滑块-您将如何制作一个简单的滑块？

Question

EDIT: I WANT MY SLIDER TO BE EASY TO YOU AND ONCE A LINK IS CLICKED THE IMAGE THAT CORRELATES WITH LINK SHOWS.

I'm looking to make a really simple "slider" that if you click a link, the img shows that correlates with it. I've been trying to find something for a bit now and things are either too flash or don't suit my needs. This came close: http://jsfiddle.net/bretmorris/ULNa2/7/

I would something a little simpler that can be applied easily to multiple images for different divs.

This is what my code looks like with just a plain img tag to it:

<div id="adobe_content" class="hdiv"><button class="x">X</button>
<img src="images/adobefront.png"><br>
<img src="images/adobeinside.png"><br>
<img src="images/adobeback.png"><br>
<h5>Adobe Brochure</h5>
<p>
I wanted to make something functional and out the box that Adobe might consider giving out. It's clean like their work and sparks interest in opening the brochure with the cut out in the center. The front flap is able to slide into a slot on the right side for a neat logo. They're now more interested in their cloud, but the information is inside is still relevant!
</p>
<b>Programs used: Adobe Illustrator, InDesign and Photoshop.</b>
</div>

The code doesn't work for me because, well I partially don't understand it, and I'm not sure how to make it suit my needs (especially if I got up to multiple images) like correlating with an image.

Answer 1

Perhaps understanding what is going on would maybe get you on the right track, so here is an explanation:

$('a.prev').click(function() {
    prevSlide($(this).parents('.slideshow').find('.slides'));
});
//clicking image goes to next slide
$('a.next, .slideshow img').click(function() {
    nextSlide($(this).parents('.slideshow').find('.slides'));
});

This part is relatively straightforward, when you click on the previous or next links, call the prevSlide or nextSlide function, passing the collection of slides as an argument.

//initialize show
iniShow();

function iniShow() {
    //show first image
    $('.slideshow').each(function() {
        $(this).find('img:first').fadeIn(500);
    })
}

Initialize the slideshow by finding each slideshow on the page and fading in the first image. $(this) refers to the <div class="slideshow"> parent, find all child image tags and take the first, fade that element in (and do it in 500 milliseconds).

function prevSlide($slides) {
    $slides.find('img:last').prependTo($slides);
    showSlide($slides);
}

function nextSlide($slides) {
    $slides.find('img:first').appendTo($slides);
    showSlide($slides);
}

The prevSlide and nextSlide functions both rearrange the order of images, this line in particular:

$slides.find('img:first').appendTo($slides);

Is moving the first image to the end of the images, so:

 <img src="http://placekitten.com/300/500" width="300" height="500" />
 <img src="http://placekitten.com/200/400" width="200" height="400" />
 <img src="http://placekitten.com/400/400" width="500" height="400" />

becomes:

  <img src="http://placekitten.com/200/400" width="200" height="400" />
  <img src="http://placekitten.com/400/400" width="500" height="400" />
  <img src="http://placekitten.com/300/500" width="300" height="500" />

$slides.find('img:last').prependTo($slides); does the inverse and moves the last image to the beginning.

function showSlide($slides) {
    //hide (reset) all slides
    $slides.find('img').hide();
    //fade in next slide
    $slides.find('img:first').fadeIn(500);
}

Finally, showSlide accepts the collection of images, hides all of them and then fades in the first image (since the collection is reordered each time, the first is a different image).

Now, if you want a link for each image that will display a corresponding image, you could do something as simple as:

<a class="image" data-src="http://placekitten.com/300/500">Kitten 1</a>
<a class="image" data-src="http://placekitten.com/200/400">Kitten 2</a>
<a class="image" data-src="http://placekitten.com/400/500">Kitten 3</a>
<div id="image-container">
   <img src="http://placekitten.com/300/500" />
</div>

and something like the following:

$('.image').on('click', function() {
  var imageSrc = $(this).data('src');
  $('#image-container img').prop('src', imageSrc);
});

Which will update the child image tag of <div id="image-container"> with the data-src attribute value in the clicked link.

http://jsfiddle.net/9sxt6n0t/

Hope this helps.

Answer 2

just a quick function to slide

function slideIt(images , prev , next){
$('.slideshow img:nth-child(1)').show();
    var imagesLength = $(images).length;
var i = 1;
$(prev).click(function() {
    $(images).hide();
    if(i !== 1){
        $(images + ':nth-child('+ (i - 1) +')').show();
        i--;
    }else{
        $(images +':nth-child('+imagesLength +')').show();
        i = imagesLength;
    }
});
//clicking image goes to next slide
$(''+next+','+images+'').on('click',function() {
    $(images).hide();
    if(i !== imagesLength){
        $(images + ':nth-child('+ (i + 1) +')').show();
        i++;
    }else{
        $(images + ':nth-child(1)').show();
        i = 1;
    }
});
}

and use like that slideIt(Images , prevArrow , nextArrow)

slideIt('.slideshow img','a.prev' , 'a.next');

DEMO HERE