Is there a Regex expression to match every that is not in quotations

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I'm working on a Regular Expression that looks at everything except information in quotations. An example would be:

Lorem ipsum "dolor" sit amet, "consectetur" adipiscing elit.

Would change to

Lorem ipsum sit amet, adipiscing elit

I've already tried

.*(?=".*").*

and

`[^"].*[^"]

neither worked so I'd appreciate help.

Heres a way ...

var str = `Lorem ipsum "dolor" sit amet, "consectetur" adipiscing elit.`

console.log(str.replace(/\"[a-z]{1,}\"/g, ''))

Double Quotes and Regular Expressions, Learn more about matlab regexp extract between quotes only. strfind() for '"', removing everything up to the first match and everything from the second match on. Which it could not be because it was only non-double-quotes in that pattern​. In the example above, it should match the spaces outside of quotations, but not the ones inside. I will use this find and replace to remove all whitespace not it quotes, in order to minify a string of JSON inside of Siri Shortcuts.

I would do something in the line of this. It handles the case where the sentence starts or ends with a quotation mark.

var str = '"Lorem" ipsum "dolor" sit amet, "consectetur", adipiscing "elit."';
var regExp = /(?:^|\W)(\".+?\")(?:\W|$)/g;
var res = str.replace(regExp, " ").trim();
console.log(res);

Extract only text between quotes of a string, We have a text and need to replace all quotes ". These common words do not make it obvious why the regexp fails, so let's elaborate how the The regular expression engine adds to the match one character after another. For more information about regular expressions, see .NET Framework Regular Expressions and Regular Expression Language - Quick Reference. You can determine whether the regular expression pattern has been found in the input string by checking the value of the returned Match object's Success property.

Try this , non-greedy (".*?") match two quotations and as few as possible between them.

var a ='Lorem ipsum "dolor" sit amet, "consectetur" adipiscing elit.';
console.log(a.replace(/".*?"/g,''));

Greedy and lazy quantifiers, A string can be classified as either matching or not matching the pattern. where the quoted string cannot contain any embedded double quotes; the pattern ". Regular expressions (regex or regexp) are extremely useful in extracting information from any text by searching for one or more matches of a specific search pattern (i.e. a specific sequence of

I hope to help you :

/[,\w]+(?![^" ]*")/g

var str = 'Lorem ipsum "dolor" sit amet, "consectetur" adipiscing elit.' ;
var res = '' ;
var patt = /[,\w]+(?![^" ]*")/g ;

while ( (arr = patt.exec(str)) != null) { res += arr[0] + ' ' ; }

console.log( res ) ;

Regular Expressions on the Computer, r/regex: Helping programmers overcome their regular expression obstacles. In this case, I do not want it to match since the = is inside a quote but I do not want to Hey there I need a regex to match any non-word character except for accent  A regular expression is a pattern that the regular expression engine attempts to match in input text. A pattern consists of one or more character literals, operators, or constructs. For a brief introduction, see .NET Regular Expressions.

Is it possible to ignore characters in between quotes? : regex, Matching these is not much more difficult if we handle the quotes separately. We end the regex with the double quote that closes the string. match. Then it should search for all pairs of double quotes and replace them with individual double  The (?! invalidates the whole match, so finding info before or after that will not occur since the match has been invalidated. I recommend that you break out the steps of this process such as 1) determine if the condition exists and if it does do what needs to be done with a separate regex(?), otherwise match what needs to be matched in a specific regex for that condition.

Regular Expressions Cookbook, It is not a tutorial, so if you're unfamiliar regular expressions, I'd recommend starting at The next step up in complexity is . , which matches any character except a newline: str_extract(x, ".a. An alternative quoting mechanism is \Q\E : all the  The regex does not need to count up an arbitrary number of quotes -- it only needs to count to 3, which regular expressions can do. As others have mentioned, you should try to write down a well-defined representation of what you expect a csv token to be – comingstorm Sep 27 '12 at 17:53 show 1 more comment 5 Answers

Regular expressions, It is heavily used for string matching / replacing in all programming languages, the single quote in the pattern, by preceding it with , so it's clear it is not part of  This is our opening chunk, which essentially matches any single or double quote, unless that quote is preceded by a backslash. That's a "Negative Lookbehind", and the (?<!) part does not consume any characters. If we didn't care to be careful about erroneously matching escaped quotes, it could simply be this:

Comments
  • Use replace i.e. str.replace(/"[^"]*"\s*/, "")
  • Or, s.split(/\s*"[^"]*"\s*/).join(" ").trim()