.

Regular expression in RegularExpressionAttribute behavior

I am using this regular expression: #"[ \]\[;\/\\\?:*""<>|+=]|^[.]|[.]$"
First part [ \]\[;\/\\\?:*""<>|+=] should match any of the characters inside the brackets.
Next part ^[.] should match if the string starts with a 'dot'
Last part [.]$ should match if the string ends with a 'dot'
This works perfectly fine if I use Regex.IsMatch() function. However if I use RegularExpressionAttribute in ASP.NET MVC, I always get invalid model. Does anyone have any clue why this behavior occurs?
Examples:
"abcdefg" should not match
".abcdefg" should match
"abc.defg" should not match
"abcdefg." should match
"abc[defg" should match
Thanks in advance!
EDIT:
The RegularExpressionAttribute Specifies that a data field value in ASP.NET Dynamic Data must match the specified regular expression..
Which means. I need the "abcdef" to match, and ".abcdefg" to not match. Basically negate the whole expression I have above.
You need to make sure the pattern matches the entire string.
In a general case, you may append/prepend the pattern with .*.
Here, you may use
.*[ \][;/\\?:*"<>|+=].*|^[.].*|.*[.]$
Or, to make it a bit more efficient (that is, to reduce backtracking in the first branch) a negated character class will perform better:
[^ \][;/\\?:*"<>|+=]*[ \][;\/\\?:*"<>|+=].*|^[.].*|.*[.]$
But it is best to put the branches matching text at the start/end of the string as first branches:
^[.].*|.*[.]$|[^ \][;/\\?:*"<>|+=]*[ \][;/\\?:*"<>|+=].*
NOTE: You do not have to escape / and ? chars inside the .NET regex since you can't use regex delimiters there.
C# declaration of the last pattern will look like
#"^[.].*|.*[.]$|[^ \][;/\\?:*""<>|+=]*[ \][;/\\?:*""<>|+=].*"
See this .NET regex demo.
RegularExpressionAttrubute:
[RegularExpression(
#"^[.].*|.*[.]$|[^ \][;/\\?:*""<>|+=]*[ \][;/\\?:*""<>|+=].*",
ErrorMessage = "Username cannot contain following characters: ] [ ; / \\ ? : * \" < > | + =")
]
Your regex is an alternation which matches 1 character out of 3 character classes, the first consisting of more than 1 characters, the second a dot at the start of the string and the third a dot at the end of the string.
It works fine because it does match one of the alternations, only not the whole string you want to match.
You could use 3 alternations where the first matches a dot followed by repeating the character class until the end of the string, the second the other way around but this time the dot is at the end of the string.
Or the third using a positive lookahead asserting that the string contains at least one of the characters [\][;\/\\?:*"<>|+=]
^\.[a-z \][;\/\\?:*"<>|+=]+$|^[a-z \][;\/\\?:*"<>|+=]+\.$|^(?=.*[\][;\/\\?:*"<>|+=])[a-z \][;\/\\?:*"<>|+=]+$
Regex demo

Use OR in Regex Expression

I have a regex to match the following:
somedomain.com/services/something
Basically I need to ensure that /services is present.
The regex I am using and which is working is:
\/services*
But I need to match /services OR /servicos. I tried the following:
(\/services|\/servicos)*
But this shows 24 matches?! https://regex101.com/r/jvB1lr/1
How to create this regex?
The (\/services|\/servicos)* matches 0+ occurrences of /services or /servicos, and that means it can match an empty string anywhere inside the input string.
You can group the alternatives like /(services|servicos) and remove the * quantifier, but for this case, it is much better to use a character class [oe] as the strings only differ in 1 char.
You want to use the following pattern:
/servic[eo]s
See the regex demo
To make sure you match a whole subpart, you may append (?:/|$) at the pattern end, /servic[eo]s(?:/|$).
In C#, you may use Regex.IsMatch with the pattern to see if there is a match in a string:
var isFound = Regex.IsMatch(s, #"/servic[eo]s(?:/|$)");
Note that you do not need to escape / in a .NET regex as it is not a special regex metacharacter.
Pattern details
/ - a /
servic[eo]s - services or servicos
(?:/|$) - / or end of string.
Well the * quantifier means zero or more, so that is the problem. Remove that and it should work fine:
(\/services|\/servicos)
Keep in mind that in your example, you have a typo in the URL so it will correctly not match anything as it stands.
Here is an example with the typo in the URL fixed, so it shows 1 match as expected.
First off you specify C# (really .Net is the library which holds regex not the language) in this post but regex101 in your example is set to PHP. That is providing you with invalid information such as needed to escape a forward slash / with \/ which is unnecessary in .Net regular expressions. The regex language is the same but there are different tools which behave differently and php is not like .Net regex.
Secondly the star * on the ( ) is saying that there may be nothing in the parenthesis and your match is getting null nothing matches on every word.
Thirdly one does not need to split the whole word. I would just extract the commonality in the words into a set [ ]. That will allow the "or-ness" you need to match on either services or servicos. Such as
(/servic[oe]s)
Will inform you if services are found or not. Nothing else is needed.

