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+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
Hi I'm struggling to understand why my Regex isn't working.
I have URL's that have DOI's on them like so:
https://link.springer.com/10.1007/s00737-021-01116-5
https://journals.sagepub.com/doi/pdf/10.1177/1078390319877228
https://journals.sagepub.com/doi/pdf/10.1177/1078390319877228
https://onlinelibrary.wiley.com/doi/10.1111/jocn.13435
https://journals.sagepub.com/doi/pdf/10.1177/1062860613484171
https://onlinelibrary.wiley.com/resolve/openurl?genre=article&title=Natural+Resources+Forum&issn=0165-0203&volume=26&date=2002&issue=1&spage=3
https://dx.doi.org/10.1108/14664100110397304?nols=y
https://onlinelibrary.wiley.com/doi/10.1111/jocn.15833
https://www.tandfonline.com/doi/pdf/10.1080/03768350802090592?needAccess=true
And I'm using for example this Regex, but it always returns empty?
print(re.findall(r'/^10.\d{4,9}/[-._;()/:A-Z0-9]+$/i', 'https://dx.doi.org/10.1108/02652320410549638?nols=y'))
Where have I gone wrong?
It looks like you come from another programming language that has the notion of regex literals that are delimited with forward slashes and have the modifiers following the closing slash (hence /i).
In Python there is no such thing, and these slashes and modifier(s) are taken as literal characters. For flags like i you can use the optional flags parameter of findall.
Secondly, ^ will match the start of the input string, but evidently the URLs you have as input do not start with 10, so that has to go. Instead you could require that the 10 must follow a word break... i.e. it should not be preceded by an alphanumerical character (or underscore).
Similarly, $ will match the end of the input string, but you have URLs that continue with URL parameters, like ?nols=y, so again the part you are interested in does not go on until the end of the input. So that has to go too.
The dot has a special meaning in regex, but you clearly intended to match a literal dot, so it should be escaped.
Finally, alphanumerical characters can be matched with \w, which also matches both lower case and capital Latin letters, so you can shorten the character class a bit and do without any flags such as i (re.I).
This leaves us with:
print(re.findall(r'\b10\.\d{4,9}/[-.;()/:\w]+',
'https://dx.doi.org/10.1108/02652320410549638?nols=y'))
Could someone help me on regex to match German words/sentences in
python? It does not work on jupyter notebook. I tried same in jsfiddle
it works fine. I tried using this below script but does not work
import re
pattern = re.compile(r'\[^a-zA-Z0-9äöüÄÖÜß]\\', re.UNICODE)
print(pattern.search(text))
Your expression will always fail:
\[^a-zA-Z0-9äöüÄÖÜß]\\
Broken down, you require
[ # literally
^ # start of the line / text
a-z # literally, etc.
The problem is that you require a [ literally right before the start of a line which can never be true (either there's nothing or a newline). So in the end, either remove the backslash to get a proper character class as in:
[^a-zA-Z0-9äöüÄÖÜß]+
But this will surely not match the words you're looking for (quite the opposite). So either use something as simple as \w+ or the solution proposed by #Wiktor in the comments section.
The square brackets define a range of characters you want to look for, however the '^' negates these characters if it appears within the character class.
If you want to specify the beginning of the line you need to put the '^' before the brackets.
Also you need to add a multiplier behind the class to search for more than just one character in this case:
r'^[a-zA-Z0-9äöüÄÖÜß]+'
One ore more characters contained in the brackets are matched as long as they are not seperated by any other character not listed between '[]'
Here's the link to the official documentation
I want the regular expression of the following pattern
Rule
The pattern should have no no punctuation except +. No blank spaces are allowed.Only alphabetical characters are allowed.
The patter should be of two or more characters
EDIT:
I am thinking something like:
'^\w\w+ |[ \w\w+ + \w\w+] $'
But it is not working.
Your question has very limited detail, however the very simple answer is:
^[+]{2,}$
However, that would only match the character "+" 2 or more times. Since you're saying punctuation, it seems to imply that you want to allow other text. In that case, I would go with:
^[\w+ ]{2,}$
Which would allow all "word characters" and spaces. In the Python string, you will need to escape the backslash with another backslash.
