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.
In .NET System.Text.RegularExpressions.Regex if ^ and $ are added to the Regex to look for exact matches, it still returns true for IsMatch if a terminating \n is appended to the string being verified.
For example, the following code:
Regex regexExact = new Regex(#"^abc$");
Console.WriteLine(regexExact.IsMatch("abc"));
Console.WriteLine(regexExact.IsMatch("abcdefg"));
Console.WriteLine(regexExact.IsMatch("abc\n"));
Console.WriteLine(regexExact.IsMatch("abc\n\n"));
returns:
true
false
true
false
What is the Regex that will return false for all of the above except the first?
Solution for the current .NET regex
You should use the very end of string anchor that is \z in .NET regex:
Regex regexExact = new Regex(#"^abc\z");
See Anchors in Regular Expressions:
$ The match must occur at the end of the string or line, or before \n at the end of the string or line. For more information, see End of String or Line.
\Z The match must occur at the end of the string, or before \n at the end of the string. For more information, see End of String or Before Ending Newline.
\z The match must occur at the end of the string only. For more information, see End of String Only.
The same anchor can be used in .net, java, pcre, delphi, ruby and php. In python, use \Z. In JavaScript RegExp (ECMAScript) compatible patterns, the $ anchor matches the very end of string (if no /m modifier is defined).
Background
see Strings Ending with a Line Break at regular-expressions.info:
Because Perl returns a string with a newline at the end when reading a line from a file, Perl's regex engine matches $ at the position before the line break at the end of the string even when multi-line mode is turned off. Perl also matches $ at the very end of the string, regardless of whether that character is a line break. So ^\d+$ matches 123 whether the subject string is 123 or 123\n.
Most modern regex flavors have copied this behavior. That includes .NET, Java, PCRE, Delphi, PHP, and Python. This behavior is independent of any settings such as "multi-line mode".
In all these flavors except Python, \Z also matches before the final line break. If you only want a match at the absolute very end of the string, use \z (lower case z instead of upper case Z). \A\d+\z does not match 123\n. \z matches after the line break, which is not matched by the shorthand character class.
In Python, \Z matches only at the very end of the string. Python does not support \z.
What is the difference between "\\w+#\\w+[.]\\w+" and "^\\w+#\\w+[.]\\w+$"? I have tried to google for it but no luck.
^ means "Match the start of the string" (more exactly, the position before the first character in the string, so it does not match an actual character).
$ means "Match the end of the string" (the position after the last character in the string).
Both are called anchors and ensure that the entire string is matched instead of just a substring.
So in your example, the first regex will report a match on email#address.com.uk, but the matched text will be email#address.com, probably not what you expected. The second regex will simply fail.
Be careful, as some regex implementations implicitly anchor the regex at the start/end of the string (for example Java's .matches(), if you're using that).
If the multiline option is set (using the (?m) flag, for example, or by doing Pattern.compile("^\\w+#\\w+[.]\\w+$", Pattern.MULTILINE)), then ^ and $ also match at the start and end of a line.
Try the Javadoc:
http://download.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
^ and $ match the beginnings/endings of a line (without consuming them)
When user pastes something like this (from notepad for example):
multi
line#email.com
into input text box, the line break dissapears and it looks like this:
multi
line#email.com
But whatever the line break is converted to does not match this regex:
'\s|\t|\r|\n|\0','i'
so this invalid character passes through js validation to the .NET application code I am working on.
It is interesting but this text editor does the same transformation, that is why I had to post original sample as code. I would like to find out what the line break got converted to, so I can add a literal to the regex but I don't know how. Many thanks!
Here is the whole snippet:
var invalidChars = new RegExp('(^[.])|[<]|[>]|[(]|[)]|[\]|[,]|[;]|[:]|([.])[.]|\s|\t|\r|\n|\0', 'i');
if (text.match(invalidChars)) {
return false;
}
Your immediate problem is escaping. You're using a string literal to create the regex, like this:
'(^[.])|[<]|[>]|[(]|[)]|[\]|[,]|[;]|[:]|([.])[.]|\s|\t|\r|\n|\0'
But before it ever reaches the RegExp constructor, the [\] becomes []; \s becomes s; \0 becomes 0; and \t, \r and \n are converted to the characters they represent (tab, carriage return and linefeed, respectively). That won't happen if you use a regex literal instead, but you still have to escape the backslash to match a literal backslash.
