此文件接受password
输入。
<html>
<form action="playvideo.php" method="post">
<input type="text" name="password"/>
<input type="submit">
</html>
此文件播放视频
<?php
if(isset($_POST["password"])){
$password=$_POST["password"];
}
else{
$password="";
}
if($password!="pass"){
echo "unauthorised";
}
else{
$local_file = 'video.mp4';
$size = filesize($local_file);
header("Content-Type: video/mp4");
header("Content-Length: ".$size);
readfile($local_file);
}
?>
此外,如果不将代码包装在if
语句中,代码也可以正常工作。
发布于 2018-08-25 02:11:20
<?php
if ((isset($_POST["password"])) && (!empty($_POST['password']))) {
$password = trim($_POST["password"]);
if ($password != "pass") {
echo "Unauthorised";
} else {
?>
<video width="320" height="240" controls autoplay> <!--Video tag introduced in HTML5-->
<source src="video.mp4" type="video/mp4">
Sorry, your browser doesn't support the video element.
</video>
<?php
}
} else {
echo 'Password is not provided';
}
?>
发布于 2018-08-25 01:25:09
需要做一些小小的改变...
你可以试着像这样
<?php
if((isset($_POST["password"])) && (!empty($_POST['password']))) {
$password=$_POST["password"]; //recommended to sanitize this input value
if($password!="pass"){
echo "unauthorised";
}else{
$local_file = 'video.mp4';
$size = filesize($local_file);
header("Content-Type: video/mp4");
header("Content-Length: ".$size);
readfile($local_file);
}
}else{
echo 'password is not provided';
}
?>
https://stackoverflow.com/questions/52009114
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