Upload video with php and html - php

Question

How can i upload a video ? i did this with a image file but is not working with a video one only showing "error"
<?php
session_start();
if( !isset($_SESSION["username"]) ){
echo "<p id='errors'>Por favor, loguese, si no es redireccionado haga click <a href='log.php'>aqui</a></p>";
header("location: log.php");
}else{
};
$img = $_GET["fileToUploadv"];
$target_dir = "video/";
$target_file = $target_dir . basename($_FILES["fileToUploadv"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUploadv"]["tmp_name"]);
if($check !== false) {
$uploadOk = 1;
} else {
echo "The file is not an image!.";
$uploadOk = 0;
}
}
// Check if file already exists
// Check file size
// Allow certain file formats
if($imageFileType != "mp4" && $imageFileType != "mp4" ) {
echo "Only use a .JPG file.";
$uploadOk = 0;
}
if (file_exists($target_file)) {
unlink("video/video.mp4");
unlink("video/video.avi");
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "Error.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["fileToUploadv"]["tmp_name"], $target_file)) {
echo "The file ". basename( $_FILES["fileToUploadv"]["name"]). " has been uploaded.";
rename("video/".basename( $_FILES["fileToUploadv"]["name"]), "video/video.mp4");
rename("video/".basename( $_FILES["fileToUploadv"]["name"]), "video/video.avi");
} else {
echo "Error.";
}
}
?>
this script have filetouploadv that is with a form input to upload mp4 or avi that is html
<form id="formimg" method="post" action="uploadvideo.php" enctype="multipart/form-data">
<label for="uploadimgg">
<p id="selectimg">Please select a MP4 or AVI file to upload.</p>
<i class="fas fa-upload"></i>
</label>
<input id="uploadimgg" type="file" name="fileToUploadv" accept="video/mp4,video/avi">
<input class="hideshowsub" id="submitpho" type="submit">
</form>
should change the name to video and upload to video/ but is not working ...move_uploaded_file because that reutrn false and do the "error" message

Answer 1

<?php
require("config.php");
$sql = "SELECT * FROM inventario";
$result = mysqli_query($link, $sql);
echo("<table>");
echo("<tr>");
echo ("<th class='midtable'>"."Username"."</th>");
echo ("<th class='midtable'>"."Password"."</th>");
echo ("<th class='midtable'>"."</th>");
echo ( "</tr>");
while ($row = mysqli_fetch_array($result))
{
echo("<tr>");
echo("<td><div class='data'>".$row['username']."</div><div class='edit'><input type='text' name='usrname' value='".$row['username']."'></div></td>"."<td><div class='data'".$row['password']."<div class='edit'><input type='text' name='password' value='".$row['password']."'></div></td>"."<td>"."<i class='fas fa-edit'></i>"."<i class='fas fa-trash-alt'></i>"."</td>" );
echo("</tr>") ;
}; ?>
and then using jQuery hide the class data and show the class edit elements in the row and a submit button to update the values in the DB... and perhaps update just that row...

Answer 2

You can't upload videos because of this:
if($check !== false)
Turn it from false to true:
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUploadv"]["tmp_name"]);
if($check !== true) {
$uploadOk = 1;
} else {
echo "The file is not an image!.";
$uploadOk = 0;
}
}