Why does .* fail to match the entire (rest of the) string in this regex?

I ran into a problem with my regular expressions, I'm using regular expressions for obtaining data from the string below:
"# DO NOT EDIT THIS MAIL BY HAND #\r\n\r\n[Feedback]:hallo\r\n\r\n# DO NOT EDIT THIS MAIL BY HAND #\r\n\r\n"
So far I got it working with:
String sFeedback = Regex.Match(Message, #"\[Feedback\]\:(?<string>.*?)\r\n\r\t\n# DO NOT EDIT THIS MAIL BY HAND #").Groups[1].Value;
This works except if the header is changed, therefore I want the regex to read from [feedback]: to the end of the string. (symbols, ascii, everything..)
I tried: \[Feedback]:(?<string>.*?)$
Above regular expression does work in some regular expression builders online but in my c# code its not working and returns a empty string. What's wrong?
The problem is that . doesn't match newlines unless you use RegexOptions.Singleline when compiling the regex or inline it using (?s):
(?s)\[Feedback\]:(.*)$
You are missing the escape character.
Also, since you are not referring to the group by name in your C# code, you could further simplify your regex to this
\[Feedback\]:(.*)$
$ in regex means:
The match must occur at the end of the string or before \n at the end of the line or string.
and . means:
Matches any single character except \n.
try to use this simple regex:
\[Feedback\]:(?<string>.*)

Excluding certain patterns in a regex

I'm working on a Regex in C# to exclude certain patterns within a string.
These are the types patterns I want to accept are: "%00" (Hex 00-FF) and any other character without a starting '%'. The patterns I would like to exclude are: "%0" (Values with a starting % and one character after) and/or characters "&<>'/".
So far I have this
Regex correctStringRegex = new Regex(#"(%[0-9a-fA-F]{2})|[^%&<>'/]|(^(%.))",
RegexOptions.IgnoreCase);
Below are examples of what I'm trying to pass and reject.
Passing String %02This is%0A%0Da string%03
Reject String %0%0Z%A&<%0a%
If a string doesn't pass all the requirements I would like to reject the whole string completely.
Any Help would be greatly appreciated!
I suggest this:
^(?:%[0-9a-f]{2}|[^%&<>'/])*$
Explanation:
^ # Start of string
(?: # Match either
%[0-9a-f]{2} # %xx
| # or
[^%&<>'/] # any character except the forbidden ones
)* # any number of times
$ # until end of string.
This ensures that % is only matched when followed by two hexadecimals. Since you're already compiling the regex with the IgnoreCase flag set, you don't need a-fA-F, either.
Hmm, given the comments so far, I think you need a different problem definition. You want to pass or fail a string, using regex, based on whether or not the string contains any invalid patterns. Im assuming a string will fail if there is ANY invalid pattern, rather than the reverse of a string passing if there is any valid pattern.
As such, I would use this regex: %(?![0-9a-f]{2})|[&<>'/]
You would then run this in such a way that a string is invalid if you GET a match, a valid string will not have any matches in this set.
A quick explanation of a rather odd regex. The format (?!) tells the regex "Match the previous symbol if the symbols in this set DONT follow it" ie: Match if suffix not present. So, what im telling it to look for is any instance of % that is not followed by 2 hex characters, or any other invalid character. The assumption is that anything that DOESN'T match this regex is a valid character entry.

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