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I am trying to use some regex to validate some input inside of Java code. I have been successful in implementing "basic" regex, but this one seems to be out of my scope of knowledge. I am working through RegEgg tutorials to learn more.
Here are the conditions that need to be validated:
Field will always have 8 characters
Can be all spaces
Or
Valid characters: a-zA-Z0-9 -!& or a space
Cannot begin with a space
If one of the special characters is used, it can be the only one used
Legal: "B-123---" "AB&& &" "A!!!!!!!"
Illegal: "B-123!!!" "AB&& -" "A-&! "
Has to have at least one alphanumeric character (Can't be all special characters ie: "!!!!!!!!"
This was my regex before additional validations were added:
^(\s{8}|[A-Za-z\-\!\&][ A-Za-z0-9\-\!\&]{7})$"
Then the additional validations for now allowing multiple of the special characters, and I am a bit stuck. I have been successful in using a positive lookahead, but stuck when trying to use the positive lookbehind. (I think the data before the lookbehind was consumed), but I am speculating as I am a neophyte with this part of regex.
using the or construct (a|b) is a large part of this, and you've begun applying it, so that's a good start.
You've made the rule that it can't start with a digit; nothing in the spec says this. also, - inside [] has special meaning, so escape it, or make sure it is first or last, because then you don't have to. That gets us to:
^(\s{8}|[A-Za-z0-9-!& -]{8})$
next up is the rule that it has to be all the same special character if used at all. Given that there are only 3 special characters, could be easier to just explicitly list them all:
^(\s{8}|[A-Za-z0-9 -]{8}|[A-Za-z0-9 !]{8}|[A-Za-z0-9 &]{8})$
Next up: Can't start with a space, and can't be all-special. Confirming the negative (that it ISNT all-special characters) gets complicated; lookahead seems like a better plan here. This:
^ is regexp-ese for: "Start of line". Note that this doesn't 'consume' a character. 1 is regexpese for 'only the exact character '1' will match here, nothinge else', but as it matches, it also 'consumes' that character, whereas ^ doesn't do that. 'start of line' is not a concept that can be consumed.
This notion of 'a match may fail, but if it succeeds, nothing is consumed' isn't limited to ^ and $; you can write your own:
(?=abc) will match if abc would match at this position, but does not consume it. Thus, the regexp ^(=abc)ab.d$ would match the input string abcd and nothing else. This is called positive lookahead. (it 'looks ahead' and matches if it sees the regular expression in the parens, failing if it does not).
(?!abc) is negative lookahead. It matches if it DOESNT see the thing in the parens. (?!abc)a.c will match the input adc but not the input abc.
(?<=abc) is positive lookbehind. It matches if the pattern you provide would match such that the match ends at the position you find yourself.
(?<!abc) is negative lookbehind.
Note that lookahead and lookbehind can be somewhat limited, in that they may not allow variable length patterns. But, fortunately, your requirements make it easy to limit ourselves to fixed size patterns here. Thus, we can introduce: (?![&!-]{8}) as a non-consuming unit in our regexp that will fail the match if we have all-8 special characters.
We can use this trick to fail on starting space too: (?! ) is all we need for that one.
Let's replace \s which is whitespace with just which is the space character (the problem description says 'space', not 'whitespace').
Putting it all together:
^( {8}|(?! )(?![&!-]{8})([A-Za-z0-9 -]{8}|[A-Za-z0-9 !]{8}|[A-Za-z0-9 &]{8}))$
Thats:
8 spaces, or...
not a space, and not all-8 special character, then,
any of the valid chars, any amount of spaces, and any amount of one of the 3 allowed special symbols, as long as we have precisely 8 of them...
.. OR the same thing as #3 but with the second of the three special symbols
.. OR with the third of the three.
Plug em in at regex101 along with your various examples of 'legal' and 'not legal' and you can play around with it some more.
NB: You can also use backreferences to attempt to solve the 'only one special character is allowed' part of this, but attempting to tackle the 'not all special characters' part seems quite unwieldy if you don't get to use (negative) lookahead.
Its a matter of asserting the right conditions at the start of the regex.
^(?=[ ]*$|(?![ ]))(?!.*([!&-]).*(?!\1)[!&-])[a-zA-Z0-9 !&-]{8}$
see -> https://regex101.com/r/tN5y4P/1
Some discussion:
^ # Begin of text
(?= # Assert, cannot start with a space
[ ]* $ # unless it's all spaces
| (?! [ ] )
)
(?! # Assert, not mixed special chars
.*
( [!&-] ) # (1)
.*
(?! \1 )
[!&-]
)
[a-zA-Z0-9 !&-]{8} # Consume 8 valid characters from within this class
$ # End of text
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub
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'm trying to do a replace with this pattern, so I need to match this:
String pattern = "[\\\\A\\\\W]its[\\\\W\\\\z]";
The way I'm interpreting my pattern is: either a beginning of the string OR a non word character like a space or comma, then an "its", then a non word character OR the end of the string.
