HTML表單輸入錯誤PHP

Question

            <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專家”交談了幾個小時，他似乎無法弄清楚我的問題。 我的輸入沒有提交給我的數據庫，標題/說明。 我可以用

$title = $_POST['dsfasfddsfa']; 

並將其插入，但是我無法插入用戶輸入的字段。 當我添加

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

我在標題和描述上得到了不確定的索引，這就是為什么我在PHP代碼頂部添加了if語句的原因。

我在一個網站上讀到了要刪除它們的信息，我以為一旦它們消失就可以了，我錯了。 當我這樣做時，什么都沒有。 我可以上傳視頻，但是這些字段不起作用。 任何幫助都很棒！ 在此先感謝，我也使用了提交作為$ _POST在兩個if語句中，如果PHP代碼不起作用的話。

Answer 1

首先嘗試制作一個簡單的大衛版本，如本例所示。

http://php.net/manual/es/tutorial.forms.php

<form action="action.php" method="post">
<p>Your name: <input type="text" name="name" /></p>
 <p>Your age: <input type="text" name="age" /></p>
 <p><input type="submit" /></p>
</form>

。

Hi <?php echo htmlspecialchars($_POST['name']); ?>.
You are <?php echo (int)$_POST['age']; ?> years old.

然后，您將逐步逐步更改所需的內容。

最后會工作。