如何使用mysql数据库将视频上传到php目录中 [英] how to upload videos to directory in php using mysql database
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问题描述
大家好，
我在这里是siddu，对PHP还是陌生的，我在这里给出了用于上传视频但不超过4MB的代码.所以你能帮我吗?
在此先感谢
hi everyone,
I am siddu here and am new to PHP, I have given code here which is used to upload video but not more than 4MB. So can you please help me out.
Thanks in advance
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title> Upload Image </title>
<?php
// if something was posted, start the process...
if(isset($_POST['upload']))
{
// connect to the database
include "connect1.php";
mysql_connect("$host", "$user", "$pass")
or die("Could not connect: " . mysql_error());
// select our database
mysql_select_db("$db") or die(mysql_error());
// the query that will add this to the database
// define the posted file into variables
echo $name = $_FILES['picture']['name'];
echo $tmp_name = $_FILES['picture']['tmp_name'];
echo $type = $_FILES['picture']['type'];
echo $size = $_FILES['picture']['size'];
// if your server has magic quotes turned off, add slashes manually
if(!get_magic_quotes_gpc()){
$name = addslashes($name);
}
// open up the file and extract the data/content from it
$extract = fopen($tmp_name, 'r');
$content = fread($extract, $size);
$content = addslashes($content);
fclose($extract);
if(!empty($_FILES))
{
$target = "D:/videos/";
$target = $target . basename( $_FILES['picture']['name']) ;
echo $target;
$addfile = "INSERT INTO video VALUES ('','$name', '$size', '$type','$target')";
mysql_query($addfile) or die(mysql_error());
echo "===========";
echo $target;
echo "===========";
$ok=1;
if(move_uploaded_file($_FILES['picture']['tmp_name'], $target))
{
echo "The file ". basename( $_FILES['picture']['name']). " has been uploaded <br/>";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
}
mysql_close();
echo "Successfully uploaded your picture!";
}else{die("No uploaded file present");
}
?>
</head>
<body>
<div align="center">
<br />
<a href="file.html">upload more images</a>
</div>
</body>
</html>
推荐答案
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