PHP错误：mysqli_stmt_bind_param（）变量数量与预准备语句中的参数数量不匹配

Question

我在我的php文件中遇到问题，错误来自第三个语句中的参数，但我找不到它在哪里。这是我的PHP文件：

$client = "test";
$resto = "test";

$st1 = mysqli_query($con, "SELECT client_id FROM Clients WHERE email = '$client'");
$res1 = mysqli_fetch_array($st1);
$clientID = $res1["client_id"];

echo "CLIENT ID = ";
echo $clientID;

$st2 = mysqli_query($con, "SELECT restaurant_id FROM Restaurants WHERE email = '$resto'");
$res2 = mysqli_fetch_array($st2);
$restaurantID = $res2["restaurant_id"];

echo "RESTO ID = ";
echo $restaurantID;

$statement = mysqli_prepare($con, "INSERT INTO reservations (resto_id, client_id) VALUES ($restaurantID,$clientID)");

mysqli_stmt_bind_param($statement,"ii",$restaurantID,$clientID);
mysqli_stmt_execute($statement);

$response = array();
$response["success"] = true;  

echo json_encode($response);

提前感谢您的帮助。

Answer 1

改变你的陈述，如下文

$statement = mysqli_prepare($con, "INSERT INTO reservations (resto_id, client_id) VALUES ( ? , ? )");