Regexp search not surrounded by

Question

I want to find all occurences of % that are not within quotation characters.

Example> "test% testing % '% hello' " would return ["%","%"]

Looking at another stack overflow thread this is what I found:

var patt = /!["'],*%,*!['"]/g

var str = "testing 123 '%' % '% ' "
var res = str.match(patt);

However this gives me null. Have you any tips of what I should do?

Demo

What is the point of such search? As far as I believe you want to simply _count_ those occurences? If so, I guess it would be easier without regex. — matewka, Dec 19 '14 at 14:55
So either quote chars? Do you need to accommodate `"bob's shoes %"` — Alex K., Dec 19 '14 at 14:55
Are quotes always balanced? Can quotes be escaped like `\'` also? — anubhava, Dec 19 '14 at 14:57
what would be the output if the input is `foo'bar%'foo"foo%bar"%` ? — Avinash Raj, Dec 19 '14 at 15:01
@Alex K. Yes, either quote characters, no quote chars within the quote. — user1506145, Dec 29 '14 at 09:51

score 2 · Accepted Answer · answered Dec 19 '14 at 14:56

2

You could try the below positive lookahead assertion based regex.

> var s = "test% testing % '% hello' "
> s.match(/%(?=(?:[^']*'[^']*')*[^']*$)/g)
[ '%', '%' ]
> var str = "testing %"
undefined
> str.match(/%(?=(?:[^']*'[^']*')*[^']*$)/g)
[ '%' ]
> var str1 = "testing '%'"
undefined
> str1.match(/%(?=(?:[^']*'[^']*')*[^']*$)/g)
null

answered Dec 19 '14 at 14:56

Avinash Raj

172,303
28
230
274

This would capture `\"$\"` – Alex K. Dec 19 '14 at 14:59
@AlexK. he wants the `%` which are present inside the single quotes. – Avinash Raj Dec 19 '14 at 15:00
His example uses `["']` – Alex K. Dec 19 '14 at 15:00
This works great! Could you please provide an explanation for your answer? – user1506145 Dec 29 '14 at 10:33
@user1506145, for explanation you might use [this online tool](https://www.regex101.com/) – Igor Savinkin Dec 29 '14 at 10:38

score 1 · Answer 2 · edited May 23 '17 at 10:26

1

Try this:

var patt=  /[^"'].*?(%).*?[^'"]/g ;
var str = "testing 123 '%' % '% ' "
var res = str.match(patt);
console.dir(res[1]); // result will be in the 1st match group: res[1]

Here is the link to the online testing.

Explanation:

[^"'] - any character except " or '
.*? any characters (except new line) any times or zero times not greedy.

Update

Actually you must check if behing and ahead of % there are no quotes. But:

JavaScript regular expressions do not support lookbehinds

So you have no way to identify " or ' preceding % sign unless more restrictions are applied.

I'd suggest to do searching in php or other language (where lookbehind is supported) or impose more conditions.

edited May 23 '17 at 10:26

Community

1
1

answered Dec 19 '14 at 14:48

Igor Savinkin

5,669
8
37
69

@matewka, do you use JS regex flavour? – Igor Savinkin Dec 19 '14 at 14:54
Thanks, what if you have this string str = "testing %". It will not be matched. – user1506145 Dec 19 '14 at 14:57
Would match `testing % '%' ` twice – Alex K. Dec 19 '14 at 14:58
@user1506145, update answer and one new version: https://www.regex101.com/r/hU6yO3/5 – Igor Savinkin Dec 19 '14 at 15:05
@Igor Savinkin This will not be matched: http://plnkr.co/edit/1dbCZhII0ggbKp0jKxQ9?p=preview – user1506145 Dec 29 '14 at 10:05
@user1506145, yes, this is not the solution. You read my update about js regex lookahead. – Igor Savinkin Dec 29 '14 at 10:40

matewka · Answer 3 · 2014-12-20T12:21:47.453

Since I'm not a big fan of regular expressions, here's my approach.
What is important in my answer, if there would be a trailing quote in the string, the other answers won't work. In other words, only my answer works in cases where there is odd number of quotes.

function countUnquoted(str, charToCount) {
    var i = 0,
        len = str.length,
        count = 0,
        suspects = 0,
        char,
        flag = false;

    for (; i < len; i++) {
        char = str.substr(i, 1);

        if ("'" === char) {
            flag = !flag;
            suspects = 0;
        } else if (charToCount === char && !flag) {
            count++;
        } else if (charToCount === char) {
            suspects++;
        }
    }

    //this way we can also count occurences in such situation
    //that the quotation mark has been opened but not closed till the end of string
    if (flag) {
        count += suspects;
    }

    return count;
}

As far as I believe, you wanted to count those percent signs, so there's no need to put them in an array.

In case you really, really need to fill this array, you can do it like that:

function matchUnquoted(str, charToMatch) {
  var res = [],
      i = 0,
      count = countUnquoted(str, charToMatch);

  for (; i < count; i++) {
    res.push('%');
  }

  return res;
}

matchUnquoted("test% testing % '% hello' ", "%");

Trailing quote

Here's a comparison of a case when there is a trailing ' (not closed) in the string.

> var s = "test% testing % '% hello' asd ' asd %"
> matchUnquoted(s, '%')
['%', '%', '%']
>
> // Avinash Raj's answer
> s.match(/%(?=(?:[^']*'[^']*')*[^']*$)/g)
['%', '%']

score 0 · Answer 4 · answered Dec 19 '14 at 18:12

Use this regex: (['"]).*?\1|(%) and the second capture group will have all the % signs that are not inside single or double quotes.

Breakdown:

(['"]).*?\1 captures a single or double quote, followed by anything (lazy) up to a matching single or double quote

|(%) captures a % only if it wasn't slurped up by the first part of the alternation (i.e., if it's not in quotes)

Regexp search not surrounded by

4 Answers4

Update

Trailing quote