How to check if the input string is a valid Regular expression?

JavascriptRegex

Javascript Problem Overview


How do you check, in JavaScript, if a string is a proper regular expression that will compile?

For example, when you execute the following javascript, it produces an error.

var regex = new RegExp('abc ([a-z]+) ([a-z]+))');
// produces:
// Uncaught SyntaxError: Invalid regular expression: /abc ([a-z]+) ([a-z]+))/: Unmatched ')'

How does one determine if a string will be a valid regex or not?

Javascript Solutions


Solution 1 - Javascript

You can use try/catch and the RegExp constructor:

var isValid = true;
try {
    new RegExp("the_regex_to_test_goes_here");
} catch(e) {
    isValid = false;
}

if(!isValid) alert("Invalid regular expression");

Solution 2 - Javascript

Here is a little function that checks the validity of both types of regexes, strings or patterns:

function validateRegex(pattern) {
    var parts = pattern.split('/'),
        regex = pattern,
        options = "";
    if (parts.length > 1) {
        regex = parts[1];
        options = parts[2];
    }
    try {
        new RegExp(regex, options);
        return true;
    }
    catch(e) {
        return false;
    }
}

The user will be able to test both test and /test/g for example. [Here][1] is a working fiddle. [1]: http://jsfiddle.net/nmuzx/

Solution 3 - Javascript

The question is solved, but if someone needs to define is the string either valid RegExp or not a RegExp at all.

You can use new Function() and templating inside of the function body with try ... catch and new RegExp() as mentioned earlier.

There is a snippet with the explanations:

const isRegExp = (string) => {
    try {
        return new Function(`
            "use strict";
            try {
                new RegExp(${string});
                return true;
            } catch (e) {
                return false;
            }
        `)();
    } catch(e) {
        return false;
    }
};

// Here the argument 'simplyString' shall be undefined inside of the function
// Function(...) catches the error and returns false
console.log('Is RegExp valid:', isRegExp('simplyString'));

// Here the argument shall cause a syntax error
// isRegExp function catches the error and returns false
console.log('Is RegExp valid:', isRegExp('string which is not a valid regexp'));

// Here the argument is not a valid RegExp, new RegExp(...) throws an error
// Function(...) catches the error and returns false
console.log('Is RegExp valid:', isRegExp('abc ([a-z]+) ([a-z]+))'));

// Valid RegExp, passed as a string
console.log('Is RegExp valid:', isRegExp('/^[^<>()[\]\\.,;:\s@\"]$/'));

// Valid RegExp, passed as a RegExp object
console.log('Is RegExp valid:', isRegExp(/^[^<>()[\]\\.,;:\s@\"]$/));

// Howewer, the code injection is possible here
console.log('Is RegExp valid:', isRegExp(');console.log("This is running inside of the Function(...) as well"'));

Solution 4 - Javascript

this function could handle the '/' char as a normal char in regex, and also consider escaping when is a common string. it will always return an Regex, null if not a good regex string.

function getRegex(regex) {
    try {
        regex = regex.trim();
        let parts = regex.split('/');
        if(regex[0] !== '/' || parts.length< 3){
          regex = regex.replace(/[.*+\-?^${}()|[\]\\]/g, '\\$&'); //escap common string
          return new RegExp(regex);
        }

        const option =parts[parts.length - 1];
        const lastIndex = regex.lastIndexOf('/');
        regex = regex.substring(1, lastIndex);
        return new RegExp(regex, option);
    } catch (e) {
        return null
    }
}

console.log(getRegex('ab/c'))
let htmlStartTag = getRegex('/<(?!/)(?!br)(.+?)(?<!/)>/mgs');
console.log(htmlStartTag)
let result = `</button><input id="warehouse-search-field-tablet"
class="form-control search-field"
 title="Warehouse Search Field" name="location" type="search" value="">content`.matchAll(htmlStartTag);
 console.log([...result])

Solution 5 - Javascript

None of the answers here satisfied my need for checking that a string is a valid regex for other languages (Mostly php), because they either ignore the flags, the delimiters or escaped special characters, so I made my own function

function isValidRegex(s) {
  try {
    const m = s.match(/^([/~@;%#'])(.*?)\1([gimsuy]*)$/);
    return m ? !!new RegExp(m[2],m[3])
        : false;
  } catch (e) {
    return false
  }
}

console.log(isValidRegex('abc')) //False
console.log(isValidRegex('/abc/')) //True
console.log(isValidRegex('/ab#\/[c]/ig')) //True
console.log(isValidRegex('@ab#\/[c]@ig')) //Special delimiters: True
console.log(isValidRegex('/ab\/[c/ig')) //False
console.log(isValidRegex('/abc/gig')) //False

You can also derive this function to transform a string into a RegExp object

function stringToRegex(s) {
   const m = s.match(/^([/~@;%#'])(.*?)\1([gimsuy]*)$/);
   return m ? new RegExp(m[2], m[3]) : new RegExp(s);
}

console.log(stringToRegex('abc'))
console.log(stringToRegex('/abc/'))
console.log(stringToRegex('/ab#\/[c]/ig'))
console.log(stringToRegex('@ab#\/[c]@ig'))
try {
  console.log(stringToRegex('/ab#\/[c/ig'))
} catch (e) {
  console.log('Not a valid regex')
}

Solution 6 - Javascript

function isRegExp(regExp){
          try {
                new RegExp(regExp);
              } catch(e) {
                return false
              }
         return true
    }

ex:
isRegExp(/@(\w+)/g) = true

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJak SamunView Question on Stackoverflow
Solution 1 - JavascriptJonView Answer on Stackoverflow
Solution 2 - JavascriptNiccolò CampolungoView Answer on Stackoverflow
Solution 3 - JavascriptEmma PaulowiczView Answer on Stackoverflow
Solution 4 - JavascriptJason SongView Answer on Stackoverflow
Solution 5 - JavascriptTofandelView Answer on Stackoverflow
Solution 6 - JavascriptRahul ShahareView Answer on Stackoverflow