Very Simple Image Slider/Slideshow with left and right button. No autoplay

Question

I'm trying to make a very, very simple image slider/slideshow with no autoplay please. I only want to go to next or prev image on click. Having some issues as you can read below.

Here's the jsFiddle http://jsfiddle.net/HsEne/12/ (i just used background colors instead of images in the fiddle as well as switched img to divs but it's the exact same problem of course)

The HTML:

<div id="slider-wrapper">

<div id="slider">
<img class="sp" src="/img1.jpg">
<img class="sp" src="/img2.jpg">
<img class="sp" src="/img3.jpg">
<img class="sp" src="/img4.jpg">
<img class="sp" src="/img5.jpg">
<img class="sp" src="/img6.jpg">
</div>

<img id="button-previous" src="/button-arrow-left.jpg">
<img id="button-next" src="/button-arrow-right.jpg">
</div>

The CSS:

#slider-wrapper {
display: block;
max-width: 790px;
margin: 5% auto;
max-height: 500px;
height: 100%; }

#slider {
display: block;
position: relative;
z-index: 99999;
max-width: 710px;
width: 100%;
margin: 0 auto; }

#button-previous { 
position: relative;
left: 0;
margin-top: 40%;
width: 40px;
height: 60px; }

#button-next {
position: relative;
margin-top: 40%;
float: right; }

.sp {
position: absolute; }

#slider .sp {
max-width: 710px;
width: 100%;
max-height: 500px; }

jQuery for next button:

jQuery("document").ready(function(){
    jQuery("#slider > img:gt(0)").hide();
     jQuery("#button-next").click(function() { 
        jQuery("#slider > img:first")
    .fadeOut(1000)
    .next()
    .fadeIn(1000)
    .appendTo("#slider");
  });
});

The above jQuery works fine, except it skips the first image if a user clicks the next button on last slide. It only displays the first image one time, all other images will keep sliding every time next is clicked. For some reason display: none gets added to the first image but I don't know where that code is coming from.

jQuery for previous button:

jQuery("document").ready(function(){
    jQuery("#slider > img:gt(0)").hide();
     jQuery("#button-previous").click(function() { 
        jQuery("#slider > img:last")
    .fadeOut(1000)
    .next()
    .fadeIn(1000)
    .appendTo("#slider");
  });
});

This jquery code above will return the just previewed image like it should. But when clicked again it doesn't do anything.

I also tried .prev() in replace of .next(). It works fine the first time you click on the previous button. Say we are viewing Image #4. When I click previous button it goes to Image #3, like it should. But when click on it again it doesn't go to Image #2, it goes back to Image #4 , and then #3 and then #4, repeating itself.

James · Accepted Answer · 2013-05-27T05:53:50.343

After reading your comment on my previous answer I thought I might put this as a separate answer.

Although I appreciate your approach of trying to do it manually to get a better grasp on jQuery I do still emphasise the merit in using existing frameworks.

That said, here is a solution. I've modified some of your css and and HTML just to make it easier for me to work with

WORKING JS FIDDLE - http://jsfiddle.net/HsEne/15/

This is the jQuery

$(document).ready(function(){
$('.sp').first().addClass('active');
$('.sp').hide();    
$('.active').show();

    $('#button-next').click(function(){

    $('.active').removeClass('active').addClass('oldActive');    
                   if ( $('.oldActive').is(':last-child')) {
        $('.sp').first().addClass('active');
        }
        else{
        $('.oldActive').next().addClass('active');
        }
    $('.oldActive').removeClass('oldActive');
    $('.sp').fadeOut();
    $('.active').fadeIn();


    });

       $('#button-previous').click(function(){
    $('.active').removeClass('active').addClass('oldActive');    
           if ( $('.oldActive').is(':first-child')) {
        $('.sp').last().addClass('active');
        }
           else{
    $('.oldActive').prev().addClass('active');
           }
    $('.oldActive').removeClass('oldActive');
    $('.sp').fadeOut();
    $('.active').fadeIn();
    });




});

So now the explanation.
Stage 1
1) Load the script on document ready.

2) Grab the first slide and add a class 'active' to it so we know which slide we are dealing with.