Your regex is also has way more brackets than it needs. I think this is what you were trying for:
/^\.|\.\.|[<>()\\,;:\s]/
That matches a dot at the beginning, two consecutive dots, or one of several forbidden characters including any whitespace character (\s matches any whitespace character, not just a space).
Ok - here it is
vbCrLF
This is what pasted line breaks are converted to. I added (vbCrLF) group and those spaces are now detected. Thanks, Dan1M
http://forums.asp.net/t/1183613.aspx?Multiline+Textbox+Input+not+showing+line+breaks+in+Repeater+Control
Why doesn't .NET regex treat \n as end of line character?
Sample code:
string[] words = new string[] { "ab1", "ab2\n", "ab3\n\n", "ab4\r", "ab5\r\n", "ab6\n\r" };
Regex regex = new Regex("^[a-z0-9]+$");
foreach (var word in words)
{
Console.WriteLine("{0} - {1}", word, regex.IsMatch(word));
}
And this is the response I get:
ab1 - True
ab2
- True
ab3
- False
- False
ab5
- False
ab6
- False
Why does the regex match ab2\n?
Update:
I don't think Multiline is a good solution, that is, I want to validate login to match only specified characters, and it must be single line. If I change the constructor for MultiLine option ab1, ab2, ab3 and ab6 match the expression, ab4 and ab5 don't match it.
If the string ends with a line break the RegexOptions.Multiline will not work. The $ will just ignore the last line break since there is nothing after that.
If you want to match till the very end of the string and ignore any line breaks use \z
Regex regex = new Regex(#"^[a-z0-9]+\z", RegexOptions.Multiline);
This is for both MutliLine and SingleLine, that doesn't matter.
The .NET regex engine does treat \n as end-of-line. And that's a problem if your string has Windows-style \r\n line breaks. With RegexOptions.Multiline turned on $ matches between \r and \n rather than before \r.
$ also matches at the very end of the string just like \z. The difference is that \z can match only at the very end of the string, while $ also matches before a trailing \n. When using RegexOptions.Multiline, $ also matches before any \n.
If you're having trouble with line breaks, a trick is to first to a search-and-replace to replace all \r with nothing to make sure all your lines end with \n only.
From RegexOptions:
Multiline mode. Changes the meaning of ^ and $ so they match at the beginning and end, respectively, of any line, and not just the beginning and end of the entire string.
So basically if you pass a RegexOptions.Multiline to the Regex constructor you are instructing that instance to treat the final $ as a match for newline characters - not simply the end of the string itself.
Use regex options, System.Text.RegularExpressions.RegexOptions:
string[] words = new string[] { "ab1", "ab2\n", "ab3\n\n", "ab4\r", "ab5\r\n", "ab6\n\r" };
Regex regex = new Regex("^[a-z0-9]+$");
foreach (var word in words)
{
Console.WriteLine("{0} - {1}", word,
regex.IsMatch(word,"^[a-z0-9]+$",
System.Text.RegularExpressions.RegexOptions.Singleline |
System.Text.RegularExpressions.RegexOptions.IgnoreCase |
System.Text.RegularExpressions.RegexOptions.IgnorePatternWhitespace));
}
Could be the ususal windows/linux line ending differences. But it's still strange that \n\n gets a false this way... Did you try with the RegexOptions.Multiline flag set?
Just to give more details to Smazy answer. This an extract from:
Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan. Copyright 2009 Jan Goyvaerts and Steven Levithan, 978-0-596-2068-7
The difference between ‹\Z› and ‹\z›
comes into play when the last
character in your subject text is a
line break. In that case, ‹\Z› can
match at the very end of the subject
text, after the final line break, as
well as immediately before that line
break. The benefit is that you can
search for ‹omega\Z› without having to
worry about stripping off a trailing
line break at the end of your subject
text. When reading a file line by
line, some tools include the line
break at the end of the line, whereas
others don’t; ‹\Z› masks this
difference. ‹\z› matches only at the
very end of the subject text, so it
will not match text if a trailing line
break follows. The anchor ‹$› is
equivalent to ‹\Z›, as long as you do
not turn on the “^ and $ match at line
breaks” option. This option is off by
default for all regex flavors except
Ruby. Ruby does not offer a way to
turn this option off. Just like ‹\Z›,
‹$› matches at the very end of the
subject text, as well as before the
final line break, if any.
Of course, I wouldn't have found it without Smazy answer.