PHP：通过id变量数据显示外键数据

Question

<?php

include 'dbconnection.php';

$value = $_POST['value'];

$sql = "SELECT * FROM users WHERE id='$value' INNER JOIN address ON users.address_id = address.id";
$results = mysqli_query($conn, $sql);

if(mysqli_num_rows($results) > 0)
{
    while($row = mysqli_fetch_array($results))
    {
        ?>
        <a>Details for: <?php echo $row['Email'];?></a><br>
        <a>Address line 1: <?php echo $row['line1'];?></a><br>
        <a>Address line 2:<?php echo $row['line2'];?></a><br>
        <a>City: <?php echo $row['city'];?></a><br>
        <a>County: <?php echo $row['county'];?></a><br>
        <a>Postcode: <?php echo $row['postcode'];?></a><br>
        <a>Country: <?php echo $row['country'];?></a><br><br>


    <?php
    }
}

?>

我试图允许管理员从表用户查看用户的详细信息，然后在表地址中显示外键数据，所有这一切都应该通过输入来完成，这些输入将存储到$价值变量。我可以让表中的每个用户一次显示但不是特定的一个。我错过了什么？