用PHP+SQLITE制作简单的视频点播程序
本文介绍怎样利用 PHPSQLITE 制作简单的视频点播程序,这是一个简单的视频点播程序，设有过多的美观界面。下面介绍一下程序文件的功能：1. init.php初始化数据库，及表文件；
<?php
$db = sqlite_open("db.sqlite"); //创建并连接数据库
$sql = "create table test (id INTEGER PRIMARY KEY,movie text,url text,content text ,timedatatime);"; //创建表
$result = sqlite_query($db, $sql);
if ($result)
{
echo "数据初始化成功";
echo "<a href='sql.php'>进入创建数据文件</a>";
}
else
{
echo "数据初始化失败，请重试！";
}
?>
2.form.php提供数据输入的表单；
<form name="form1" method="post" action="add.php">
<table width="48%" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="36%" height="38"><div align="right">影片名称：</div></td>
<td width="64%"> <input type="text" name="movie"> </td>
</tr>
<tr>
<td height="39"><div align="right">影片地址：</div></td>
<td><input type="text" name="url"></td>
</tr>
<tr>
<td height="33"><div align="right">影片简介：</div></td>
<td><textarea name="content"></textarea></td>
</tr>
<tr>
<td height="53" colspan="2"><div align="center">
<input type="submit" name="Submit" value="确 定">
<input type="reset"value="重 填">
</div></td>
</tr>
</table>
</form>
3.add.php添加表数据；
<?php
$movie = $_POST['movie'];
$url = $_POST['url'];
$content = $_POST['content'];
$now = date("Y- m- dH:i:s");
if (trim($movie) == "" || trim($url) == "" || trim($content) == "")
{
echo "请填写完整数据再提交！";
exit();
}
else
{
$db = sqlite_open("db.sqlite");
$sql = "insert into test (movie,url,content,time) values ( '$movie','$url','$content','$now')";
$result = sqlite_query($db, $sql);
if ($result)
{
echo "数据添加成功";
echo "点击<a href='sql.php'>该处</a>继续添加";
}
else
{
echo "数据添加失败，请检查后重试";
}
}
?>
4.index.php视频点播主界面；
<?php
$db = sqlite_open("db.sqlite"); //连接数据库
$sql = "select * from test ";
$query = sqlite_query($db, $sql); //选出表中数据
?>
<html>
<head>
<title>视频点播</title>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
</head>
<body>
<table width="100%" border="1">
<tr>
<?php
while ($res = sqlite_fetch_array($query))
{
// show the data
echo "<td>";
echo "<a href=red.php?id=" . $res['id'] . ">" . $res['name'] . "</a>";
echo "</td>";
}
?>
<tr>
</table>
</body>
</html>
5.show.php视频点播播放界面；
<?php
$id = $_GET['id']; //取得Id的值
$db = sqlite_open("db.sqlite"); //连接数据库
$sql = "select * from test where id='$id'";
$query = sqlite_query($db, $sql);
$res = sqlite_fetch_array($query);//取出符合要求的数据
?>
<HTML>
<HEAD>
<TITLE><?php echo $res['movie'] ;?></TITLE>
<META http-equiv=Content-Type content="text/html; charset=gb2312">
<META content="MSHTML 6.00.2800.1106" name=GENERATOR>
</HEAD>
<BODY bgcolor="#F4FFF4">
<table width="576" height="418" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td height="62" colspan="3"><img src="vod1.jpg" width="590" height="62"></td>
</tr>
<tr>
<td width="21%"><img src="vod2.jpg" width="130" height="290"></td>
<td width="55%">
<object ID=RVOCX classid="clsid:CFCDAA03-8BE4-11cf-B84B-0020AFBBCCFA" width=326 height=245>
<param name="src" value="rtsp://localhost:554/<?php echo $res['url'] ;?>">
<param name="controls" value="ImageWindow">
<param name="autostart" value="true">
<param name="console" value="_master">
<embed width="326" height="245" src="rtsp://localhost:554/<?php echo $res['url'] ;?>" controls="ImageWindow" console="_master" >
</embed>
</object>
<object ID=RVOCX classid="clsid:CFCDAA03-8BE4-11cf-B84B-0020AFBBCCFA" width=326 height=40>
<param name="controls" value="ControlPanel">
<param name="console" value="_master">
<embed width="326" height="40" src="cabin_w_layout.rpm"
controls="ControlPanel" console="_master" >
</embed>
</object></td>
<td width="24%"><img src="vod4.jpg" width="135" height="289"></td>
</tr>
<tr>
<td colspan="3"><img src="vod3.jpg" width="590" height="66"></td>
</tr>
</table>
</body>
</html> 不知道再那裏學PHP+MYSQL啊？ 有机会一定要试下SQLite
呵呵
http://blog.csdn.net/hfly2005/archive/2005/07/07/416378.aspx 引用第3楼esnak于2006-03-07 16:26发表的“”:
http://blog.csdn.net/hfly2005/archive/2005/07/07/416378.aspx
呵呵!
