I am attempting to write some regex that detects a valid string (with single quotes). I specifically mean that it is looking for valid string literals, as in don't match this text 'but do match this text'.
This is being done in a TextMate grammar (.tmLanguage.json file) which uses Oniguruma regex (used by Ruby and some versions of PHP). Because it is a TextMate grammar, it also means I can use begin and end keys to have regex that matches the beginning and end of what I want to match, and everything in between is automatically matched. I want to ensure that multiline stings will work, but only if they have a backslash before the new line. Because of how these grammar files work, if I want to have multiline matches, I must use begin and end.
Overall, here are some examples of strings that should/should not match:
'match this'
not this 'match this' not this
'match this\
and this'
'dont match this
and dont match this'
'match this \'and this too'
"dont match this"
'dont match this"
I tried using lookbehinds and lookaheads, but it was difficult to make it work with the begin/end structure. I was able to write a regex string as you would with normal regex ("match": "'(\\'|\\\n|[^'\n])*'"), but no matter what I do it cannot work for multiline matches in a TextMate grammar file. When searching online, I was unable to get anything about matching actual literal strings, I could only find how to tell if my regex matches a string (as in just a piece of text), and nothing about begin/end regex.
To sum up, how would I create regex that matches single quote strings, which can be multiline if they have a backslash before the newline, using begin/end regex expressions?
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 need a regex to replace this text:
("{scriptid}_*0123_00000000_ABC-Description*");
With this text:
("{scriptid}_*0123_00000000_ABC-Description*"));
Or in other words: *I have to add a second closing brace.
Explanation:
("{scriptid}_*0123_00000000_ABC-Description*");
("{scriptid}_4-digits_8-alphanumeric_X-alphanumeric-description");
I tried some expressions but it doesn't really work.
Could some one please help me?
A generic regex that matches all the lines ending with ');' (match also if there are trailing spaces) can be this:
s/\);\s*$/\)\);/
Test online here.
UPDATE: generic regex that adds a closing parenthesis, matches only lines starting with lr_start_transaction:
s/^\s*(lr_start_transaction.*\))\s*;\s*$/\1\);/
Test online here.
Anyway, if we are not talking about generic regex you have to also specify the language you are coding with (cause each regex engine is thinly different).
Consider also to include some lines that does not have to match.
Based on your initial string, you can select it with this:
/(\("\{scriptid\}_[0-9]{4}_[a-z0-9]{8}_[-a-z0-9]+"\));$/i
The $ matches the end of the line, and the i flag is for case-insensitive.
You can then replace with $1); (To simplify the replacement, I left the ; out of the capture group in the match.
http://refiddle.com/refiddles/5625058d75622d65a01c0000
I have the following:
itemid=44'>Red Flower</a>
I need it to be this:
_ITEMID_START_44_ITEMID_END_
Can this be done with regular expressions? I need to keep the id (44 in the example), and replace everything on the left with _ITEMID_START_and everything on the right with _ITEMID_END_.
Note: The itemid is one digit or two but never no more than two.
I found something about tagged regular expressions and backreferences which seems like it would work but the syntax is killing me.
I tried this (and other attempts):
Find What: ^(\bitemid=\b)^([0-9][0-9]^)\b'>\b[a-z]+\b</a>\b)
Replace With: ^(\b_ITEMID_START_\b^2^(\b_ITEMID_END_\b
I am using UltraEdit to do the find and replace in over 20,000 *.html files. Any help would be very much appreciated.
The solution of Casimir et Hippolyte and also first solution of Avinash Raj work both in UltraEdit with selecting Perl as regular expression engine. The second search string of Avinash Raj requires removing backslash left of character ' in search string to work in UltraEdit.
UltraEdit has 3 regular expression engines: UltraEdit, Unix and Perl.
The search string in the question is a mixture of UltraEdit and Perl regular expression syntax and therefore does not work.
With UltraEdit reguar expression engine:
Find what: itemid=^([0-9]+^)*</a>
Replace with: _ITEMID_START_^1_ITEMID_END_
With Unix or Perl regular expression engine:
Find what: itemid=([0-9]+).*</a>
Replace with: _ITEMID_START_\1_ITEMID_END_
More secure because non greedy, but only with Perl regex engine:
Find what: itemid=(\d+).*?</a>
Replace with: _ITEMID_START_\1_ITEMID_END_
IDM published the power tips tagged expressions for UltraEdit regex engine and Perl regular expressions: Backreferences for Perl regex engine.
You can try this:
Find What: \bitemid=([0-9][0-9]?)'>[^<]*</a>
Replace With: _ITEMID_START_\1_ITEMID_END_
A replacement string is a normal string, and all the regex special characters (except for the backreference) loose their special meaning.
\b the word boundary is the limit between a character that come from the \w character class (a shortcut for [A-Za-z0-9_]) and an other character.
Note: I can't try it with ultraedit, if you obtain a literal \1, replace it with $1
The below regex would match everything and capture only the digits which was just after to the itemid=. And in the replacement part, the whole line is replaced with _ITEMID_START_\1_ITEMID_END_ (\1 represents the first captured group. It may vary for different languages)
.*(?<=\bitemid=)([0-9]{1,2}).*
And the substitution would be,
_ITEMID_START_\1_ITEMID_END_
DEMO
If you just want to replace only,
itemid=44'>Red Flower</a>
with
_ITEMID_START_44_ITEMID_END_
Then your regex would be,
\bitemid=([0-9]{1,2})\'>[^<]*<\/a>
And the substitution would be,
_ITEMID_START_\1_ITEMID_END_
I am still not so used to the vim regex syntax. I have this code:
rename_column :keywords, :textline_two_id_4, :textline_two_id_4
I would like to match the last id with a positive lookahead in VIMs regex syntax.
How would you do this?
\id#=_\d$
This does not work.
This perl syntax works:
id(?=_\d$)
Edit - the answer:
/id\(_\d$\)\#=
Can someone explain the syntax?
If you check the vim help, there is not much to explain: (:h \#=)
\#= Matches the preceding atom with zero width. {not in Vi}
Like "(?=pattern)" in Perl.
Example matches
foo\(bar\)\#= "foo" in "foobar"
foo\(bar\)\#=foo nothing
This should match the last id:
/id\(_\d$\)\#=
save some back slashes with "very magic":
/\vid(_\d$)#=
actually, it looks more straightforward to use vim's \zs \ze:
id\ze_\d$