HTML表单输入错误PHP

            <form action="index.php" name="Submit_Form" method="post" enctype="multipart/form-data" id="upload" class="upload">
        <fieldset>
            <legend>Upload</legend><br/>
            Title:  <input type="text" name="title" id="name" class="name" required> <br/><br/>
            <textarea name="description" rows="6" cols="35" maxlength="120"></textarea><br/>
            <input type="file" id="file" name="file[]" required multiple><br/>
            <input type="submit" id="submit" name="submit" value="Upload">
        </fieldset> 

        <div class="bar">
            <span class="bar-fill" id="pb"><span class="bar-fill-text" id="pt"></span></span>
        </div>
        <div id="uploads" class="uploads"> 
        Uploaded file links will appear here.   
        </div>
<?php
if (isset($_POST['title'])) {
$title = $_POST['title'];
}
if (isset($_POST['description'])) {
$description = $_POST['description'];
}
$dbhost     = "localhost";
$dbname     = "blog";
$dbuser     = "root";
$dbpass     = "";

// database connection
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// query
$stmt = $conn->prepare("INSERT INTO videos (title,description) VALUES (?,?)");
$stmt->bind_param("ss", $title, $description);
$stmt->execute();
$stmt->close();
$conn->close();
 ?> 
 <script src="upload.js"></script>
        <script>
            document.getElementById('submit').addEventListener('click', function(e) {
                e.preventDefault();
                var f = document.getElementById('file'),
                    pb = document.getElementById('pb'),
                    pt = document.getElementById('pt');

                    blog.uploader({
                    files: f,
                    progressBar: pb,
                    progressText: pt,
                    processor: 'uploads.php',
                    finished: function(data) {
                        var uploads = document.getElementById('uploads'),
                        succeeded = document.createElement('div'),
                        failed = document.createElement('div'),
                        anchor,
                        span,
                        x;
                    if(data.failed.length) {
                        failed.innerHTML = '<p>This item failed to upload:</p>';
                    }
                    uploads.textContent = '';
                    for(x = 0; x < data.succeeded.length; x = x + 1) {
                        anchor = document.createElement('a');
                        anchor.href = 'uploads/' + data.succeeded[x].file;
                        anchor.textContent = data.succeeded[x].name;
                        anchor.target = '_blank';
                        succeeded.appendChild(anchor);  
                    }
                    for(x = 0; x < data.failed.length; x = x + 1) {
                        span = document.createElement('span');
                        span.textContent = data.failed[x].name;
                        failed.appendChild(span);
                    }
                    uploads.appendChild(succeeded);
                    upload.appendChild(failed);
                },
                    error: function() {
                        console.log('Not working'); 
                    }
                });
            });
        </script>
    </form>
</body>
</html> 

我和"PHP专家"谈了几个小时，他似乎无法解决我的问题。我的输入没有提交到我的数据库，标题/描述。我可以用

$title = $_POST['dsfasfddsfa']; 

，它会插入，但是我不能插入用户输入的内容。当我添加

$title = $_POST['title']; & $description = $_POST['description']; 

我在标题和描述上得到了未识别的索引，这就是为什么我在php代码的顶部添加了if语句。

我在一个网站上看到它应该删除它们，我以为一旦它们消失就会起作用，我错了。当我这样做的时候，没有任何效果。我可以上传视频，但田地坏了。任何帮助都是伟大的!提前感谢。如果php代码不工作，我还在开始的两个if语句中使用submit作为$_POST。