JQuery .hasClass for multiple values in an if statement

Question

I have a simple if statement as such:

if ($(\'html\').hasClass(\'m320\')) {

// do stuff 

}

This works as expected. However, I want to add

Answer 1

This may be another solution:

if ($('html').attr('class').match(/m320|m768/)) {  
  // do stuff   
}  

according to jsperf.com it's quite fast, too.

Answer 2

For fun, I wrote a little jQuery add-on method that will check for any one of multiple class names:

$.fn.hasAnyClass = function() {
    for (var i = 0; i < arguments.length; i++) {
        if (this.hasClass(arguments[i])) {
            return true;
        }
    }
    return false;
}

Then, in your example, you could use this:

if ($('html').hasAnyClass('m320', 'm768')) {

// do stuff 

}

You can pass as many class names as you want.

---

Here's an enhanced version that also lets you pass multiple class names separated by a space:

$.fn.hasAnyClass = function() {
    for (var i = 0; i < arguments.length; i++) {
        var classes = arguments[i].split(" ");
        for (var j = 0; j < classes.length; j++) {
            if (this.hasClass(classes[j])) {
                return true;
            }
        }
    }
    return false;
}

if ($('html').hasAnyClass('m320 m768')) {
    // do stuff 
}

Working demo: http://jsfiddle.net/jfriend00/uvtSA/

Answer 3

var classes = $('html')[0].className;

if (classes.indexOf('m320') != -1 || classes.indexOf('m768') != -1) {
    //do something
}

Answer 4

You could use is() instead of hasClass():

if ($('html').is('.m320, .m768')) { ... }

Answer 5

Try this:

if ($('html').hasClass('class1 class2')) {

// do stuff 

}

Answer 6

Here is a slight variation on answer offered by jfriend00:

$.fn.hasAnyClass = function() {
    var classes = arguments[0].split(" ");
    for (var i = 0; i < classes.length; i++) {
        if (this.hasClass(classes[i])) {
            return true;
        }
    }
    return false;
}

Allows use of same syntax as .addClass() and .removeClass(). e.g., .hasAnyClass('m320 m768') Needs bulletproofing, of course, as it assumes at least one argument.