Why doesn't it match on this "its" inside this string?
its about time
The idea of what this is supposed to do it's supposed to detect incorrectly written words like "its" and fix them to "it's".
Also why do I need so many escape characters in order for the pattern to be accepted by the vm at all?
\\A and \\z are boundary matches. They cannot go inside character classes. If you use them properly, i.e. with two slashes instead of four, regex pattern compiler would throw an exception, because \A or \z cannot go inside [] blocks.
Use straight | syntax with non-capturing groups instead:
String pattern = "(?:\\A|\\W)its(?:\\W|\\z)";
Demo.
I need to validate input: valid variants are either number or empty string. What is the correspondent regular expression?
String pattern = "\d+|<what shoudl be here?>";
UPD: dont suggest "\d*" please, I'm just curious how to tell "empty string" in regexp.
In this particular case, ^\d*$ would work, but generally speaking, to match pattern or an empty string, you can use:
^$|pattern
Explanation
^ and $ are the beginning and end of the string anchors respectively.
| is used to denote alternates, e.g. this|that.
References
regular-expressions.info/Anchors and Alternation
Related questions
Regular expression which matches a pattern, or is an empty string
Note on multiline mode
In the so-called multiline mode (Pattern.MULTILINE/(?m) in Java), the ^ and $ match the beginning and end of the line instead. The anchors for the beginning and end of the string are now \A and \Z respectively.
If you're in multiline mode, then the empty string is matched by \A\Z instead. ^$ would match an empty line within the string.
Examples
Here are some examples to illustrate the above points:
String numbers = "012345";
System.out.println(numbers.replaceAll(".", "<$0>"));
// <0><1><2><3><4><5>
System.out.println(numbers.replaceAll("^.", "<$0>"));
// <0>12345
System.out.println(numbers.replaceAll(".$", "<$0>"));
// 01234<5>
numbers = "012\n345\n678";
System.out.println(numbers.replaceAll("^.", "<$0>"));
// <0>12
// 345
// 678
System.out.println(numbers.replaceAll("(?m)^.", "<$0>"));
// <0>12
// <3>45
// <6>78
System.out.println(numbers.replaceAll("(?m).\\Z", "<$0>"));
// 012
// 345
// 67<8>
Note on Java matches
In Java, matches attempts to match a pattern against the entire string.
This is true for String.matches, Pattern.matches and Matcher.matches.
This means that sometimes, anchors can be omitted for Java matches when they're otherwise necessary for other flavors and/or other Java regex methods.
Related questions
Regex doesn't work in String.matches()
/^\d*$/
Matches 0 or more digits with nothing before or after.
Explanation:
The '^' means start of line. '$' means end of line. '*' matches 0 or more occurences. So the pattern matches an entire line with 0 or more digits.
To explicitly match the empty string, use \A\Z.
You can also often see ^$ which works fine unless the option is set to allow the ^ and $ anchors to match not only at the start or end of the string but also at the start/end of each line. If your input can never contain newlines, then of course ^$ is perfectly OK.
Some regex flavors don't support \A and \Z anchors (especially JavaScript).
If you want to allow "empty" as in "nothing or only whitespace", then go for \A\s*\Z or ^\s*$.
Just as a funny solution, you can do:
\d+|\d{0}
A digit, zero times. Yes, it does work.
There shouldn't be anything wrong with just "\d+|"
One of the way to view at the set of regular language as the closure of the below things:
Special < EMPTY_STRING > is the regular language
Any symbol from alphaphet is the valid regular language
Any concatentation and union of two valid regexps is the regular language
Any union of two valid regular language is the regular language
Any transitive closure of the regexp is the regular language
Concreate regular language is concrete element of this closure.
I didn't find empty symbol in POSIX standard to express regular language idea from step (1).
But it is exist extra thing like question mark there which is by posix definition is the following:
(regexp|< EMPTY_STRING >)
So you can do in the following manner for bash, perl, and python:
echo 9023 | grep -E "(1|90)?23"
perl -e "print 'PASS' if (qq(23) =~ /(1|90)?23/)"
python -c "import re; print bool(re.match('^(1|90)?23$', '23'))"
To make any pattern that matches an entire string optional, i.e. allow a pattern match an empty string, use an optional group:
^(pattern)?$
^^ ^^^
See the regex demo
If the regex engine allows (as in Java), prefer a non-capturing group since its main purpose is to only group subpatterns, not keep the subvalues captured:
^(?:pattern)?$
The ^ will match the start of a string (or \A can be used in many flavors for this), $ will match the end of string (or \z can be used to match the very end in many flavors, and Java, too), and the (....)? will match 1 or 0 (due to the ? quantifier) sequences of the subpatterns inside parentheses.
A Java usage note: when used in matches(), the initial ^ and trailing $ can be omitted and you can use
String pattern = "(?:\d+)?";