Question

交换后不工作变焦在缩放显示旧图像时单击附加图像我究竟做错了什么交换后不工作变焦在缩放显示旧图像时单击附加图像我做错了什么 “＆GT;

“＆gt;” 中data-zoom-image =“”title =“”alt =“”/＆gt;“＆gt;

“title =”“＆gt;”title =“”alt =“”/＆gt;

    <script>
    $(document).ready(function(){

    // Get main image info
    var img = $('#zoom_01').attr('src'); // get img popup href
    if (img != undefined) {
    var imgWidth = img.substring(img.lastIndexOf('-') + 1, img.lastIndexOf('x'));
    var imgHeight = img.substring(img.lastIndexOf('x') + 1, img.lastIndexOf('.'));
    }

    $('.image-additional img').hover(function() {
    var newsrc = $(this).parent().attr('href');

    $('#zoom_01').attr({
    src: newsrc,
    title: $(this).attr('title'),
    alt: $(this).attr('alt'),
    width: imgWidth,
    height: imgHeight
    });

    $('#zoom_01').parent().attr('href', newsrc);
    });
    });
    </script>

Answer 1

<div class="<?php echo $class; ?>">
  <?php if ($thumb || $images) { ?>
  <ul class="thumbnails">
    <?php if ($thumb) { ?>
    <li><div class="thumbnail" title="<?php echo $heading_title; ?>"><img id="zoom_01" src="<?php echo $thumb; ?>" data-zoom-image="<?php echo $popup; ?>" title="<?php echo $heading_title; ?>" alt="<?php echo $heading_title; ?>" /><a href="<?php echo $popup; ?>"><i class="fa fa-search"></i></a></div></li>
    <?php } ?>
    <?php if ($images) { ?>
    <?php foreach ($images as $image) { ?>
    <li class="image-additional"><a class="thumbnail" href="<?php echo $image['popup']; ?>" title="<?php echo $heading_title; ?>"> <img src="<?php echo $image['thumb']; ?>" title="<?php echo $heading_title; ?>" alt="<?php echo $heading_title; ?>" /></a></li>
    <?php } ?>
    <?php } ?>
  </ul>
  <?php } ?>
</div>

交换后缩放

1 个答案: