How to include optional parameters in jquery plugin

Question

(function ( $ ) {
    $.fn.greenify = function( options ) {
        // This is the easiest way to have default options.
        var settings = $.extend({
            // These are the defaults.
            color: "#556b2f",
            backgroundColor: "white"
        }, options );
        // Greenify the collection based on the settings variable.
        return this.css({
            color: settings.color,
            backgroundColor: settings.backgroundColor
        });
    };
}( jQuery ));

Taken from Docs the above code shows how to make a simple jquery plugin. What if I only supply one of multiple parameters listed in a plugin that accepts multiple? Which ones are applied?

If I have a plugin that can take multiple parameters, which ones will be applied in something like this:

$.doStuff({color:"black"},{color:"blue"});

What set of options will be "extended" when only one set of parameters if supplied?

E.x. $.doStuff({color:"black"});

well, typically you won't have multiple parameters, instead you'll have one object with multiple properties. — user400654
– user400654, Commented Jan 22, 2014 at 22:13

Barmar · Accepted Answer · 2014-01-22 22:13:31Z

1

If you call a function with more parameters than there are argument variables, and the function doesn't use the special arguments variable to access variable-length arguments, the extra parameters will be ignored.

So in your example:

$.greenify({color: "black"}, {color: "blue"});

the options variable will be set to {color: "black"}, and the {color: "blue"} parameter will be ignored.

answered Jan 22, 2014 at 22:13

Barmar

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3 Comments

Craig Lafferty Over a year ago

And does that logic follow most cases? So if I have 5 parameters and the plugin accepts 6 the last one is ignored?

Barmar Over a year ago

Yes. Doesn't that fit the description "more parameters than argument variables"?

Craig Lafferty Over a year ago

I haven't slept in days. please excuse the brain farts. Thanks.

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