PHP if语句中的else语句不提供输出也不显示任何错误

Question

我希望代码在其中roommate为null的Student_room表中插入一行。例如，当我尝试在roomate_id的地方插入room_id= Block 1/R2时，应该在The room is fully occupied....的地方插入roomate_id并打印room_id = Block 1/R1。

else {
    $sql2 = $mysqli->query("SELECT * FROM `student_room` WHERE room_id = '$room_id' AND roomate_id IS NULL AND academic_year = '$academic_year'");

    if($sql2->num_rows == 0) {
        echo "<p2>The double room is fully occupied please check other rooms</p2>";
    } else {
        $assign = "UPDATE `student_room` SET roomate_id = $student_id WHERE room_id = '$room_id'";

        if (mysqli_query($mysqli, $assign)) {
            $update = "UPDATE `room` SET room_status = 'occupied' WHERE room_id = '$room_id'";                
        }

        if (mysqli_query($mysqli, $update)) {
            $app_no = $mysqli->query("SELECT app_id FROM `application` WHERE student_id = $student_id");
            $row = mysqli_fetch_array($app_no);
            $app_id = $row['app_id'];
            $username = $_SESSION['username'];

            $sql3 ="INSERT INTO `staff_application` (username, app_no, academic_year) VALUES ('$username', $app_id, '$academic_year')";
        }

        if (mysqli_query($mysqli, $sql3)) {
            echo "<p2>Room successfully assigned</p2>";