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View Full Version : Uploading, displaying and playing videos.


ak47
06-22-2017, 10:18 AM
I wish to Upload, display and play videos in my website. Here I am able to upload video's url and id into my table. But the problem is video's name is not inserting into table. The video are not uploading into specified directory and I hope because of this Videos are not playing. Please help me to overcome these problems.

My Code is:

dashboard.php


<div class="row">
<?php
$conn = mysqli_connect("localhost", "root", "", "videos") or die("Connection Server" . mysqli_error());
$query=mysqli_query($conn, "SELECT * FROM vid");
$id = (isset($_GET['id']) ? $_GET['id'] : '');
if(isset($_POST['submit']))
{
echo "<br />".$id."has been uploaded";
}
?>
<div>
<h3>List Of Videos</h3><br />
<?php
$conn = mysqli_connect("localhost", "root", "", "videos") or die("Connection Server" . mysqli_error());
$query=mysqli_query($conn, "SELECT * FROM vid");
while($row=mysqli_fetch_assoc($query))
{
$id = $row['id'];
$name = $row['name'];
$url = $row['url'];
echo "<a href='watch.php id=$id'>&nbsp;$id&nbsp;&nbsp;$name&nbsp;&nbsp;$url</a> <br />";
}
?>
</div>

index.php

<?php
$conn = mysqli_connect("localhost", "root", "", "videos") or die("Connection Server" . mysqli_error());
if(isset($_POST['submit']))
{
$id = (isset($_FILES['file']['id']) ? $_FILES['file']['id'] : '');
$name = (isset($_FILES['file']['name']) ? $_FILES['file']['name'] : '');
$temp = (isset($_FILES['file']['tmp_name']) ? $_FILES['file']['tmp_name'] : '');
move_uploaded_file($temp,"uploaded/".$name);
$url = "127.0.0.1/PHP/video%20upload%20and%20playback/uploaded*​/…";
mysqli_query($conn, "INSERT INTO vid VALUE('$id','$name','$url') ");
header("location:dashboard.php");
}
?>

watch.php

<?php
$conn = mysqli_connect("localhost", "root", "", "videos") or die("Connection Server" . mysqli_error());
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title>Watch videos</title>
</head>
<body>
<?php
$conn = mysqli_connect("localhost", "root", "", "videos") or die("Connection Server" . mysqli_error());
if(isset($_GET['id']))
{
$id=(isset($_GET['id']) ? $_GET['id'] : '');
$query=mysqli_query($conn, "SELECT * FROM vid WHERE id='$id'");
while($row=mysqli_fetch_assoc($query))
{
$id = $row['id'];
$name = $row['name'];
$url = $row['url'];
echo "You are watching ".$name."<br />";
echo "<embed src='$id' width='560' height='315'></embed>";
}
}
else
{
echo "Error !";
}
?>
</body>
</html>