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testingvideo.php
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testingvideo.php
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<?php
session_start();
require('php/database.php');
//zoek de id in de url
$id = $_GET['id'];
//query voor de om de video op te halen
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<?php require('components/head.php');?>
<title>Social Guys</title>
<link rel="stylesheet" href="css/kanaal.css">
</head>
<body class="bg-back">
<div class="d-flex">
<?php require("components/sidebar.php"); ?>
<div id="page-content-wrapper">
<?php require("components/navigation.php");?>
<div class="col">
<?php
$query = "SELECT * FROM `video` WHERE `KanaalID` = ". $id;
//query om de kanaal te zoeken
$queryKanaal = "SELECT * FROM kanaal WHERE Kanaal_ID = ". $id;
//subscriber count
$subscribers = "SELECT COUNT(KanaalID) FROM `subscriptions` WHERE Subscription = ". $id;
//resultaat voor de video query
$result = mysqli_query($conn, $query);
$foutmelding = "";
$path = 'uploads/videos/';
//checked of de url wel een id ontvangt en geen string
if ((int) $id === $id)
{
$foutmelding .= "<p>ERROR: You're trying to find a user by name! Not by ID.</p>";
echo $foutmelding;
exit;
}
if ($result)
{
//resultaat voor de kanaal query
$resultaat = mysqli_query($conn, $queryKanaal);
//checked of de resultaat goed is gegaan als het niet zo is dan krijg je een foutmelding
if (!$resultaat)
{
$foutmelding .= "<p>ERROR: Something went wrong trying to connect to the database.";
echo $foutmelding;
exit;
}
//loop over de resultaat voor van de resultaten.
foreach($resultaat as $kanaal)
{
$subscribeQuery = "SELECT * FROM `subscription` WHERE `KanaalID` = '$id'";
// echo "";
// echo "<img class='img-fluid ml-1' style='border-radius: 150px;' src='uploads/profile/".$kanaal['ProfielPhoto']."' alt='Profilephoto for this channel'>";
// echo "<h2 class='text-white'>".$kanaal['Naam']."</h2>";
?>
<div class="col-md-12">
<img src="uploads/<?php echo $kanaal['ProfielPhoto']; ?>" >
<p><?php echo $kanaal['Naam']; ?></p>
<?php
//om de subscriber count te laten zien
//de row['COUNT(KanaalID)'] is de tabel hoofd output van de sql query
//dit is nu de beste manier om dit te doen en het werkt. Dit zelfde kan je ook gebruiken voor view count.
$subscribercount = mysqli_query($conn, $subscribers);
if ($subscribercount->num_rows > 0) {
// output data of each row
while($row = $subscribercount->fetch_assoc()) {
echo "<label> ".$row['COUNT(KanaalID)']." Subscribers</label>";
}
} else {
echo "<label> 0 Subscribers</label>";
}
?>
</div>
<form action="php/subscribe.php" method="post">
<input type="hidden" name="kanaalID" value="<?php echo $kanaal['Kanaal_ID']; ?>">
<input type="submit" class="btn btn-outline-info" name="subscribe" value="Subscribe"></input>
</form>
<form action="php/unsubscribe.php" method="post">
<input type="hidden" name="kanaalID" value="<?php echo $kanaal['Kanaal_ID']; ?>">
<input type="submit" class="btn btn-outline-info" name="subscribe" value="Unsubscribe"></input>
</form>
<?php
foreach($result as $video)
{
//query voor de video categorie
$queryC = "SELECT Naam FROM `categorie` WHERE Categorie_ID = ". $video['CategorieID'];
$resultC = mysqli_query($conn, $queryC);
$filename = $video['File'];
$filepath = $path.$filename;
$fileExtension = substr($filename, -3);
//echo de video titel
// echo "<h3 class='text-white'>".$video['Naam']."</h3>";
// echo "<video witdth='320' height='240' poster='uploads/thumbnails/".$video['Thumbnail']."' controls muted>";
// echo "<source src='".$filepath."' type='video/".$fileExtension."'>";
// echo "Your browser doesn't support the video tag!";
// echo "</video>";
foreach($resultC as $catagorie)
{
//echo de categorie naam door de query
//echo "<p class='text-white'>".$catagorie['Naam']."</p>";
//om de meest bekeken video te vinden SELECT COUNT(`VideoID`) AS `value_occurrence` FROM `views` GROUP BY `VideoID` ORDER BY `value_occurrence` DESC LIMIT 1
//deze is misschien beter SELECT `VideoID` FROM `views` GROUP BY `VideoID` ORDER BY COUNT(*) DESC LIMIT 1 limit kan veranderd worden naar van alles
$views = "SELECT COUNT(VideoID) FROM `views` WHERE VideoID = ". $video['Video_ID'];
?>
<a href="video.php?id=<?php echo $video['Video_ID']; ?>">
<div class="col-md-4">
<div class="card mb-4 text-white bg-dark">
<img class="card-img-top" src="uploads/thumbnails/<?php echo $video['Thumbnail'] ?>" >
<div class="card-body">
<h5 class="card-title"><?php echo $video['Naam']; ?></h5>
<div class="info-section">
<?php
//om de subscriber count te laten zien
//de row['COUNT(KanaalID)'] is de tabel hoofd output van de sql query
//dit is nu de beste manier om dit te doen en het werkt. Dit zelfde kan je ook gebruiken voor view count.
$viewcount = mysqli_query($conn, $views);
if ($viewcount->num_rows > 0) {
// output data of each row
while($rows = $viewcount->fetch_assoc()) {
echo "<label> ".$rows['COUNT(VideoID)']." Views</label>";
}
} else {
echo "<label> 0 Views</label>";
}
?>
<span> <?php echo $catagorie['Naam']; ?> </span>
</div>
</div>
</div>
</a>
<?php
}
}
}
} else
{
$foutmelding .= "<p>ERROR: Something went wrong trying to connect to the database.";
echo $foutmelding;
exit;
}
?>
</div>
</div>
<?php require("components/comments.php");?>
</div>
<?php require("components/scripts.php");?>
</body>
</html>