Return first occurrence of Regex not matched

Question

I have a code like this https://jsfiddle.net/tgvtceg3/

var text = "abc(de192/+£,€.&";
var pattern = new RegExp(/^[0-9a-zA-Z\-\,+&.\/]+$/);
var res = pattern.test(text)
alert(res);

I want that code return first occurrence not matched in regex... for example in thi case I want "("

Any suggest are welcome

It's really not clear what you're asking. Please give more information on the result you're expecting and why this does not work for you — Rory McCrossan, Mar 09 '17 at 08:33
Possible duplicate of [Regular expression to stop at first match](http://stackoverflow.com/questions/2503413/regular-expression-to-stop-at-first-match) — gyre, Mar 09 '17 at 08:54
It is not that dupe. This question is not about matching up to the leftmost occurrence of a pattern. — Wiktor Stribiżew, Oct 05 '19 at 00:11

score 2 · Answer 1 · answered Mar 09 '17 at 08:35

You can have a negative regex to define and then just use string.match(regex) to get all matches.

To get first mismatch, just do result[0]

Regex: /[^0-9a-zA-Z\-\,+&.\/]/g

This will match all characters except mentioned ones.

var text = "abc(de192/+£,€.&";
var pattern = new RegExp(/[^0-9a-zA-Z\-\,+&.\/]/g);
var res = text.match(pattern)
console.log(res);
console.log(res[0]);

Wiktor Stribiżew · Accepted Answer · 2017-03-09T10:49:47.213

You can only use another regex to check the text that made your previous validation regex fail to match. Use something like this:

var text = "abc(de192/+£,€.&";
var pattern = /^[0-9a-zA-Z,+&.\/-]+$/;
var res = pattern.test(text)
if (!res) {
   var m=text.match(/[^0-9a-zA-Z,+&.\/-]+/) || [""];
   console.log(m[0]);
}

The /[^0-9a-zA-Z,+&.\/-]+/ regex will find the first occurrence of the char other than those defined in your original regex.

EDIT: Using a dynamic approach to pattern building:

var text = "abc(de192/+£,€.&";
var block = "0-9a-zA-Z,+&./-"; // Define ranges/chars here
var pattern = new RegExp("^[" + block + "]+$"); // Use constructor notation to build the validation regex
var res = pattern.test(text)
if (!res) {
   var m=text.match(new RegExp("[^" + block + "]+")) || [""];
   console.log(m[0]);
}

Thanks... Do you know how I can change dinamically from this /^[0-9a-zA-Z,+&.\/-]+$/ to this /[^0-9a-zA-Z,+&.\/-]+/... replace? — Francesco G., Mar 09 '17 at 10:45

score 0 · Answer 3 · edited May 23 '17 at 12:00

As described in this question

You need to make your regular expression non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".

So I hope this will help, by adding '?' to the end of your regex to make it nongreedy, and move the ^ to the character section in the regex. for debug regex, I use regex101.com which is great.

var text = "abc(de192/+£,€.&";
var pattern = new RegExp(/[^0-9a-zA-Z\-\,+&.\/]?/);
var res = pattern.test(text)
alert(res);

Moishe Lipsker · Answer 4 · 2017-03-09T08:39:54.460

0

You can just add a ^ into the beginning of the character case to look for any character not inside it:

var text = "abc(de192/+£,€.&";
var pattern = new RegExp(/([^0-9a-zA-Z\-\,+&.\/]+)/);
var res = pattern.exec(text);
console.log(res ? res[0] : undefined);

edited Mar 09 '17 at 08:39

answered Mar 09 '17 at 08:37

Moishe Lipsker

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Sorry but please check other answers as this has been covered in more than 1 answer. – Rajesh Mar 09 '17 at 08:38
@Rajesh This answer also briefly explains the logic as well as handles a case where `res` has no matches (besides for using `exec` instead and omitting the `g` modifier which isn't needed). BTW I wrote this before seeing your answer. – Moishe Lipsker Mar 09 '17 at 08:44

Return first occurrence of Regex not matched

4 Answers4