"Поворот"

Автор: Super User. Опубликовано в Uncategorised

{source}

<head>
<script type="text/javascript" src="/https://ajax.googleapis.com/ajax/libs/jquery/1.4.3/jquery.min.js"></script>
<script type="text/javascript" src="/js/jquery-1.9.1.js"></script>
<script type="text/javascript" src="/js/jquery-ui-1.10.3.custom.min.js"></script>
<script src="/http://code.jquery.com/jquery-1.9.0.js"></script>
<script src="/http://code.jquery.com/jquery-migrate-1.0.0.js"></script>
<script src="/http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<script src="/http://ajax.googleapis.com/ajax/libs/jquery/1.5/jquery.min.js"></script>
<script src="/http://ajax.googleapis.com/ajax/libs/jqueryui/1.8/jquery-ui.min.js"></script>
<script type="text/javascript" src="/http://jqueryrotate.googlecode.com/svn/trunk/jQueryRotate.js"></script>
<script src="/js/jquery.Jcrop.min.js" type="text/javascript"></script>
<link rel="stylesheet" href="/css/jquery.Jcrop.css" type="text/css">
<link rel="stylesheet" type="text/css" href="/css/style.css">
<link rel="stylesheet" type="text/css" href="/css/jquery-ui-1.10.3.custom.min.css">
<style>
#dinamic_block
{
margin-top: <?php echo (800 - $_GET['g_size'][1])/2; ?>px;
margin-left: <?php echo (800 - $_GET['g_size'][0])/2; ?>px;
width: <?php echo $_GET['g_size'][0]; ?>px;
height: <?php echo $_GET['g_size'][1]; ?>px;
}
</style>
</head>
<body>

<h1>Поворот фото</h1>

<br /><br /><br /><br />


<img id="photo" src="/<?php echo $_GET['g_path'];?>"/>
<p class="transition" id="rotate_link">Далее</p>
<p id="previous" class="transition">Назад</p>
<img src="/images/left.png" id="rotate_left" class="transition">
<img src="/images/right.png" id="rotate_right" class="transition">
</body>
<script type="text/javascript" src="/js/rotate.js"></script>
{/source}

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