这是我通过PHP和Bootstrap图像库的代码。它的工作很好，但我想得到弹出窗口的图像，当用户点击那个特定的图像。是否有任何功能，我可以添加在这段代码?
Thanks in advance
<h3>Photo Gallery</h3>
<ul id="categories">
<li>
<a href="javascript:void(0);" data-filter=".resort" onClick="makeCall('Resort');" class="Resort active">1</a>
</li>
<li class="rooms">
<a href="javascript:void(0);" data-filter=".beach" onClick="makeCall('Rooms');" class="Rooms">2</a>
</li>
<li>
<a href="javascript:void(0);" data-filter=".beach" onClick="makeCall('Dinning');" class="Dinning">3</a>
</li>
<li>
<a href="javascript:void(0);" data-filter=".beach" onClick="makeCall('Swiming-Pool');" class="Swiming-Pool">4</a>
</li>
<li class="last">
<a href="javascript:void(0);" data-filter=".beach" onClick="makeCall('all');" class="all">View all</a>
</li>
</ul>
<div class="row gallery gallery-section" id="demoLightbox" class="lightbox hide fade" tabindex="-1" role="dialog" aria-hidden="true">
<div class='lightbox-content'>
<?php
$path = 'images/gallery/'; // Create initial "Folders" array
$filetype = '*.*';
$_POST['type']='all';
if($_POST['type']!='all')
{
$classloop[] = $_POST['type'];
$folders[]=ucfirst($_POST['type']);
$start=0;
$foldercount = count($folders)-1;
}
else
{
if ($dir = opendir($path))
{
$j = 0;
while (($file = readdir($dir)) !== false)
{
if ($file != '.' && $file != '..' && is_dir($path.$file))
{
$j++;
$folders[$j] = $file;
}
}
}
$classloop = array("","","resort","envato");
$start=1;
$foldercount = count($folders);
}
for ($i = $start; $i <= $foldercount ; $i++)
{
$folderpath = $path.$folders[$i];
$images = glob($folderpath."/".$filetype);
$count = count($images);
if($count>0)
{
for ($j = 0; $j < $count; $j++)
{ ?>
<?php
$z = $j%3;
echo '<div class="'.$folders[$i].' photo-item hover test" style="float:left;">';
echo '<a href="'.$images[$j].'" class="image-box">';
echo '<div class="photo">';
echo '<span class="text"><span class="anchor"></span></span>';
echo '</div>';
echo '<img src="'.$images[$j].'" alt="" width="260" height="170"/>';
echo ' </a></div>';
echo '<div class="col-md-4 col-sm-4 col-xs-6"><a href="#demoLightbox" class="gphoto" data-toggle="lightbox">';
echo '<img src="images/delux.jpg" class="img-responsive" alt="" />';
echo '</a></div>';
}
}
else
{
echo '<div align="center" style="width: 100%;">No image(s) found</div>';
}
}
?>
</div>
</div>
</div>
</div>
</section>
Yeah..
放入css
<style type="text/css">
/* popup_box DIV-Styles*/
#popup_box {
display:none; /* Hide the DIV */
position:fixed;
_position:absolute; /* hack for internet explorer 6 */
height:auto;
width:600px;
/*background:#FFFFFF;*/
left: 450px;
top:63px;
z-index:100; /* Layering ( on-top of others), if you have lots of layers: I just maximized, you can change it yourself */
margin-left: 0px;
/* additional features, can be omitted */
/*border:2px solid #ff0000; */
padding:0px;
font-size:15px;
-moz-box-shadow: 0 0 15px #FF3333;
-webkit-box-shadow: 0 0 15px #0033FF;
box-shadow: 0 0 25px #990000;
}
a{
cursor: pointer;
text-decoration:none;
}
/* This is for the positioning of the Close Link */
#popupBoxClose {
font-size:20px;
line-height:15px;
right:5px;
top:5px;
position:absolute;
color:#6fa5e2;
font-weight:500;
}
</style>
创建一个div
在按钮上附加一个javascript函数，点击
打开弹出窗口
<button type="button" class="open" data-dismiss="modal" aria-label="Open" onclick="return loadPopupBox();"><span aria-hidden="true">Open</span></button>
关闭弹出窗口
<button type="button" class="close" data-dismiss="modal" aria-label="Close" onclick="return unloadPopupBox();"><span aria-hidden="true">×</span></button>
写点javascript
function unloadPopupBox() {
$('#yourdiv').hide();
}
function loadPopupBox(data) { // To Load the Popupbox
$('#yourdiv').show();
});
确保在加载页面时隐藏div
如果你是指点击时出现的模态，是的有，我这里有一个模板，你可以使用(或不)，你可以玩