Related
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.
For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
I am building a parser, and I would like to remove comments from various lines. For example,
variable = "some//thing" ////actual comment
Comment marker is //. In this case, variable would contain "some//thing" and everything else would be ignored. I plan to do it using regex replace. Currently I am using (".*"|[ \t])*(\/\/.*) as regex. However replacing it replaces "some//thing" ////actual comment entirely.
I can not figure out the regex which I should use instead. Thanks for any help.
Additional info - I am using C# with netcoreapp 1.1.0
Edit - some cases might be of a line with just comment like //line comment. Strings also might contain escaped quotes.
Here is the ugly regex pattern. I believe it will work well. I have tried it with every pathological example I can think of, including lines that contain syntax errors. For example, a quoted string that has too many quotes, or too few, or has a double escaped quote, which is, therefore, not escaped. And with quoted strings in the comments, which I have been known to do when I want to remind myself of alternatives.
The only time that it trips up is if there is a double slash inside a seemingly quoted string and somehow that string is malformed and the double slash ends up legally outside the properly quoted portion. Syntactically that makes it a valid comment, even though not the programmer's intention. So, from the programmer's perspective it's wrong, but by the rules, it's really a comment. Meaning, the pattern only appears to trip up.
When used the pattern will return the non-comment portion of the line(s). The pattern has a newline \n in it to allow for applying it to an entire file. You may need to modify that if you system interprets newlines in some other fashion, for example as \r or \r\n. To use it in single line mode you can remove that if you choose. It is at characters 17 and 18 in the one-liner and is on the fifth line, 6th and 7th printing characters in the multi-line version. You can safely leave it there, however, as in single-line mode it makes no difference, and in multi-line mode it will return a newline for lines of code that are either blank, or have a comment beginning in the first column. That will keep the line numbers the same in the original version and the stipped version if you write the results to a new file. Makes comparison easy.
One major caveat for this pattern: It uses a grouping construct that has varying level of support in regex engines. I believe as used here, with a lookaround, it's only the .NET and PCRE engines that will accept it YMMV. It is a tertiary type: (?(_condition_)_then_|_else_). The _condition_ pattern is treated as a zero-width assertion. If the pattern matches, then the _then_ pattern is used in the attempted match, otherwise the _else_ pattern is used. Without that construct, the pattern was growing to uncommon lengths, and was still failing on some of my pathological test cases.
The pattern presented here is as it needs to be seen by the regex engine. I am not a C# programmer, so I don't know all the nuances of escaping quoted strings. Getting this pattern into your code, such that all the backslashes and quotes are seen properly by the regex engine is still up to you. Maybe C# has the equivalent of Perl's heredoc syntax.
This is the one-liner pattern to use:
^((?:(?:(?:[^"'/\n]|/(?!/))*)(?("(?=(?:\\\\|\\"|[^"])*"))(?:"(?:\\\\|\\"|[^"])*")|(?('(?=(?:\\\\|\\'|[^'])*'))(?:'(?:\\\\|\\'|[^'])*')|(?(/)|.))))*)
If you want to use the ignore pattern whitespace option, you can use this version:
(?x) # Turn on the ignore white space option
^( # Start the only capturing group
(?: # A non-capturing group to allow for repeating the logic
(?: # Capture either of the two options below
[^"'/\n] # Capture everything not a single quote, double quote, a slash, or a newline
| # OR
/(?!/) # Capture a slash not followed by a slash [slash an negative look-ahead slash]
)* # As many times as possible, even if none
(?(" # Start a conditional match for double-quoted strings
(?=(?:\\\\|\\"|[^"])*") # Followed by a properly closed double-quoted string
) # Then
(?:"(?:\\\\|\\"|[^"])*") # Capture the whole double-quoted string
| # Otherwise
(?(' # Start a conditional match for single-quoted strings
(?=(?:\\\\|\\'|[^'])*') # Followed by a properly closed single-quoted string
) # Then
(?:'(?:\\\\|\\'|[^'])*') # Capture the whole double-quoted string
| # Otherwise
(?([^/]) # If next character is not a slash
.) # Capture that character, it is either a single quote, or a double quote not part of a properly closed
) # end the conditional match for single-quoted strings
) # End the conditional match for double-quoted strings
)* # Close the repeating non-capturing group, capturing as many times as possible, even if none
) # Close the only capturing group
This allows for your code to explain this monstrosity so that when someone else looks at it, or in a few months you have to work on it yourself, there's no WTF moment. I think the comments explain it well, but feel free to change them any way you please.
As mentioned above, the conditional match grouping has limited support. One place it will fail is on the site you linked to in an earlier comment. Since you're using C#, I choose to do my testing in the .NET Regex Tester, which can handle those constructs. It includes a nice Reference too. Given the proper selections on the side, you can test either version above, and experiment with it as well. Considering its complexity, I would recommend testing it, somewhere, against data from your files, as well as any edge cases and pathological tests you can dream up.
Just to redeem this small pattern, there is a much bigger pattern for testing email address that is 78 columns by 81 lines, with a couple dozen characters to spare. (Which I do not recommend using, or any other regex, for testing email addresses. Wrong tool for the job.) If you want to scare yourself, have a peek at it on the ex-parrot site. I had nothing to do with that!!
"[^"\\]*(?:\\[\W\w][^"\\]*)*"|(\/\/.*)
Flags: global
Matches full strings or a comment.
Group 1: comment.
So if there's no comment, replace with the same matching text. Otherwise, do your thing on the comment itself.
I want to parse the 2 digits in the middle from a date in dd/mm/yy format but also allowing single digits for day and month.
This is what I came up with:
(?<=^[\d]{1,2}\/)[\d]{1,2}
I want a 1 or 2 digit number [\d]{1,2} with a 1 or 2 digit number and slash ^[\d]{1,2}\/ before it.
This doesn't work on many combinations, I have tested 10/10/10, 11/12/13, etc...
But to my surprise (?<=^\d\d\/)[\d]{1,2} worked.
But the [\d]{1,2} should also match if \d\d did, or am I wrong?
On lookbehind support
Major regex flavors have varying supports for lookbehind differently; some imposes certain restrictions, and some doesn't even support it at all.
Javascript: not supported
Python: fixed length only
Java: finite length only
.NET: no restriction
References
regular-expressions.info/Flavor comparison
Python
In Python, where only fixed length lookbehind is supported, your original pattern raises an error because \d{1,2} obviously does not have a fixed length. You can "fix" this by alternating on two different fixed-length lookbehinds, e.g. something like this:
(?<=^\d\/)\d{1,2}|(?<=^\d\d\/)\d{1,2}
Or perhaps you can put both lookbehinds as alternates of a non-capturing group:
(?:(?<=^\d\/)|(?<=^\d\d\/))\d{1,2}
(note that you can just use \d without the brackets).
That said, it's probably much simpler to use a capturing group instead:
^\d{1,2}\/(\d{1,2})
Note that findall returns what group 1 captures if you only have one group. Capturing group is more widely supported than lookbehind, and often leads to a more readable pattern (such as in this case).
This snippet illustrates all of the above points:
p = re.compile(r'(?:(?<=^\d\/)|(?<=^\d\d\/))\d{1,2}')
print(p.findall("12/34/56")) # "[34]"
print(p.findall("1/23/45")) # "[23]"
p = re.compile(r'^\d{1,2}\/(\d{1,2})')
print(p.findall("12/34/56")) # "[34]"
print(p.findall("1/23/45")) # "[23]"
p = re.compile(r'(?<=^\d{1,2}\/)\d{1,2}')
# raise error("look-behind requires fixed-width pattern")
References
regular-expressions.info/Lookarounds, Character classes, Alternation, Capturing groups
Java
Java supports only finite-length lookbehind, so you can use \d{1,2} like in the original pattern. This is demonstrated by the following snippet:
String text =
"12/34/56 date\n" +
"1/23/45 another date\n";
Pattern p = Pattern.compile("(?m)(?<=^\\d{1,2}/)\\d{1,2}");
Matcher m = p.matcher(text);
while (m.find()) {
System.out.println(m.group());
} // "34", "23"
Note that (?m) is the embedded Pattern.MULTILINE so that ^ matches the start of every line. Note also that since \ is an escape character for string literals, you must write "\\" to get one backslash in Java.
C-Sharp
C# supports full regex on lookbehind. The following snippet shows how you can use + repetition on a lookbehind:
var text = #"
1/23/45
12/34/56
123/45/67
1234/56/78
";
Regex r = new Regex(#"(?m)(?<=^\d+/)\d{1,2}");
foreach (Match m in r.Matches(text)) {
Console.WriteLine(m);
} // "23", "34", "45", "56"
Note that unlike Java, in C# you can use #-quoted string so that you don't have to escape \.
