jQuery extract specified substring and following numbers from string

Question

I need to extract a specified substring and then the numbers that follow that substring.

For example, I need to extract cid=159 from the string: http://website.com/ProductCats.asp?cid=159&otherjunk=1. Even though cid= remains contant, the numbers that follow can change, and can be anywhere between 1-4 digits.

Is there any way that I can extract that pattern?

Thanks in advanced.

Yes, first split it on cid=, then split the second half of that by & and get the first half. — user400654
– user400654, Commented Feb 22, 2013 at 19:55

Iron3eagle · Accepted Answer · 2013-02-22 19:57:52Z

2

This might help. I can't remember where I found it, but it's a function that gets the URL and assigns the variables to an array that can be referenced. Also this is pure javascript.

function getUrlVars() {
    var vars = [], hash;
    var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
    for(var i = 0; i < hashes.length; i++)     {
        hash = hashes[i].split('=');
        vars.push(hash[0]);
        vars[hash[0]] = hash[1];
    }
    return vars;
}

So you would do

var myURLArray = getUrlVars();
var cid = myURLArray['cid'];

It's helped me a lot. Again I didn't make this, but it's useful.

answered Feb 22, 2013 at 19:57

Iron3eagle

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Tim Withers · Accepted Answer · 2013-02-22 20:03:42Z

0

Using just javascript for any url, not just the current window URL if needed. I borrowed the getLocation function from a previous SO question related to this.

var getLocation = function(href) {
    var l = document.createElement("a");
    l.href = href;
    return l;
};
var getUriObject = function(href){
    var l = getLocation(href);
    var queryArray=l.search.substr(1).split('&');
    var uriObject={};
    for(var i=0;i<queryArray.length;i++){
        uriObject[queryArray[i].split('=')[0]]=queryArray[i].split('=')[1];
    }
    return uriObject;
}
alert(getUriObject("http://website.com/ProductCats.asp?cid=159&otherjunk=1").cid);

answered Feb 22, 2013 at 20:03

Tim Withers

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