If you want to experiment with regex strings, I would highly recommend the website http://regex101.com
EDIT: I have now seen your updated question, and to only have alphabetical characters and the plus symbol, you will want
^[A-Za-z+]{2,}$
I am seeing the following phenomenon, couldn't seem to figure it out, and didn't find anything with some search through archives:
if I type in:
>>> if re.search(r'\n',r'this\nis\nit'):<br>
... print 'found it!'<br>
... else:<br>
... print "didn't find it"<br>
...
I will get:
didn't find it!
However, if I type in:
>>> if re.search(r'\\n',r'this\nis\nit'):<br>
... print 'found it!'<br>
... else:<br>
... print "didn't find it"<br>
...
Then I will get:
found it!
(The first one only has one backslash on the r'\n' whereas the second one has two backslashes in a row on the r'\\n' ... even this interpreter is removing one of them.)
I can guess what is going on, but I don't understand the official mechanism as to why this is happening: in the first case, I need to escape two things: both the regular expression and the special strings. "Raw" lets me escape the special strings, but not the regular expression.
But there will never be a regular expression in the second string, since it is the string being matched. So there is only a need to escape once.
However, something doesn't seem consistent to me: how am I supposed to ensure that the characters REALLY ARE taken literally in the first case? Can I type rr'' ? Or do I have to ensure that I escape things twice?
On a similar vein, how do I ensure that a variable is taken literally (or that it is NOT taken literally)? E.g., what if I had a variable tmp = 'this\nis\nmy\nhome', and I really wanted to find the literal combination of a slash and an 'n', instead of a newline?
Thanks!Mike
re.search(r'\n', r'this\nis\nit')
As you said, "there will never be a regular expression in the second string." So we need to look at these strings differently: the first string is a regex, the second just a string. Usually your second string will not be raw, so any backslashes are Python-escapes, not regex-escapes.
So the first string consists of a literal "\" and an "n". This is interpreted by the regex parser as a newline (docs: "Most of the standard escapes supported by Python string literals are also accepted by the regular expression parser"). So your regex will be searching for a newline character.
Your second string consists of the string "this" followed by a literal "\" and an "n". So this string does not contain an actual newline character. Your regex will not match.
As for your second regex:
re.search(r'\\n', r'this\nis\nit')
This version matches because your regex contains three characters: a literal "\", another literal "\" and an "n". The regex parser interprets the two slashes as a single "\" character, followed by an "n". So your regex will be searching for a "\" followed by an "n", which is found within the string. But that isn't very helpful, since it has nothing to do with newlines.
Most likely what you want is to drop the r from the second string, thus treating it as a normal Python string.
re.search(r'\n', 'this\nis\nit')
In this case, your regex (as before) is searching for a newline character. And, it finds it, because the second string contains the word "this" followed by a newline.
Escaping special sequences in string literals is one thing, escaping regular expression special characters is another. The row string modifier only effects the former.
Technically, re.search accepts two strings and passes the first to the regex builder with re.compile. The compiled regex object is used to search patterns inside simple strings. The second string is never compiled and thus it is not subject to regex special character rules.
If the regex builder receives a \n after the string literal is processed, it converts this sequence to a newline character. You also have to escape it if you need the match the sequence instead.
All rationale behind this is that regular expressions are not part of the language syntax. They are rather handled within the standard library inside the re module with common building blocks of the language.
The re.compile function uses special characters and escaping rules compatible with most commonly used regex implementations. However, the Python interpreter is not aware of the whole regular expression concept and it does not know whether a string literal will be compiled into a regex object or not. As a result, Python can't provide any kind syntax simplification such as the ones you suggested.
Regexes have their own meaning for literal backslashes, as character classes like \d. If you actually want a literal backslash character, you will in fact need to double-escape it. It's really not supposed to be parallel since you're comparing a regex to a string.
Raw strings are just a convenience, and it would be way overkill to have double-raw strings.