CODEWITHSUNDEEP
javascriptjquery
Alexander Solonik
Alexander Solonik

Reputation: 10230

Call not working as a chained function in Jquery

I have the following snippet of code : 

$('#move').click(function () {
    $('#block').animate({
        'left': '+=50px'
    }, 500);
});

function get_fx() {
    var store = console.log($(this).queue('fx').length);
    return store;
}

function trigger_it(param) {
    $(param).trigger('click');
}

call_it = setInterval(function () {
    trigger_it('#move')
},
2000);

Fiddle Here

Now, after the animate function has been run on $('#block'), I would like the get_fx() function to return the number of animations still in the queue for $('#block'), so I have the following line of code, after the animate function : 

.call(this , get_fx())

Now this actually seems to work, but let me show you the result I get in the console : 

0 // 0 is displayed , which i guess is the correct result .. I am not sure . 
TypeError: $(...).animate(...).call is not a function // I get this error , I don't know why .

Now can somebody tell me why am I getting that error and is my call function doing what its suppose to do ?

EDIT:: I don't want to go with the callback option of the animate function instead i want to use the call function on get_fx() .

Upvotes: 2

Views: 135

Answers (1)

Tintu C Raju
Tintu C Raju

Reputation: 2700

    animate()

function support call back function. You can utilise this. So you can use like this

        $('#move').click(function(){
            $('#block').animate({ 'left' : '+=50px' } , 500, function(){

                 get_fx();

            });

Upvotes: 1

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