How to insert JavaScript Data into mysql database? - php

Question

The Data i am trying to insert is a blob or a file
This is for a school project and the lecturer said to insert it into the database for now.
This is what i have right now
function stop() {
cancelAnimationFrame(rafId);
endTime = Date.now();
$('#stop-me').disabled = true;
document.title = ORIGINAL_DOC_TITLE;
toggleActivateRecordButton();
console.log('frames captured: ' + frames.length + ' => ' +
((endTime - startTime) / 1000) + 's video');
embedVideoPreview();
};
function embedVideoPreview(opt_url) {
var url = opt_url || null;
var video = $('#video-preview video') || null;
var downloadLink = $('#video-preview a[download]') || null;
if (!video) {
video = document.createElement('video');
video.autoplay = true;
video.controls = true;
video.loop = true;
//video.style.position = 'absolute';
//video.style.top = '70px';
//video.style.left = '10px';
video.style.width = canvas.width + 'px';
video.style.height = canvas.height + 'px';
$('#video-preview').appendChild(video);
downloadLink = document.createElement('a');
downloadLink.download = 'capture.webm';
downloadLink.textContent = '[ download video ]';
downloadLink.title = 'Download your .webm video';
var p = document.createElement('p');
p.appendChild(downloadLink);
$('#video-preview').appendChild(p);
} else {
window.URL.revokeObjectURL(video.src);
}
if (!url) {
var webmBlob = Whammy.fromImageArray(frames, 1000 / 60);
url = window.URL.createObjectURL(webmBlob);
}
video.src = url;
downloadLink.href = url;
And this is how i am inserting it into the database from the same page i am also not to sure on where the video blob is also created.
<?php
require("connect.php");
$namey = video;
$up = mysql_query("INSERT INTO video VALUES ($namey)");
?>

Answer 1

Ok, so say you have an input for a user to upload a video...
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="VideoToUpload" />
<input type="submit" value="Upload Video" />
</form>
Then in your new upload.php you will want to process and save the file
$allowedExts = array("mp4", "mov");
$extension = end(explode(".", $_FILES["VideoToUpload"]["name"]));
if (($_FILES["VideoToUpload"]["type"] == "video/mp4") || ($_FILES["VideoToUpload"]["type"] == "video/quicktime")):
if($_FILES["VideoToUpload"]["error"] > 0):
echo "Error: " . $_Files["VideoToUpload"]["error"];
else:
move_uploaded_file($_FILES["VideoToUpload"]["tmp_name"], dirname(__FILE__) . "/FolderWhereIWantMyVideoSaved/" . $_FILES["VideoToUpload"]["name"]);
endif;
endif;
$VideoURL = "http://domain/and/path/to/FolderWhereIWantMyVideoSaved/" . $_FILES["VideoToUpload"];
$mysqli = new mysqli('ip.of.data.base', 'DatabaseName', 'Password', 'Username');
$stmt = $mysqli->prepare("INSERT INTO Videos (Name, Type, URL) VALUES (?,?,?)") or die ($mysqli->error);
$stmt->bind_param('sss', $_FILES["VideoToUpload"]["name"],$_FILES["VideoToUpload"]["type"], $VideoURL);
$stmt->execute() or die ($mysqli->error);
$stmt->close();
$mysqli->close();
Now the file is saved on the server, and the URL to that is in the database. So to display the video all you would then do is.
$mysqli = new mysqli('ip.of.data.base', 'DatabaseName', 'Password', 'Username');
$stmt = $mysqli->prepare("SELECT Name, Type, URL FROM Videos WHERE ID=?") or die ($mysqli->error);
$stmt->execute() or die($mysqli->error);
$stmt->bind_result($Name, $Type, $URL);
$stmt->store_result();
$stmt->fetch();
...
<video width="320" height="240" controls>
<source src="<?= $URL ?>" type="<?= $Type ?>">
Your browser does not support the video tag.
</video>
...
$stmt->close();
$mysqli->close();