For completeness, here's how you'd use the capturing group option in C#:
Regex r = new Regex(#"(?m)^\d+/(\d{1,2})");
foreach (Match m in r.Matches(text)) {
Console.WriteLine("Matched [" + m + "]; month = " + m.Groups[1]);
}
Given the previous text, this prints:
Matched [1/23]; month = 23
Matched [12/34]; month = 34
Matched [123/45]; month = 45
Matched [1234/56]; month = 56
Related questions
How can I match on, but exclude a regex pattern?
Unless there's a specific reason for using the lookbehind which isn't noted in the question, how about simply matching the whole thing and only capturing the bit you're interested in instead?
JavaScript example:
>>> /^\d{1,2}\/(\d{1,2})\/\d{1,2}$/.exec("12/12/12")[1]
"12"
To quote regular-expressions.info:
The bad news is that most regex
flavors do not allow you to use just
any regex inside a lookbehind, because
they cannot apply a regular expression
backwards. Therefore, the regular
expression engine needs to be able to
figure out how many steps to step back
before checking the lookbehind.
Therefore, many regex flavors,
including those used by Perl and
Python, only allow fixed-length
strings. You can use any regex of
which the length of the match can be
predetermined. This means you can use
literal text and character classes.
You cannot use repetition or optional
items. You can use alternation, but
only if all options in the alternation
have the same length.
In other words your regex does not work because you're using a variable-width expression inside a lookbehind and your regex engine does not support that.
In addition to those listed by #polygenelubricants, there are two more exceptions to the "fixed length only" rule. In PCRE (the regex engine for PHP, Apache, et al) and Oniguruma (Ruby 1.9, Textmate), a lookbehind may consist of an alternation in which each alternative may match a different number of characters, as long as the length of each alternative is fixed. For example:
(?<=\b\d\d/|\b\d/)\d{1,2}(?=/\d{2}\b)
Note that the alternation has to be at the top level of the lookbehind subexpression. You might, like me, be tempted to factor out the common elements, like this:
(?<=\b(?:\d\d/|\d)/)\d{1,2}(?=/\d{2}\b)
...but it wouldn't work; at the top level, the subexpression now consists of a single alternative with a non-fixed length.
The second exception is much more useful: \K, supported by Perl and PCRE. It effectively means "pretend the match really started here." Whatever appears before it in the regex is treated as a positive lookbehind. As with .NET lookbehinds, there are no restrictions; whatever can appear in a normal regex can be used before the \K.
\b\d{1,2}/\K\d{1,2}(?=/\d{2}\b)
But most of the time, when someone has a problem with lookbehinds, it turns out they shouldn't even be using them. As #insin pointed out, this problem can be solved much more easily by using a capturing group.
EDIT: Almost forgot JGSoft, the regex flavor used by EditPad Pro and PowerGrep; like .NET, it has completely unrestricted lookbehinds, positive and negative.
I am updating some code that I didn't write and part of it is a regex as follows:
\[url(?:\s*)\]www\.(.*?)\[/url(?:\s*)\]
I understand that .*? does a non-greedy match of everything in the second register.
What does ?:\s* in the first and third registers do?
Update: As requested, language is C# on .NET 3.5
The syntax (?:) is a way of putting parentheses around a subexpression without separately extracting that part of the string.
The author wanted to match the (.*?) part in the middle, and didn't want the spaces at the beginning or the end from getting in the way. Now you can use \1 or $1 (or whatever the appropriate method is in your particular language) to refer to the domain name, instead of the first chunk of spaces at the beginning of the string
?: makes the parentheses non-grouping. In that regex, you'll only pull out one piece of information, $1, which contains the middle (.*?) expression.
What does ?:\s* in the first and third registers do?
It's matching zero or more whitespace characters, without capturing them.
The regex author intends to allow trailing whitespace in the square-bracket-tags, matching all DNS labels following the "www." like so:
[url]www.foo.com[/url] # foo.com
[url ]www.foo.com[/url ] # same
[url ]www.foo.com[/url] # same
[url]www.foo.com[/url ] # same
Note that the regex also matches:
[url]www.[/url] # empty string!
and fails to match
[url]stackoverflow.com[/url] # no match, bummer
You may find this Regular Expressions Cheat Sheet very helpful (hopefully). I spent ages trying to learn Regex with no luck. And once I read this cheat-sheet - I immediately understood what I previously failed to learn.
http://krijnhoetmer.nl/stuff/regex/cheat-sheet/