can't remove parameter using javascript

Question

i have a url like this

test.html?dir=asc&end_date=2016-09-23&order=created_at&start_date=2016-08-14

i want to remove the parameter using the following javascript

function removeParam(uri) {

   uri =  uri.replace(/([&\?]start_date=*$|start_date=*&|[?&]start_date=(?=#))/, '');

   return uri.replace(/([&\?]end_date=*$|end_date=*&|[?&]end_date=(?=#))/, '');
}

but it didn't work, anyone know what's wrong with that?

What is the final output you want to get? `test.html`? Just find the index of `?` and get the substring before it. — Wiktor Stribiżew, Sep 26 '16 at 09:05
in regex ... =* is 0 or more `=` ... perhaps you meant `=.*` — Jaromanda X, Sep 26 '16 at 09:05
Sidenote: Escaping `?` inside character class is unnecessary. — hindmost, Sep 26 '16 at 09:08
I used to describe it the same but actually it is more or zero @JaromandaX — revo, Sep 26 '16 at 09:10
Well, it's not about how I and you describe it but about how it works in fact. @JaromandaX — revo, Sep 26 '16 at 09:20
That's why I said *I used to...* since it's a common phrase across articles and docs around Regular Expressions and is easily understandable and imaginal for everyone, mostly beginners, who refers to them in the first place. And since we are talking programmatically, in programming context their order differs a lot. @JaromandaX — revo, Sep 26 '16 at 09:36
Oh my god, you made me sad by removing your comments. @JaromandaX — revo, Sep 26 '16 at 09:41

score 0 · Answer 1 · answered Sep 26 '16 at 09:15

in modern browsers you can do this quite simply

var x = new URL(location.origin + '/test.html?dir=asc&end_date=2016-09-23&order=created_at&start_date=2016-08-14');
x.searchParams.delete('start_date');
x.searchParams.delete('end_date');
var uri = x.pathname.substr(1) + x.search; // substr(1) because you don't have a leading / in your original uri

at least, I think it's simpler

score 0 · Answer 2 · answered Sep 26 '16 at 09:34

0

Your RegExp is no match!

If you want remove end_date, you should:

uri.replace(/(end_date=)([^*&]+)/, 'end_date=')

And so on.

answered Sep 26 '16 at 09:34

Fry

1

can't remove parameter using javascript

2 Answers2