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+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
I have two regex that should be matched:
"^[a-z0-9\\!#\\$\\^&\\-\\+%\\=_\\(\\)\\{\\}\\<\\>'\";\\:/\\.,~`\\|\\\\]+$"
and
".*(g[o0]+gle).*"
The first one accept any alpha numeric character (with few more extras). Like helloworld123. The second one should reject any string that contain the word "google" (in diffrent forms - like: gooo0gle).
Allowed:
hello
helloworld
helloworld123
Disallowed:
hellogoogle
google
...
I want to use the RegularExpression to match this string. Thought about something like:
[RegularExpression("^[a-z0-9\\!#\\$\\^&\\-\\+%\\=_\\(\\)\\{\\}\\<\\>'\";\\:/\\.,~`\\|\\\\]+$|.*(g[o0]+gle).*"]
But it's not working since the second part (.*(g[o0]+gle).*) should be NOT.
How to do it right?
Thanks.
You can use your second regex by placing it in a negative look ahead and use the first regex as character set and combine both to get following regex that you can use,
^(?!.*g[o0]+gle)[-a-z0-9!#$^&+%=_(){}<>'";:\/.,~`|]+$
Here, this (?!.*g[o0]+gle) negative look ahead will reject any strings that contains google or any variation as supported by your regex, and this character set [-a-z0-9!#$^&+%=_(){}<>'";:\/.,~|]+` will match one or more characters allowed by it.
Also, you don't need to escape most special characters while they are in character set, hence I have unescaped most of them except / and also always place the hyphen - either as the very first character or very last character in the character set, else depending upon the regex dialects, you may see weird behavior.
Regex Demo
I am new to both .NET (C#) and regular expressions.
I need a regular expression to match against a url:
If url string contains "/id/Whatever_COMES_HERE_EVERY_CHAR_ACCEPTED/" : return true
If url string contains only "/id/" : return false
I have tried the following but it only returns true if url is http:// localhost/id/
This is my script:
string thisUrl = HttpContext.Current.Request.Url.AbsolutePath;
Match match = Regex.Match(thisUrl, #"/id/*$");
What am i doing wrong?
You have this:
/id/*$
What this is doing is matching the literal string /id/ and then you have the quantifier * which means 0 or more times. Then you have $ which means end of the string.
You are looking for repetition of the literal / which is not what you want. (So this: http:// localhost/id/////////////////// should have matched too with your original regex)
What you need is something like this:
/id/.+$
This will match the literal /id/ followed by the . which in regex means any character which is quantified with the + which means 1 or more.
You could tighten it up and use \S instead of . which means non-whitespace characters (since a URL shouldn't have whitespace)
Also note: there are a variety of online regex tools which are really useful when trying to figure out and test a regex. A couple of examples:
http://rubular.com/
http://regex101.com/
http://www.regxlib.com/
And even extension for visual studio you can use:
https://visualstudiogallery.msdn.microsoft.com/bf883ae3-188b-43bc-bd29-6235c4195d1f
When you use the start it signals that 0 or more of the preceding char shall be present. You will want to use
"/id/.+" to signal that at least one more char must come after the /
If you're just looking for true/false solution, you should use IsMatch() function. The other issue is that * (zero or more) and + (one or more) are quantifiers and must be preceeded by a character class or group. Dot (.) is a character class that represents ANY character. So the correct solution for your problem would be:
Regex.IsMatch(thisUrl, #"/id/.+$");
Considering that the input is a URL, this regex can be improved upon by restricting character classes to valid URL characters only, but for your purpose the above should be sufficient.
I have a username validator IsValidUsername, and I am testing "baconman" but it is failing, could someone please help me out with this regex?
if(!Regex.IsMatch(str, #"^[a-zA-Z]\\w+|[0-9][0-9_]*[a-zA-Z]+\\w*$")) {
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'm having a hard time finding a solution to this and am pretty sure that regex supports it. i just can't recall the name of the concept in the world of regex.
i need to search and replace a string for a specific pattern but the patterns can be different and the replacement needs to "remember" what it's replacing.
For example, say i have an arbitrary string: 134kshflskj9809hkj
and i want to surround the numbers with parentheses,
so the result would be: (134)kshflskj(9809)hkj
Finding numbers is simple enough, but how to surround them?
Can anyone provide a sample or point me in the right direction?
In some various langauges:
// C#:
string result = Regex.Replace(input, #"(\d+)", "($1)");
// JavaScript:
thestring.replace(/(\d+)/g, '($1)');
// Perl:
s/(\d+)/($1)/g;
// PHP:
$result = preg_replace("/(\d+)/", '($1)', $input);
The parentheses around (\d+) make it a "group" specifically the first (and only in this case) group which can be backreferenced in the replacement string. The g flag is required in some implementations to make it match multiple times in a single string). The replacement string is fairly similar although some languages will use \1 instead of $1 and some will allow both.
Most regex replacement functions allow you to reference capture groups specified in the regex (a.k.a. backreferences), when defining your replacement string. For instance, using preg_replace() from PHP:
$var = "134kshflskj9809hkj";
$result = preg_replace('/(\d+)/', '(\1)', $var);
// $result now equals "(134)kshflskj(9809)hkj"
where \1 means "the first capture group in the regex".
Another somewhat generic solution is this:
search : /([\d]+)([^\d]*)/g
replace: ($1)$2
([\d]+): match a set of one or more digits and retain them in a group
([^\d]*): match a set of non-digits, and retain them as well. \D could work here, too.
g: indicate this is a global expression, to work multiple times on the input.
($1): in the replace block, parens have no special meaning, so output the first group, surrounding it with parens.
$2: output the second group
I used a pretty good online regex tool to test out my expression. The next step would be to apply it to the language that you are using, as each has its own implemention nuance.
Backreferences (grouping) are not necessary if you're just looking to search for numbers and replace with the found regex surrounded by parens. It is simpler to use the whole regex match in the replacement string.
e.g for perl
$text =~ s/\d+/($&)/g;
This searches for 1 or more digits and replaces with parens surrounding the match (specified by $&), with trailing g to find and replace all occurrences.
see http://www.regular-expressions.info/refreplace.html for the correct syntax for your regex language.
Depending on your language, you're looking to match groups.
So typically you'll make a pattern in the form of
([0-9]{1,})|([a-zA-Z]{1,})
Then, you'll iterate over the resulting groups in (specific to your language).