node.js - Removing utm_* parameters from URL in javascript with a regex - Stack Overflow

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I did not find any good answer to this question so I share what I found and works

if you want to remove all the google analytics terms from an URL, you mostly want to keep the other parameters and get a clean valid URL at the end

url = url.replace(/(\&|\?)utm([_a-z0-9=+\-]+)/igm, "$1");

with a url like this ;utm_medium=affiliation&utm_content=catalogue-RDC&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249

you will get this ?&&&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249

this url is already valid but we have some dupe & signs if you remove the $1 from the first request you will with only a & sign and not the ? that you should have in the beginning

so next clean up we keep the first ? sign => $1 and remove the other leading &

url = url.replace(/(\?)\&+/igm, "$1");

here we have a nice clean URL

full version :

url = url.replace(/(\&|\?)utm([_a-z0-9=+\-]+)/igm, "$1");
url = url.replace(/(\?)\&+/igm, "$1");

if you can find a one liner you're wele

Edit : the resulting URL should be this one :

I did not find any good answer to this question so I share what I found and works

if you want to remove all the google analytics terms from an URL, you mostly want to keep the other parameters and get a clean valid URL at the end

url = url.replace(/(\&|\?)utm([_a-z0-9=+\-]+)/igm, "$1");

with a url like this https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?utm_source=325483&utm_medium=affiliation&utm_content=catalogue-RDC&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249

you will get this https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?&&&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249

this url is already valid but we have some dupe & signs if you remove the $1 from the first request you will with only a & sign and not the ? that you should have in the beginning

so next clean up we keep the first ? sign => $1 and remove the other leading &

url = url.replace(/(\?)\&+/igm, "$1");

here we have a nice clean URL

full version :

url = url.replace(/(\&|\?)utm([_a-z0-9=+\-]+)/igm, "$1");
url = url.replace(/(\?)\&+/igm, "$1");

if you can find a one liner you're wele

Edit : the resulting URL should be this one : https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249

• javascript
• node.js
• regex
• url
• clean-urls

Share Improve this question Follow edited Jul 5, 2018 at 10:04 benraay asked Jul 5, 2018 at 9:13 benraaybenraay 86399 silver badges1515 bronze badges 3

• What would be the resulting correct url? – Jorge.V Commented Jul 5, 2018 at 9:36
• Good point I have edited my question – benraay Commented Jul 5, 2018 at 10:05
• @benraay /(?<=&|\?)utm_.*?(&|$)/igm does not remove the trailing ? if the query string only contains utm params. It is also not portable across JS environments that do not support ECMAScript 2018 standard. – Wiktor Stribiżew Commented Jul 5, 2018 at 10:19

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 7

I think it could be as simple as: url = url.replace(/(?<=&|\?)utm_.*?(&|$)/igm, "");

You do not need to escape &

(?<=&|\?) = positive lookbehind

.*? = everything, but "not greedy"

You may use a single regex patible with all JS versions that will

• match and capture ? that is followed by 1 or more utm param that are followed with a param other than utm one and replace with $1 to restore that ? since it is necessary
• or, match any ? with 1 or more utm params in the query string where no params other than utm are present (so, $1 will be empty, and ? will get removed)
• or, just match all utm params to remove them.

The regex will look like

.replace(/(\?)utm[^&]*(?:&utm[^&]*)*&(?=(?!utm[^\s&=]*=)[^\s&=]+=)|\?utm[^&]*(?:&utm[^&]*)*$|&utm[^&]*/gi, '$1')

See the regex demo

Details

• (\?)utm[^&]*(?:&utm[^&]*)*&(?=(?!utm[^\s&=]*=)[^\s&=]+=) - ?utm (with ? inside a capturing group later referenced with $1), 0+ chars other than &, and then 0 or more repetitions of &utm followed with 0+ chars other than & and then a & that is followed with 0+ chars other than whitespace, & and = and then = that is not utm param
• | - or
• \?utm[^&]*(?:&utm[^&]*)*$ - ?utm, 0+ chars other than &, and then 0 or more repetitions of &utm followed with 0+ chars other than & and then the end of the string
• | - or
• &utm[^&]* - a &, utm and then 0+ chars other than &

JS demo:

var urls = ['https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?utm_source=325483&utm_medium=affiliation&utm_content=catalogue-RDC&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249', 'https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?t=55&utm_source=325483&utm_medium=affiliation&utm_content=catalogue-RDC&awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249','https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?awc=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249&utm_tt=78', 'https://www.somewebsite.fr/produit/yi-camera-3600-noir-vr-33705370/offre-81085802?utm=6901_1530705916_88ef12642ad61dfc5239ba01bbbe5249&utm=ewe'];

var u = 'utm[^&]*';
var rx = new RegExp("(\\?)"+u+"(?:&"+u+")*&(?=(?!utm[^\s&=]*=)[^\s&=]+=)|\\?"+u+"(?:&"+u+")*$|&"+u, "ig");
for (var url of urls) {
  console.log(url, "=>", url.replace(rx, '$1'));
}

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