正确!
用PHP+SQLITE制作简单的视频点播程序
本文介绍怎样利用 PHPSQLITE 制作简单的视频点播程序,这是一个简单的视频点播程序，设有过多的美观界面。下面介绍一下程序文件的功能：1. init.php初始化数据库，及表文件；
<?php
$db = sqlite_open("db.sqlite"); //创建并连接数据库
$sql = "create table test (id INTEGER PRIMARY KEY,movie text,url text,content text ,timedatatime);"; //创建表
$result = sqlite_query($db, $sql);
if ($result)
{
echo "数据初始化成功";
echo "<a href='sql.php'>进入创建数据文件</a>";
}
else
{
echo "数据初始化失败，请重试！";
}
?>
2.form.php提供数据输入的表单；
<form name="form1" method="post" action="add.php">
<table width="48%" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="36%" height="38"><div align="right">影片名称：</div></td>
<td width="64%"> <input type="text" name="movie"> </td>
</tr>
<tr>
<td height="39"><div align="right">影片地址：</div></td>
<td><input type="text" name="url"></td>
</tr>
<tr>
<td height="33"><div align="right">影片简介：</div></td>
<td><textarea name="content"></textarea></td>
</tr>
<tr>
<td height="53" colspan="2"><div align="center">
<input type="submit" name="Submit" value="确 定">
<input type="reset"value="重 填">
</div></td>
</tr>
</table>
</form>
3.add.php添加表数据；
<?php
$movie = $_POST['movie'];
$url = $_POST['url'];
$content = $_POST['content'];
$now = date("Y- m- dH:i:s");
if (trim($movie) == "" || trim($url) == "" || trim($content) == "")
{
echo "请填写完整数据再提交！";
exit();
}
else
{
$db = sqlite_open("db.sqlite");
$sql = "insert into test (movie,url,content,time) values ( '$movie','$url','$content','$now')";
$result = sqlite_query($db, $sql);
if ($result)
{
echo "数据添加成功";
echo "点击<a href='sql.php'>该处</a>继续添加";
}
else
{
echo "数据添加失败，请检查后重试";
}
}
?>
4.index.php视频点播主界面；
<?php
$db = sqlite_open("db.sqlite"); //连接数据库
$sql = "select * from test ";
$query = sqlite_query($db, $sql); //选出表中数据
?>
<html>
<head>
<title>视频点播</title>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
</head>
<body>
<table width="100%" border="1">
<tr>
<?php
while ($res = sqlite_fetch_array($query))
{
// show the data
echo "<td>";
echo "<a href=red.php?id=" . $res['id'] . ">" . $res['name'] . "</a>";
echo "</td>";
}
?>
<tr>
</table>
</body>
</html>
5.show.php视频点播播放界面；
<?php
$id = $_GET['id']; //取得Id的值
$db = sqlite_open("db.sqlite"); //连接数据库
$sql = "select * from test where id='$id'";
$query = sqlite_query($db, $sql);
$res = sqlite_fetch_array($query);//取出符合要求的数据
?>
<HTML>
<HEAD>
<TITLE><?php echo $res['movie'] ;?></TITLE>
<META http-equiv=Content-Type content="text/html; charset=gb2312">
<META content="MSHTML 6.00.2800.1106" name=GENERATOR>
</HEAD>
<BODY bgcolor="#F4FFF4">
<table width="576" height="418" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<td height="62" colspan="3"><img src="vod1.jpg" width="590" height="62"></td>
</tr>
<tr>
<td width="21%"><img src="vod2.jpg" width="130" height="290"></td>
<td width="55%">
<object ID=RVOCX classid="clsid:CFCDAA03-8BE4-11cf-B84B-0020AFBBCCFA" width=326 height=245>
<param name="src" value="rtsp://localhost:554/<?php echo $res['url'] ;?>">
<param name="controls" value="ImageWindow">
<param name="autostart" value="true">
<param name="console" value="_master">
<embed width="326" height="245" src="rtsp://localhost:554/<?php echo $res['url'] ;?>" controls="ImageWindow" console="_master" >
</embed>
</object>
<object ID=RVOCX classid="clsid:CFCDAA03-8BE4-11cf-B84B-0020AFBBCCFA" width=326 height=40>
<param name="controls" value="ControlPanel">
<param name="console" value="_master">
<embed width="326" height="40" src="cabin_w_layout.rpm"
controls="ControlPanel" console="_master" >
</embed>
</object></td>
<td width="24%"><img src="vod4.jpg" width="135" height="289"></td>
</tr>
<tr>
<td colspan="3"><img src="vod3.jpg" width="590" height="66"></td>
</tr>
</table>
</body>
</html>
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