How do I pass arguments to my JavaScript function?

Question

I have this JavaScript function:

function prefixOneDropDown(){

        var source = $('#prefix-one');
        var selected = source.find('option[selected]');
        var options = $('option', source);  
        var DefinitionList = '<dl id="dd-prefix-one" class="dropdown f-left"><dt><a href="#">' + selected.text() + '<span class="value">' + selected.val() + '</span></a></dt><dd><ul></ul></dd></dl>';
        $(DefinitionList).insertAfter(source);

        options.each(function(){ $("#dd-prefix-one dd ul").append('<li><a href="#"><span class="number-plate">' + $(this).text() + '</span><span class="value">' + $(this).val() + '</span></a></li>'); });
    }

I was wondering if there is a way declaring var selected, options, DefinitionList outside the function...then call them in as arguments?

I've tried a number of different approaches but I'm missing something.

The function is called in my js like so : prefixOneDropDown();

Any help would be Greatly Appreciated, Thanks

The reason why I want the variables declared outside is that there are five of these functions...so the are declared only once — Nasir
– Nasir, Commented Jan 20, 2011 at 14:12

hellatan · Accepted Answer · 2011-01-20 14:33:55Z

2

did you try this:

function prefixOneDropDown(source, options, dl){

    $(dl).insertAfter(source);

    options.each(function(){ $("#dd-prefix-one dd ul").append('<li><a href="#"><span class="number-plate">' + $(this).text() + '</span><span class="value">' + $(this).val() + '</span></a></li>'); });
}

var source = $('#prefix-one'),
    selected = source.find('option[selected]'),
    options = $('option', source),
    DefinitionList = '<dl id="dd-prefix-one" class="dropdown f-left"><dt><a href="#">' + selected.text() + '<span class="value">' + selected.val() + '</span></a></dt><dd><ul></ul></dd></dl>';

prefixOneDropDown(source, options, DefinitionList);

edited Jan 20, 2011 at 14:33

answered Jan 20, 2011 at 14:17

hellatan

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3 Comments

pimvdb Over a year ago

I might be wrong, but select and option are not used in the function

hellatan Over a year ago

ah, no, you're right @pimvdb. select is only referenced outside, and i have a typo with option - should be options, good catch =)

hellatan Over a year ago

def np, nasir. glad i could help

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