Javascript 在MySQL中将表的一行链接到另一个表的多行

我有以下表格：

提交表单后，个人详细信息将插入个人表，书籍详细信息将插入书籍表。我想通过一个唯一的id将一个个人表行链接到book表中的多行（因为它是用同一个查询插入的），这样我可以在以后确定这些书籍与此人相关

这是我的PHP和MySQL脚本：

$mysqli = new mysqli($host,$user,$password,$database);

$users_firstname = $_POST['firstname'];
$users_middlename = $_POST['middlename'];
$users_lastname = $_POST['lastname'];
$users_gender= $_POST['gender'];
$users_location= $_POST['location'];
$users_email= $_POST['email'];
$users_mobile= $_POST['mobile'];

$query = "INSERT INTO personaldetails(FirstName ,MiddleName,LastName,
Gender,Location,Email,Mobile) VALUES ('$users_firstname',
'$users_middlename', '$users_lastname', '$users_gender','$users_location','$users_email','$users_mobile');";

foreach($_POST['booktitle'] as $key => $bookTitle) {
    $bookTitle = mysqli_real_escape_string($mysqli, $bookTitle);
    $bookGenre = mysqli_real_escape_string($mysqli, $_POST['bookgenre'][$key]);
    $bookWriter = mysqli_real_escape_string($mysqli, $_POST['bookwriter'][$key]);
    $bookDescription = mysqli_real_escape_string($mysqli, $_POST['bookdescription'][$key]);

    $query .= "INSERT INTO bookdetails(BookTitle ,BookGenre,BookWriter,
                BookDescription) VALUES('$bookTitle',
             '$bookGenre', '$bookWriter', '$bookDescription');";
}

$result = mysqli_multi_query($mysqli, $query);

---

一个可能的解决方案是使用

但是您还必须在book table中创建一个额外的列来存储用户id，以便将一本书与一个用户关联起来，让它成为user\u id（它将存储我们将通过

mysqli\u insert\u id（）

获得的新创建用户的id）

您还必须单独执行查询以获取新插入的用户Id

因此，代码将类似于-

$mysqli = new mysqli($host,$user,$password,$database);

    $users_firstname = $_POST['firstname'];
    $users_middlename = $_POST['middlename'];
    $users_lastname = $_POST['lastname'];
    $users_gender= $_POST['gender'];
    $users_location= $_POST['location'];
    $users_email= $_POST['email'];
    $users_mobile= $_POST['mobile'];

    $user_query = "INSERT INTO personaldetails(FirstName ,MiddleName,LastName,
    Gender,Location,Email,Mobile) VALUES ('$users_firstname',
    '$users_middlename', '$users_lastname', '$users_gender','$users_location','$users_email','$users_mobile');";

//execute the user query
$result = mysqli_query($mysqli, $user_query);
//get the user id of newly inserted user
$user_id = mysqli_insert_id($mysqli);

    foreach($_POST['booktitle'] as $key => $bookTitle) {
        $bookTitle = mysqli_real_escape_string($mysqli, $bookTitle);
        $bookGenre = mysqli_real_escape_string($mysqli, $_POST['bookgenre'][$key]);
        $bookWriter = mysqli_real_escape_string($mysqli, $_POST['bookwriter'][$key]);
        $bookDescription = mysqli_real_escape_string($mysqli, $_POST['bookdescription'][$key]);

//use the user id here to relate it with the book
    $book_query = "INSERT INTO bookdetails(BookTitle ,BookGenre,BookWriter,
                BookDescription, user_id) VALUES('$bookTitle',
             '$bookGenre', '$bookWriter', '$bookDescription', '$user_id');";
//execute the query for book
$result = mysqli_query($mysqli, $book_query);

}

---

显示此错误，解析错误：语法错误，意外的“foreach”（T_foreach）非常感谢@manoj:）虽然它不是唯一的，但我现在有了基本的想法，我将这样做。谢谢