3) Hide all slides and show active slide. So now slide #1 is display block and all the rest are display:none;

Stage 2 Working with the button-next click event.
1) Remove the current active class from the slide that will be disappearing and give it the class oldActive so we know that it is on it's way out.

2) Next is an if statement to check if we are at the end of the slideshow and need to return to the start again. It checks if oldActive (i.e. the outgoing slide) is the last child. If it is, then go back to the first child and make it 'active'. If it's not the last child, then just grab the next element (using .next() ) and give it class active.

3) We remove the class oldActive because it's no longer needed.

4) fadeOut all of the slides

5) fade In the active slides

Step 3

Same as in step two but using some reverse logic for traversing through the elements backwards.

It's important to note there are thousands of ways you can achieve this. This is merely my take on the situation.

Thanks for the detailed response. Worked perfectly and I added some animate effects! — bob, May 28 '13 at 14:46

score 11 · Answer 2 · answered May 27 '13 at 00:31

11

Why try to reinvent the wheel? There are more lightweight jQuery slideshow solutions out there then you could poke a stick at, and someone has already done the hard work for you and thought about issues that you might run into (cross-browser compatability etc).

jQuery Cycle is one of my favourite light weight libraries.

What you want to achieve could be done in just

    jQuery("#slideshow").cycle({
    timeout:0, // no autoplay
    fx: 'fade', //fade effect, although there are heaps
    next: '#next',
    prev: '#prev'
    });

JSFIDDLE TO SEE HOW EASY IT IS

answered May 27 '13 at 00:31

James

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I marked your comment up because its useful. But I'm trying to learn jquery, so would prefer to know how to do it manually. I don't think it would be that hard, but maybe it is. – bob May 27 '13 at 03:16
That's understandable, I've added another answer for you to look over that uses just jQuery. – James May 27 '13 at 05:49

score 1 · Answer 3 · answered Jun 07 '14 at 13:09

Very simple code to make jquery slider Here is two div first is the slider viewer and second is the image list container. Just copy paste the code and customise with css.

    <div class="featured-image" style="height:300px">
     <img id="thumbnail" src="01.jpg"/>
    </div>

    <div class="post-margin" style="margin:10px 0px; padding:0px;" id="thumblist">
    <img src='01.jpg'>
    <img src='02.jpg'>
    <img src='03.jpg'>
    <img src='04.jpg'>
    </div>

    <script type="text/javascript">
            function changeThumbnail()
            {
            $("#thumbnail").fadeOut(200);
            var path=$("#thumbnail").attr('src');
            var arr= new Array(); var i=0;
            $("#thumblist img").each(function(index, element) {
               arr[i]=$(this).attr('src');
               i++;
            });
            var index= arr.indexOf(path);
            if(index==(arr.length-1))
            path=arr[0];
            else
            path=arr[index+1];
            $("#thumbnail").attr('src',path).fadeIn(200);
            setTimeout(changeThumbnail, 5000);  
            }
            setTimeout(changeThumbnail, 5000);
    </script>

score -1 · Answer 4 · answered Jul 08 '14 at 12:44

<script type="text/javascript">
                    $(document).ready(function(e) {
                        $(".mqimg").mouseover(function()
                        {
                            $("#imgprev").animate({height: "250px",width: "70%",left: "15%"},100).html("<img src='"+$(this).attr('src')+"' width='100%' height='100%' />"); 
                        })
                        $(".mqimg").mouseout(function()
                        {
                            $("#imgprev").animate({height: "0px",width: "0%",left: "50%"},100);
                        })
                    });
                    </script>
                    <style>
                    .mqimg{ cursor:pointer;}
                    </style>
                    <div style="position:relative; width:100%; height:1px; text-align:center;">`enter code here`
                    <div id="imgprev" style="position:absolute; display:block; box-shadow:2px 5px 10px #333; width:70%; height:0px; background:#999; left:15%; bottom:15px; "></div>
<img class='mqimg' src='spppimages/1.jpg' height='100px' />
<img class='mqimg' src='spppimages/2.jpg' height='100px' />
<img class='mqimg' src='spppimages/3.jpg' height='100px' />
<img class='mqimg' src='spppimages/4.jpg' height='100px' />
<img class='mqimg' src='spppimages/5.jpg' height='100px' />

Very Simple Image Slider/Slideshow with left and right button. No autoplay

4 Answers4

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