我想如果没有输入,例如http://bloxhotel.nl/video?watch=YxIiPLVR6NA
不会弹出一个框
<?php
ob_start();
include 'includes/inc.bootstrap.php';
$page = Array('title' => ((isset($user)) ? $user->username : ''), 'onload' => ((isset($_GET['utm_source']) && $_GET['utm_source'] == 'welcome') ? 'Pixel.View.OpenUrlOverlay(\'/pixlett/Video.php\');' : ''), 'tab' => Array('me', 'home'), 'access' => Array(true, false));
include 'content/header.php';
include 'content/menu.php';
?>
<center>
<div class="c_box" style="overflow: hidden; width: 58%; height: 10%;">
<div class="heading blue">Ga naar Video</div>
<tr style="padding: 4px;">
<html>
<body>
<form action="welcome_get.php" method="get">
Youtube Link: <input type="text" name="link"><br>
<input type="submit">
</form>
</body>
</html>
</div>
<center>
<div class="c_box" style="overflow: hidden; width: 58%; height: 10%;">
<div class="heading blue">Video</div>
<?php
$watch = $_GET['watch'];
$embed = '<embed src="http://bloxhotel.nl/Player7.swf" flashVars="video_id='.$watch.'" style="width: 100%; height: 57%;" allowScriptAccess="always" autoplay="false" allowFullscreen="true" type="application/x-shockwave-flash" wmode="transparent" id="player_demo"></embed></object>';
echo $embed;
?>
</div>
</center>
这
<?php
$watch = $_GET['watch'];
$embed = '<embed src="http://bloxhotel.nl/Player7.swf" flashVars="video_id='.$watch.'" style="width: 100%; height: 57%;" allowScriptAccess="always" autoplay="false" allowFullscreen="true" type="application/x-shockwave-flash" wmode="transparent" id="player_demo"></embed></object>';
echo $embed;
?>
不需要显示所以如果视频中没有输入?watch=该框需要保持隐藏
有人可以帮忙吗