Там 3 различных фильтров: books, authors и stores (select списки), и я могу использовать их все сразу или только один или два из них, поэтому я использую UNION, чтобы объединить все запросыСвязывание нескольких массивов в п.д.о.
require('database.php');
if(isset($_POST['books'])){
$books_ids = $_POST["books"];
}
if(isset($_POST['authors'])){
$authors_ids = $_POST["authors"];
}
if(isset($_POST['stores'])){
$stores_ids = $_POST["stores"];
}
$query = "";
if(!empty($books_ids))
{
$books_ids_in = implode(',', array_fill(0, count($books_ids), '?'));
$query .= "SELECT
b.id,
b.`name`,
b.`year`,
GROUP_CONCAT(DISTINCT a.`name`) AS author_names,
GROUP_CONCAT(DISTINCT s.`name`) AS store_names,
'book' as param
FROM
books AS b
LEFT JOIN books_authors AS b_a ON b.id = b_a.book_id
LEFT JOIN authors AS a ON a.id = b_a.author_id
LEFT JOIN books_stores AS b_s ON b.id = b_s.book_id
LEFT JOIN stores AS s ON s.id = b_s.store_id
WHERE
b.id IN (". $books_ids_in .")
GROUP BY b.id
ORDER BY b.id";
}
if(!empty($authors_ids))
{
$authors_ids_in = implode(',', array_fill(0, count($authors_ids), '?'));
if (!empty($query)) {
$query .= " UNION ";
}
$query .= "SELECT
b.id,
b.`name`,
b.`year`,
GROUP_CONCAT(DISTINCT a.`name`) AS author_names,
GROUP_CONCAT(DISTINCT s.`name`) AS store_names,
'author' as param
FROM
books AS b
LEFT JOIN books_authors AS b_a ON b.id = b_a.book_id
LEFT JOIN authors AS a ON a.id = b_a.author_id
LEFT JOIN books_stores AS b_s ON b.id = b_s.book_id
LEFT JOIN stores AS s ON s.id = b_s.store_id
WHERE
b.id IN (
SELECT DISTINCT book_id FROM books_authors WHERE author_id IN (". $authors_ids_in .")
)
GROUP BY b.id
ORDER BY b.id";
}
if(!empty($stores_ids))
{
$stores_ids_in = implode(',', array_fill(0, count($stores_ids), '?'));
if (!empty($query)) {
$query .= " UNION ";
}
$query .= "SELECT
b.id,
b.`name`,
b.`year`,
GROUP_CONCAT(DISTINCT a.`name`) AS author_names,
GROUP_CONCAT(DISTINCT s.`name`) AS store_names,
'store' as param
FROM
books AS b
LEFT JOIN books_authors AS b_a ON b.id = b_a.book_id
LEFT JOIN authors AS a ON a.id = b_a.author_id
LEFT JOIN books_stores AS b_s ON b.id = b_s.book_id
LEFT JOIN stores AS s ON s.id = b_s.store_id
WHERE
b.id IN (
SELECT DISTINCT book_id FROM books_stores WHERE store_id IN (". $stores_ids_in .")
)
GROUP BY b.id
ORDER BY b.id";
}
if(!empty($query)) {
$stmt = $conn->prepare($query);
if(!empty($books_ids))
{
foreach ($books_ids as $k => $id) {
$stmt->bindValue(($k+1), $id);
}
}
if(!empty($authors_ids))
{
foreach ($authors_ids as $k => $id) {
$stmt->bindValue(($k+1), $id);
}
}
if(!empty($stores_ids))
{
foreach ($stores_ids as $k => $id) {
$stmt->bindValue(($k+1), $id);
}
}
$stmt->execute();
$results = $stmt->fetchAll();
echo json_encode($results);
}
$conn = null;
код прекрасно работает, когда я использую только один фильтр, но когда я пытаюсь использовать 2 или более, я получаю ошибку
Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid parameter number: number of bound variables does not match number of tokens' in C:\xampp\htdocs\bookstore\filter.php:123 Stack trace: #0 C:\xampp\htdocs\bookstore\filter.php(123): PDOStatement->execute() #1 {main} thrown in C:\xampp\htdocs\bookstore\filter.php on line 123
Я думаю, что-то не так с использованием bindValue, но я не знаю, как это исправить?
UPD var_dump($query) (3 книги и 2 авторы выбрали)
string(1097) "SELECT b.id, b. имени , b. года , GROUP_CONCAT(DISTINCT a. имени ) AS author_names, GROUP_CONCAT(DISTINCT s. имени ) AS store_names, 'book' as param FROM books AS b LEFT JOIN books_authors AS b_a ON b.id = b_a.book_id LEFT JOIN authors AS a ON a.id = b_a.author_id LEFT JOIN books_stores AS b_s ON b.id = b_s.book_id LEFT JOIN stores AS s ON s.id = b_s.store_id WHERE b.id IN (?,?,?) GROUP BY b.id ORDER BY b.id UNION SELECT b.id, b. имени , b. года , GROUP_CONCAT(DISTINCT a. имени ) AS author_names, GROUP_CONCAT(DISTINCT s. имени ) AS store_names, 'author' as param FROM books AS b LEFT JOIN books_authors AS b_a ON b.id = b_a.book_id LEFT JOIN authors AS a ON a.id = b_a.author_id LEFT JOIN books_stores AS b_s ON b.id = b_s.book_id LEFT JOIN stores AS s ON s.id = b_s.store_id WHERE b.id IN (SELECT DISTINCT book_id FROM books_authors WHERE author_id IN (?,?)) GROUP BY b.id ORDER BY b.id" 01201
Ошибка означает, что у вас меньше/больше значений привязки, чем ожидалось в запросе. Я бы выделил количество циклов bindValue, чтобы найти виновника. –
Нет, количество аргументов такое же, как в запросе, я уже проверил. Я выбрал 3 книги и 2 автора и в моем запросе 'WHERE b.id IN (?,?,?)' И 'WHERE author_id IN (?,?))' – Heidel
var_dump ($ query); и показать результат, пожалуйста, –