Adding function to jquery

Question

I am writing a function called serializePost.

The function will never be called on a set of objects. only ever one object.

hence my question is;

do I need to do this.each or can I simply use this

Will this work:

(function($) {
    $.fn.serializePost = function() {
        var data = {};
        var formData = $(this).serializeArray();
        for (var i = formData.length; i--;) {
            var name = formData[i].name;
            var value = formData[i].value;
            var index = name.indexOf('[]');
            if (index > -1) {
                name = name.substring(0, index);
                if (!(name in data)) {
                    data[name] = [];
                }
                data[name].push(value);
            }
            else
                data[name] = value;
        }

        return data;
    };
})(jQuery);

user113716 · Accepted Answer · 2011-02-21 20:50:11Z

2

No need to use .each(), and no need to wrap it either.

var formData = this.serializeArray();

Inside the plugin, this is already a reference to the jQuery object.

answered Feb 21, 2011 at 20:50

user113716

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1 Comment

Hailwood Over a year ago

Cheers for the quick answer :) will accept asap, and cheers for the point about wrapping this.

Brad Christie · Accepted Answer · 2011-02-21 20:51:16Z

2

You can observe the source of the .val() function which basically pops the first element off the selection chain and works with it:

var elem = this[0]

answered Feb 21, 2011 at 20:51

Brad Christie

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3 Comments

Hailwood Over a year ago

An interesting idea. But, If the user was infact passing in a chain of objects they are obviously making an error somewhere and would not get the desired result. hence is this not a bit of a waste?

Brad Christie Over a year ago

@Hailwood: If this is something you'll solely be using, the end answer is: code it however you'd like. If this could and will potentially be used by others, code it to be "bulletproof", expecting every possible anomaly and acting accordingly. Most functions of jQuery that deal with single elements pop the first element off the chain and work with it, so it's not far-fetched to use the same behavior.

user113716 Over a year ago

@Hailwood: Brad makes a good point here. Although it is extracting the DOM element, which isn't what you want, it does ensure that you're only dealing with one element in case multiple were mistakenly passed. You may want to use this.eq(0).serialize... for safety. @Brad +1

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