使用按钮将图像从一张桌子移动到另一张桌子

Question

I am creating a button that will allow an admin to verify the image from users:

I don't know how to get the image ID when I click accept or reject.

Here's my code:

<?php

    while($row = mysqli_fetch_array($result)) {
      echo "<div class='grid-item'><img src='unimages/{$row['un_image']}' 
 onclick=onClick(this) style='width:98%' class='verifyimage' />
            <form method='post' action='adminverify.php'>
            <input class='button1' type='submit' name='accept' value='✓'>  
            <input class='button2' type='submit' name='reject' value='✘'>  
            </form>
            </div>
            ";

    }

mysqli_close($db);
?> 
</div>

If the admin accepts, then the image should move from table2 to table1.

I know using INSERT INTO and DELETE will work, but how do I get the id for my picture.

Table 1:

Table 2:

Answer 1

Mmmh, you can use GET method for easier script:

while($row = mysqli_fetch_array($result)) {
  echo "
    <a href='?&action=accept&id={$row['un_id']}'>Accept</a>
    <a href='?&action=reject&id={$row['un_id']}'>Reject</a> 
  ";
}

And you use it like :

if(isset($_GET['action']) && isset($_GET['id'])) {
  $image_id = $_GET['id']
  // Then check if you must accept or reject with $_GET['action'] value
}

Answer 2

I'd add a hidden input that will be sent with the form:

echo "<div .....
      <form....>
      <input type='hidden' name='imageId' value='{$row['un_id']}'>
      ....
      </form></div>";

You'll then have it in your receiving php script as

$image_id = $_POST['imageId'];