Php Mysql：在创建/编辑图书表时，从另一个表的列表中选择值（作者）

很难解释我需要做什么，但我想知道是否有人能帮助我，我对Mysql很陌生，刚刚开始学习。 所以我有4页，书单和作者名单也增加了新的作者和新书。其中一项任务是在添加新书或edting时，我可以从该书的作者列表中选择一位作者，也可以将其留空。使用

创建了以下3个表：

create table books(
    id INTEGER PRIMARY KEY AUTO_INCREMENT,
    title varchar(255),
    author_id integer,
    grade integer,
    is_read varchar(255) // when using bool, gives an error
);

到目前为止，我已经这样做了，但在添加或编辑书籍数据时，我仍然无法从作者列表中选择作者

function getBooks()
{
    $connection = getConnection();
    $books = [];

    $getBooks = $connection->prepare("SELECT books.id, books.title, author.first_name, author.last_name,
    books.grade, books.is_read FROM books LEFT JOIN books_author INNER JOIN author ON
    books_author.author_id = author.id ON books.id = books_author.book_id WHERE title is not null ");
    $getBooks ->execute();
    foreach ($getBooks as $book) {
        $books[] = ["id" => $book['id'], "title" => urldecode($book['title']),"grade" => trim($book['grade']), "isRead" => trim($book['is_read'])];
    }
    return $books;
}

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左连接。。。在…上你能扩展的正常语法是否不能-发生了什么？在我看来，这个查询不正确——如果在phpmyadmin中执行它，它是否可以正常工作？是的，如上所述，我认为您已经混合了联接，并在第一个联接中附加了一个联接。有关进一步的帮助，请参阅：

create table books_author(
    author_id integer,
    book_id integer
);

function getBooks()
{
    $connection = getConnection();
    $books = [];

    $getBooks = $connection->prepare("SELECT books.id, books.title, author.first_name, author.last_name,
    books.grade, books.is_read FROM books LEFT JOIN books_author INNER JOIN author ON
    books_author.author_id = author.id ON books.id = books_author.book_id WHERE title is not null ");
    $getBooks ->execute();
    foreach ($getBooks as $book) {
        $books[] = ["id" => $book['id'], "title" => urldecode($book['title']),"grade" => trim($book['grade']), "isRead" => trim($book['is_read'])];
    }
    return $books;
}