我试图实现与php东西。我有我的管理表中的两个地址，我会想，以显示我的用户。我想显示的地址之一（ID等于3）一些特定的用户ID和其他地址（ID = 4）的每一个其他用户ID。检查下面我的代码

<?php
$servername = "localhost";
$username = "user";
$password = "password";
$dbname = "user";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$id = "SELECT * FROM users WHERE id = '".$_SESSION['user']."'";

if ( ($id != 11303) && ($id != 27)) {
    $sql = "SELECT address FROM admin WHERE id='3'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            echo "" . $row["address"]. "";
        }
    } else {
        echo "0 results";
    }
}
else {
    $sql = "SELECT address FROM admin WHERE id='4'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            echo "" . $row["address"]. "";
        }
    } else {
        echo "0 results";
    }
}
$conn->close();
?>

我试图从他们的session获取用户ID，然后我用"if"检查，如果他们的session ID从管理表等于某些特定ID，然后select id为3的地址，并显示它"else"它显示的地址与ID 4.但它显示地址ID为3到​​每一个人。请谁能帮帮忙？