我有一个PHP表单，如何为输入字段提供当前page_id的值？

Question

I have a PHP form where I have a hidden input field. I need it to have a value of current page_id, that goes to my SQL table in a column called page_id.

My images table: images = image_id, page_id, image_text, image_file; My pages table: pages = page_id, page_name;

Example of my page URL: mywebsite/page.php?id=4

Form (page.php):

<form method="POST" action="page.php" enctype="multipart/form-data">
     <input type="hidden" name="page_id" value="<?php echo $page['page_id']; ?>">
        <div>
          <input type="file" name="image">
        </div>
        <div>
          <textarea 
            id="text" 
            cols="40" 
            rows="4" 
            name="image_text" 
            placeholder="Write something here.."></textarea>
        </div>
        <div> 
            <button type="submit" name="upload">POST</button>
        </div>
</form>

classes.php file:

class Page {
    public function fetch_all(){
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM pages");
        $query->execute();

        return $query->fetchAll();
    }    
    public function fetch_data($page_id) {
        global $pdo;

        $query = $pdo->prepare("SELECT * FROM pages WHERE page_id = ?");
        $query->bindValue(1, $page_id);
        $query->execute();

        return $query->fetch();
    }
}

With the code I have right now, it gives me a random page_id=6, no idea where the 6 comes from..

What I want: image page_id = page_id of the page where the image was posted.

Any help is appreciated!

Answer 1

$page on your form isn't defined.

Use $_GET to pull the id from the URL.

Your hidden input should change to the following:

<input type="hidden" name="page_id" value="<?php echo $_GET['id'] ?>" />