CODEWITHSUNDEEP
jqueryhtmldelayfadeinfadeout
Dijam
Dijam

Reputation: 167

Problem with showing/hiding divs

I am trying to show some random images in a div. My HTML code is like this:

<div class="sponsors">
<div id="sponsorsContent">
    <div class="spacer"></div>
    <div class="sponsorSlide" id="imgageSlide_1">
        <img alt="" src="<?php echo $this->baseUrl(); ?>/img/Sponsors/img_1.jpg" /><br />
        <span class="thumbnail-text">Image Text</span>
    </div>
    <div class="spacer"></div>
    <div class="sponsorSlide" id="imgageSlide_2">
        <img alt="" src="<?php echo $this->baseUrl(); ?>/img/Sponsors/img_2.jpg" /><br />
        <span class="thumbnail-text">Image Text</span>
    </div> ... </div></div>

I am trying to first shuffle "sponsorSlide" divs and then select 7 of them randomly. I want to show it with fadeIn and fadeOut. Generally, I'm trying this code:

$('#sponsorsContent').addClass('horizontal');
$('#sponsorsContent div').addClass('hidden').removeClass('visible');
$('#sponsorsContent .sponsorSlide').rand(7).removeClass('hidden');
setInterval(function(){
    $('#sponsorsContent').fadeOut(500);
    $('#sponsorsContent .sponsorSlide').delay(500).addClass('hidden').removeClass('visible').shuffle();
    $('#sponsorsContent .sponsorSlide').rand(7).removeClass('hidden').addClass('visible');                  
    $('#sponsorsContent').fadeIn(1500);
}, 5000);

The main problem is, when the code runs, exactly before the div fades out, you can see that the images are changing! But I want to randomize them when they are not visible! I used different ways:

FYI, I am trying to have this concept:

1- showing some random images 2- fade out 3- shuffle images 4- Select 7 random divs 5- fade in 6- go 2

Does anybody have any idea that what is my main mistake? Am I doing something in a wrong way? I got confused and I really want to find out, which way I should try and how I can manage it to work as the way I want?

ps: It does work without fadings! But I need to fade them in and out.

Upvotes: 6

Views: 241

Answers (1)

MeLight
MeLight

Reputation: 5565

You should be using a callback function in order to prevent the change from showing. If you want to hide -> change -> show, you should do it like this:

$('#image_div').fadeOut(500, function() { //first fade out!
    $('#image_div img').attr('src', 'image.jpg'); //only then change the picture
    $('#image_div').fadeIn(500);   //and fade in
});

What comes in the function(){//code} will happen ONLY when the fadeOut() is finished.

Upvotes: 5

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