Question

I am trying to create very simple image browser with jQuery. I am doing it only for learning, it will not be used in any project at this time.

I have images and thumbnails as follows:

<div class="images">
  <img src="http://www.dummyimage.com/400x200/222/fff.jpg" alt="" data-idi="1"/>
  <img src="http://www.dummyimage.com/400x200/534/fff.jpg" alt="" data-idi="2"/>
  <img src="http://www.dummyimage.com/400x200/345/fff.jpg" alt="" data-idi="3"/>
  <img src="http://www.dummyimage.com/400x200/fed/111.jpg" alt="" data-idi="4"/>
</div>

<div class="thumbs">
  <button data-idb="1"><img src="http://www.dummyimage.com/50x50/222/fff.jpg" alt="" /></button>
  <button data-idb="2"><img src="http://www.dummyimage.com/50x50/534/fff.jpg" alt="" /></button>
  <button data-idb="3"><img src="http://www.dummyimage.com/50x50/345/fff.jpg" alt="" /></button>
  <button data-idb="4"><img src="http://www.dummyimage.com/50x50/fed/111.jpg" alt="" /></button>   
</div>

Hovering on thumbnail should replace main image above.

I have managed to do it with one, let’s call it ”set” of images, as presented on jsfiddle.net.

Here is simple script used to achieve this:

  $('#images img').hide().first().show();
  $('#thumbs button').on( "mouseover", function(e) {
     var idb = $(this).data("idb");
     $('#images img').hide();
     var img = $('#images').find("[data-idi='" + idb + "']").show();
  });

Now I want to add another set of images, and wrap the script with .each() function. But I am having a problem on very begining, when I am trying to show only first image:

var imgs = $('.images').hide();
$.each(imgs, function( index, val ) {
    var firstImage = $(this).children(":first");
    console.log(firstImage);
    firstImage.show();
});

Proper image is being consoled, but it is not showing, as excepted. Live example on jsfiddle.net. Could you help me with this problem? TIA

Answer 1

You’ve already hidden the parent .images element.

After calling .show(), you have a visible element inside a hidden parent.

You need to select the <img> elements themselves by writing $('.images img').