如何将多个值提取到选择表单中，然后再次将它们添加到新表中？

问：

//Connects to the database
$mysqli = new mysqli(DB_HOST,DB_USER,DB_PASSWORD,DB_DB);
if($mysqli->connect_errno){
    echo "Connection error " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}

if (isset($_POST['points']) & isset($_POST['user'])) {
    if(!($stmt = $mysqli->prepare("INSERT INTO penalty (user_id, first_name, last_name, name, value) VALUES (?, ?, ?, ?, ?);"))){
        echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
    }

    if(!($stmt->bind_param("sssss",$_POST['points'],$_POST['user']))){
        echo "Bind failed: " . $stmt->errno . " " . $stmt->error;
    }

    if(!$stmt->execute()){
        echo "Execute failed: " . $stmt->errno . " " . $stmt->error;
    } else {
        $stmt->close();
        header('Location: GivePoints1.php');
        echo "Added successfully";
        exit();
    }

    $stmt->close();

}

<form class="form-horizontal" action="GivePoints1.php" method="post">

    <div class="form-group">
        <label class="control-label col-sm-2">Penalty:</label>
        <select name="points[]" id="points">
            <?php

            //Prepare SELECT statement for user's name
            if(!($stmt = $mysqli->prepare("SELECT name, value FROM points"))){
                echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
            }

          
            //Execute the SELECT statement
            if(!$stmt->execute()){
                echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }

            //Bind values to variables
            if(!$stmt->bind_result($name, $value)){
                echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }

            //Output name
            while($stmt->fetch()){
                echo'<option value=" '.$name.' "> '.$name.' - ' . $value .' Pts </option>';
            }

            $stmt->close();
            ?>
            </select>
    </div>

    <div class="form-group">
         <label class="control-label col-sm-2">Give To:</label>
        <select name="user[]" id="user">
            <?php

            //Prepare SELECT statement for user's name
            if(!($stmt = $mysqli->prepare("SELECT id, first_name, last_name FROM award_user"))){
                echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
            }

          
            //Execute the SELECT statement
            if(!$stmt->execute()){
                echo "Execute failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }

            //Bind values to variables
            if(!$stmt->bind_result($id, $first_name, $last_name)){
                echo "Bind failed: "  . $mysqli->connect_errno . " " . $mysqli->connect_error;
            }

            //Output name
            while($stmt->fetch()){
                echo'<option value=" '.$id.' "> '.$first_name.' ' . $last_name .' </option>';
            }

            $stmt->close();
            ?>
            </select>
    </div>

    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-10">
            <button type="submit" class="btn btn-lg btn-primary ">Submit</button>
        </div>
    </div>
</form>

有人可以引导我走上正确的道路吗？

我需要将这些值添加到新表中，其中 id 将user_id（从表中提取award_user id 更改为惩罚表中的user_id）

我需要将 id - 作为 user_id、first_name、last_name、名称和值放入表惩罚中。它打印选择没有问题，但不会向新表添加任何内容

我现在想做一段时间，但没有运气。我会做一些解释