Change variable value if document.URL contains array element

Question

I have a javascript/jquery function that displays a notification. I want to make this function display a different mode if they aren't on x, y and z page.

This is how I tried to achieve this:

function display_alert(message, type, delay, mode)
{
    type    = (typeof type === "undefined") ? "danger" : type;
    delay   = (typeof delay === "undefined") ? 3000 : delay;
    mode    = (typeof mode === "undefined") ? 'normal' : mode;

    var current_location = document.URL;
    var home_locations = ['home', 'remote', 'zip'];

    for (var i = 0; i < home_locations.length; i++)
    {
        if (current_location.toString().indexOf(home_locations[i]) == -1)
        {
            // alert(home_locations[i]); return;

            mode = 'top';
            break;
        }   
    }

    ...

So if document.URL doesn't contain one of the array elements, then I want the mode variable to become top.

I think this is a simple problem, but I just can't see how to fix it.

Sylvain · Accepted Answer · 2014-02-10 23:40:37Z

1

You could use a regular expression built with your home_locations array :

var home_regex = new RegExp('('+home_locations.join('|')+')');
// home_regex = /(home|remote|zip)/;
if (!home_regex.test(document.URL)) mode = 'top';

answered Feb 10, 2014 at 23:40

Sylvain

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Comments

Anthony Chu · Accepted Answer · 2014-02-10 23:38:41Z

0

You need to reverse the logic in your loop...

var current_location = document.URL;
var home_locations = ['home', 'remote', 'zip'];

var found = false;

for (var i = 0; i < home_locations.length; i++)
{
    if (current_location.toString().indexOf(home_locations[i]) >= 0)
    {
        found = true;
        break;
    }   
}

if (!found) {
    mode = 'top';
}

answered Feb 10, 2014 at 23:38

Anthony Chu

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