jQuery, function called without parameters, but there is parameters

Question

One thing i can't understand about jQuery, maybe someone can explain me this.

for example: HTML:

<div>
<img class="class1" src="file.jpg" />
</div>
<div>
<img class="class2" src="file.jpg" />
</div>

jQuery:

(function ($) {

    $(document).ready(function () {
        $('.class1, .class2').myfunction();
    });

    function myfunction(param1, param2) {
        var img = $('img', param1);
    }

})(jQuery);

So the question is what exactly in param1 and in param2, if function has been called like that(without passing parameters to the function)?

You can always console.log(arguments) in a function body to see the parameters being passed. — Bryan Downing
– Bryan Downing, Commented Sep 24, 2013 at 18:01
you executed a function that doesn't exist. myFunction is not a property of $.fn — user400654
– user400654, Commented Sep 24, 2013 at 18:03
Even if you did define myfunction properly, there would be no params because you didn't pass any! — user400654
– user400654, Commented Sep 24, 2013 at 18:04

Renato Vianna · Accepted Answer · 2013-09-24 18:13:32Z

1

The function call should be as follows:

View:

<div>
    <img id="img1" src="file.jpg" />
</div>
<div>
    <img id="img2" src="file.jpg" />
</div>

JQuery:

(function ($) {
    $(document).ready(function () {
        var path1 = $('#img1').attr('src');
        var path2 = $('#img2').attr('src');
        myfunction(path1, path2);
    });

    function myfunction(param1, param2) {
        // your code
    }
})(jQuery);

answered Sep 24, 2013 at 18:13

Renato Vianna

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1 Comment

user2336995 Over a year ago

Yeah, thank you. Or $.fn.myfunction = function(){myfunction($('div'),'param2')} then my example will be right too. But thanks.

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