CODEWITHSUNDEEP
javascriptregex
Yves Calaci
Yves Calaci

Reputation: 1129

Extract double quoted strings from a string, without the quotes itself

Suppose I have a string like 'there goes "something here", and "here" and I have "nothing else to say"'. I need a regex to retrieve exactly ['something here', 'here', 'nothing else to say']. Note that the results aren't quote-escaped, this is because I don't want to make a second call for replace(/"/g, '') on each match.

For reference, this:

'there goes "something here", and "here" and I have "nothing else to say"'.match(/".*?"/g)

Gives me this:

['"something here"', '"here"', '"nothing else to say"']

However, like I said, I dont want the results to be quote-scaped and make a replace call on each result.

Upvotes: 1

Views: 459

Answers (3)

tonitone120
tonitone120

Reputation: 2290

For when there should be a \w after first quotation mark: 

string = 'there goes "something here", and "here" and I have "nothing else to say"';
regexp = /(?<=")\w.*?(?=")/g
result = string.match(regexp);
console.log(result);

Accounting for there to be spacing between quotation mark and content:

let string = 'there goes " something here ", and "here" and I have "nothing else to say"';

function everyOther(string) {

let regexp = /(?<=").*?(?=")/g

let answerArray = [];

for (; ; ) {
    if (regexp.lastIndex !== 0) {
        regexp.lastIndex += 2;
    }
    let matchArr = regexp.exec(string);
    if (regexp.lastIndex === 0) { 
        break;
    } 
    let match = matchArr[0]
    answerArray.push(match);
}

console.log(answerArray);

}

everyOther(string);

Upvotes: 1

AMD
AMD

Reputation: 203

By using some of the latest JS features

let reg = /"(.*?)"/gi;
let results = str.matchAll(reg);

// results - is not an array, but an iterable object
// convert results to an Array
results =  Array.from(results); 

// using map method to get second items from each array item
// which is a captured group
results = results.map(item => item[1]);

More info => javascript.info ::: matchAll with groups

Upvotes: 0

Nadhir Falta
Nadhir Falta

Reputation: 5257

This should work "([^"]*)"
and here is an example test: https://regex101.com/r/CjK2fp/1

Upvotes: 1

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