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+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
I try to match with java 8 regex string form any language
as long as it includes letters, digits and . or -
String s = "בלה בלה";
String pattern= "^[\\p{L}\\p{Digit}_.-]*$";
return s.matches(pattern);
what am i missing as this code returns null for hebrew valid string.
You may add a whitespace to your pattern, and use \w instead of \p{L}\p{Digit}_ while passing the Pattern.UNICODE_CHARACTER_CLASS flag:
String s = "בלה בלה";
String pattern= "(?U)[\\w\\s.-]*";
System.out.println(s.matches(pattern));
// => true
See the Java demo
Since the pattern is used inside String#matches() method, the ^ and $ anchors are not necessary. If you plan to use the pattern with the Pattern#find() method, enclose the pattern within anchors as in the original code ("^(?U)[\\w\\s.-]*$").
Pattern details:
(?U) - the Pattern.UNICODE_CHARACTER_CLASS embedded modifier flag that makes shorthand character classes Unicode aware (you may see what \w matches in this mode)
[\\w\\s.-]* - zero or more:
\w - word chars (letters, digits, _ and some more)
\s - whitespaces
. - a dot (no need to escape it inside a character class)
- - a hyphen (no need as it is at the end of the character class)
i am using regex to check correctness of the string in my application. I want to check if string has a following pattern: x=y&a=b&... x,y,a,b etc. can be empty.
Example of correct strings:
abc=def&gef=cda&pdf=cdf
=&gef=def
abc=&gef=def
=abc&gef=def
Example of incorrect strings:
abc=def&gef=cda&
abc=def&gef==cda&
abc=defgef=cda&abc=gda
This is my code showing current solution:
String pattern = "[[a-zA-Z0-9]*[=]{1}[a-zA-Z0-9]*[&]{1}]*";
if(!Pattern.matches(pattern, s)){
throw new IllegalArgumentException(s);
}
This solution is bad because it accepts strings like:
abc=def&gef=def&
Can anyone help me with correct pattern?
You may use the following regex:
^[a-zA-Z0-9]*=[a-zA-Z0-9]*(?:&[a-zA-Z0-9]*=[a-zA-Z0-9]*)*$
See the regex demo
When used with matches(), the ^ and $ anchors may be omitted.
Details:
^ - start of string
[a-zA-Z0-9]* - 0+ alphanumeric chars (may be replaced with \p{Alnum})
= - a = symbol
[a-zA-Z0-9]* - 0+ alphanumeric chars
= - a = symbol
(?: - start of a non-capturing group matching sequences of...
& - a & symbol
[a-zA-Z0-9]*=[a-zA-Z0-9]* - same as above
)* - ... zero or more occurrences
$ - end of string
NOTE: If you want to make the pattern more generic, you may match any char other than = and & with a [^&=] pattern that would replace a more restrictive [a-zA-Z0-9] pattern:
^[^=&]*=[^=&]*(?:&[^=&]*=[^=&]*)*$
See this regex demo
I believe you want this.
([a-zA-Z0-9]*=[a-zA-Z0-9]*&)*[a-zA-Z0-9]*=[a-zA-Z0-9]*
This matches any number of repetitions like x=y, with a & after each one; followed by one repetition like x=y without the following &.
Here you go:
^\w*=\w*(?:&(?:\w*=\w*))*$
^ is the starting anchor
(\w*=\w*) is to represent parameters like abc=def
\w matches a word character [a-zA-Z0-9_]
\w* represents 0 or more characters
& represents tha actual ampersand literal
(&(\w*=\w*))* matches any subsequents parameters like &b=d etc.
$ represents the ending anchor
Regex101 Demo
EDIT: Made all groups non-capturing.
Note: As #WiktorStribiżew has pointed out in the comments, \w will match _ as well, so above regex should be modified to exclude underscores if they are to be avoided in the pattern, i.e [A-Za-z0-9]
I need to help in writing regex pattern to remove only part of the matcher from original string.
Original String: 2017-02-15T12:00:00.268+00:00
Expected String: 2017-02-15T12:00:00+00:00
Expected String removes everything in milliseconds.
My regex pattern looks like this: (:[0-5][0-9])\.[0-9]{1,3}
i need this regex to make sure i am removing only the milliseconds from some time field, not everything that comes after dot. But using above regex, I am also removing the minute part. Please suggest and help.
You have defined a capturing group with (...) in your pattern, and you want to have that part of string to be present after the replacement is performed. All you need is to use a backreference to the value stored in this capture. It can be done with $1:
String s = "2017-02-15T12:00:00.268+00:00";
String res = s.replaceFirst("(:[0-5][0-9])\\.[0-9]{1,3}", "$1");
System.out.println(res); // => 2017-02-15T12:00:00+00:00
See the Java demo and a regex demo.
The $1 in the replacement pattern tells the regex engine it should look up the captured group with ID 1 in the match object data. Since you only have one pair of unescaped parentheses (1 capturing group) the ID of the group is 1.
Change your pattern to (?::[0-5][0-9])(\.[0-9]{1,3}), run the find in the matcher and remove all it finds in the group(1).
The backslash will force the match with the '.' char, instead of any char, which is what the dot represents in a regex.
The (?: defines a non-capturing group, so it will not be considered in the group(...) on the matcher.
And adding a parenthesis around what you want will make it show up as group in the matcher, and in this case, the first group.
A good reference is the Pattern javadoc: http://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html
Use $1 and $2 variable for replace
string.replaceAll("(.*)\\.\\d{1,3}(.*)","$1$2");
I am creating a regular expression to match the string
#servername:port:databasename
and through https://regex101.com/ I came up with
\#(((\w+.*-*)+)?\w+)(:\d+)(:\w+)
which matches
e.g. #CORA-PC:1111:databasename or #111.111.1.111:111:databasename
However when I use this regular expression to pattern match in my java code the String #CORA-PC:1111:database=name is also matched.
Why is \w matching the = equals sign? I also tried [0-9a-zA-Z] but it also matched the = equals sign?
Can anyone help me with this?
Thanks!
The .* is a greedy dot matching subpattern that matches the whole line and then backtracks to accommodate for the subsequent subpatterns. That is why the pattern can match a = symbol (see demo - Group 3 matches that part with =) .
Your pattern is rather fragile, as the first part contains nested quantifiers with optional subpatterns that slows down the regex execution and causes other issues. You need to make it more linear.
#(\w+(?:[-.]\w+)*)?(:\d+)(:\w+)
See the regex demo
The regex will match
# - # symbol
(\w+(?:[-.]\w+)*)? - an optional group matching
\w+ - 1+ word chars
(?:[-.]\w+)* - 0+ sequences of a - or . ([-.]) followed with 1+ word chars
(:\d+) - a : symbol followed with 1+ digits
(:\w+) - a : symbol followed with 1+ word chars
If you need to avoid partial matching, use String#matches() (see demo).
NOTE: In Java, backslashes must be doubled.
Code example (Java):
String s = "#CORA-PC:1111:databasename";
String rx = "#(?:\\w+(?:[-.]\\w+)*)?:\\d+:\\w+";
System.out.println(s.matches(rx));
Code example (JS):
var str = '#CORA-PC:1111:databasename';
alert(/^#(?:\w+(?:[-.]\w+)*)?:\d+:\w+$/.test(str));