How to get URL parameters from custom input with JQuery

Question

I have a view that I want to store some URL parameters and have tried everything. Here is my current code.

Sample URL: https://www.facebook.com?username=""&password=""(I'm trying to collect the parameter values)

HTML:

<input id="urlInput" type="text" placeholder="Enter URL" class="form-control" />

Javascript:

var url = $("#urlInput").val();//This pulls the value of the input/URL with parameters
var getURLUser = url.substring(url.indexOf("?"));
$("#urluser").html(getURLUser);

What is wrong? Thanks.

Check this link jquerybyexample.net/2012/06/…

Arun Kumar
– Arun Kumar

2018-09-11 15:47:38 +00:00
Commented Sep 11, 2018 at 15:47 — Arun Kumar
– Arun Kumar, Commented Sep 11, 2018 at 15:47
Also this might be help you.

uedemir
– uedemir

2018-09-11 16:04:05 +00:00
Commented Sep 11, 2018 at 16:04 — uedemir
– uedemir, Commented Sep 11, 2018 at 16:04

Leo · Accepted Answer · 2018-09-11 15:58:43Z

3

Currently, you're getting a String from the beginning of ? to the end of the String.

You could split the string by &password=, and get the right side of it:

var url = "https://www.facebook.com?username=123&password=(I'm trying to collect the parameter values)";
var getURLUser = url.split("&password=")[1];
// results in "(I'm trying to collect the parameter values)"
console.log(getURLUser);

or

var url = "https://www.facebook.com?username=123&password=(I'm trying to collect the parameter values)";
var getURLUser = url.split("&password=");
console.log(getURLUser[0]);
// results in "https://www.facebook.com?username=123"
console.log(getURLUser[1]);
// results in "(I'm trying to collect the parameter values)"

edited Sep 11, 2018 at 15:58

Leo

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answered Sep 11, 2018 at 15:46

dustytrash

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3 Comments

Mathew Dodson Over a year ago

I've tried adjusting the numbers, but to no avail. How would you get just the username in the middle? Right now, I have "var geturlUser = url.split("?username=")[1] && url.split("&password=")[0];"//After the username and before the password?? Same Example applies.

dustytrash Over a year ago

@MathewDodson url.split('?username=')[1].split('&password')[0];

Mathew Dodson Over a year ago

Works like a charm. Thank you.

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