PHP錯誤：mysqli_stmt_bind_param（）變量數量不匹配準備語句中的參數數量

Question

我面臨一個問題，在我的PHP文件中，錯誤來自第三條語句中的參數，但我無法找到哪裏是嗎。這裏是我的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); 

非常感謝您的幫助。

Answer 1

更改您的發言像波紋管

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