converting RegExp to String then back to RegExp

Question

So I have a RegExp regex = /asd/

I am storing it as a as a key in my key-val store system.

So I say str = String(regex) which returns "/asd/".

Now I need to convert that string back to a RegExp.

So I try: RegExp(str) and I see /\/asd\//

this is not what I want. It is not the same as /asd/

Should I just remove the first and last characters from the string before converting it to regex? That would get me the desired result in this situation, but wouldn't necessarily work if the RegExp had modifiers like /i or /g

Is there a better way to do this?

Does (new RegExp(regex)).source() help before storage? developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — jedifans
– jedifans, Commented Aug 25, 2016 at 20:34
@jedifans I see what you're saying, yeah that should work. That way I'm not storing the slashes in the string — max pleaner
– max pleaner, Commented Aug 25, 2016 at 20:38
In ES6 there is also flags along with source, but it's not widely supported yet. — Xotic750
– Xotic750, Commented Aug 25, 2016 at 21:15
Well, to be fair, I didn't read any of the answers. From like 15 years of Perl, I've gleaned that pre-compiled regexes of the form /../xxx, can't contain all the modifiers available at the time of execution. If it can't do that, it's better to have it all available in a hash, or if it applies to all of them (like 'g') then it doesn't matter. Usually, you'd take the path of least resistance, but if you don't know if it works for every thing, it's better to be conservative and slightly basic. The foundation of good code. — user557597
– user557597, Commented Aug 25, 2016 at 21:15

castletheperson · Accepted Answer · 2016-08-25 21:48:02Z

75

If you don't need to store the modifiers, you can use Regexp#source to get the string value, and then convert back using the RegExp constructor.

var regex = /abc/g;
var str = regex.source; // "abc"
var restoreRegex = new RegExp(str, "g");

If you do need to store the modifiers, use a regex to parse the regex:

var regex = /abc/g;
var str = regex.toString(); // "/abc/g"
var parts = /\/(.*)\/(.*)/.exec(str);
var restoredRegex = new RegExp(parts[1], parts[2]);

This will work even if the pattern has a / in it, because .* is greedy, and will advance to the last / in the string.

If performance is a concern, use normal string manipulation using String#lastIndexOf:

var regex = /abc/g;
var str = regex.toString(); // "/abc/g"
var lastSlash = str.lastIndexOf("/");
var restoredRegex = new RegExp(str.slice(1, lastSlash), str.slice(lastSlash + 1));

edited Aug 25, 2016 at 21:48

answered Aug 25, 2016 at 20:42

castletheperson

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6 Comments

castletheperson Over a year ago

I suppose it's also worth mentioning you could just do var restoredRegex = eval(str); but that will be slower, and has the ability to execute arbitrary JavaScript.

Xotic750 Over a year ago

How about regex.toString().split('/')

castletheperson Over a year ago

@Xotic750 That will fail if a / is in the regex pattern itself

Pegasus Epsilon Over a year ago

So close to greatness -- var restoredRegex = RegExp.apply(RegExp, str.match(/\/(.*)\/(.*)/));

Mörre Over a year ago

A different problem: Using new RegExp('\n', 'g') gives me a different string result than /\n/g. Looks like the backslash has to be escaped in the string. Also, I think one can also use var regex = /abc\n/g; var str = regex.source; var flags = regex.flags; var restoredRegex = new RegExp(str, flags); -- I think this is cleaner than the string parsing. Of course, it is two strings instead of just one.

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Paul Draper · Accepted Answer · 2020-09-15 13:08:46Z

24

const regex = /asd/gi;

converting RegExp to String

const obj = {flags: regex.flags, source: regex.source};
const string = JSON.stringify(obj);

then back to RegExp

const obj2 = JSON.parse(string);
const regex2 = new RegExp(obj2.source, obj2.flags);

Requires ES6+.

answered Sep 15, 2020 at 13:08

Paul Draper

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Comments

jedifans · Accepted Answer · 2016-08-25 20:46:15Z

2

You can use the following before storage of your regex literal:

(new RegExp(regex)).source

See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source

Example:

regex = /asd/

string = (new RegExp(regex)).source
// string is now "asd"

regex = RegExp(string)
// regex has the original value /asd/

edited Aug 25, 2016 at 20:46

answered Aug 25, 2016 at 20:40

jedifans

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4 Comments

Wiktor Stribiżew Over a year ago

And if it is regex = /asd/ig? Also, it should be source, not source().

castletheperson Over a year ago

Why are you calling new RegExp? It's already a regex. Wouldn't you just do string = regex.source?

jedifans Over a year ago

True @WiktorStribiżew, source does not take account of that, though I suppose .flags could be stored too. @4castle, I wasn't sure if a regex literal was actually a RegExp instance.

castletheperson Over a year ago

Ah, well yes. A regex literal is syntactic sugar for making a RegExp instance.

Gilad Novik · Accepted Answer · 2018-07-10 12:24:04Z

1

let rx = RegExp.apply(RegExp, str.match(/\/(.*)\/(.*)/).slice(1));

A modified version of @PegasusEpsilon answer

edited Jul 10, 2018 at 12:24

answered Jul 9, 2018 at 20:28

Gilad Novik

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1 Comment

tsh Over a year ago

ES6+ syntax should be RegExp(...str.match(/\/(.*)\/(.*)/).slice(1))

Kingsley · Accepted Answer · 2020-08-21 17:42:19Z

StackOverflow saves the day again, thanks @4castle! I wanted to store some regex rules in a JS file, and some in a DB, combine them into an array of objects like so:

module.exports = {
    [SETTINGS.PRODUCTION_ENV]: [

        {
            "key": /<meta name="generator"[\s\S]*?>/gmi,
            "value": "",
            "regex": true
        },

        ...
    ]
}

Then, loop through each environment's objects and apply it to a string of text. This is for a node/lambda project, so I wanted to use ES6. I used @4castle's code, with some destructuring, and I ended up with this:

    let content = body;
    const regexString = replacement.key.toString();
    const regexParts = /\/(.*)\/(.*)/.exec(regexString);
    const {1: source, 2: flags} = regexParts;
    const regex = new RegExp(source, flags);
    content = content.replace(regex, replacement.value);
    
    return content;

Works a treat!

vinays84 · Accepted Answer · 2023-12-29 15:45:57Z

0

Another option based on previous answers:

const strToRegex = str => new RegExp(str.slice(1,str.lastIndexOf('/')), str.slice(str.lastIndexOf('/')+1));
let initRegex = /asd/i;
let newRegex = strToRegex(initRegex.toString());

answered Dec 29, 2023 at 15:45

vinays84

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