如何通过PHP将数据插入到嵌套表中？

Question

我认为这与PDO有关。

这是我patientinfo表

patientid | name | age | email | address 

，这是我的发言表

patientid | remarksid | date | description 

我想将数据插入到patientinfo和言论表，其中两个表patientid将被同步。

问题是我不知道如何查询这个。这是我做的，但它给了我一个错误。

$query = "INSERT INTO patientinfo (name, age, email, address) 
       VALUES (:name, :age, :email, :address);"; 

    $query_params = array(
      ':name'  => $_POST['name'], 
      ':age'  => $_POST['age'], 
      ':email' => $_POST['email'], 
      ':address' => $_POST['address'], 
    ); 

    $query = "INSERT INTO remarks (patient_id, description) VALUES (:patient_id, :remarks) WHERE remarks.patient_id = patientinfo.patient_id;"; 
    $query_params = array(':remarks' => $_POST['remarks']); 

    try{ 
     $stmt = $dbname->prepare($query); 
     $result = $stmt->execute($query_params); 
    } 

    catch(PDOException $ex){ 
      $response["success"] = 0; 
      $response["message"] = $ex ; 

      die(json_encode($response)); 
    } 

我在patientinfo AUTOINCREMENT做出patientid。请致电 ！非常感谢你的帮助！

Answer 1

$query = "INSERT INTO patientinfo (name, age, email, address) 
       VALUES (:name, :age, :email, :address);"; 

$query_params = array(
     ':name'  => $_POST['name'], 
     ':age'  => $_POST['age'], 
     ':email' => $_POST['email'], 
     ':address' => $_POST['address'], 
); 

try{ 
    $stmt = $dbname->prepare($query); 
    $stmt->execute($query_params); 

    $patient_id = $dbname->lastInsertId(); 

    $query = "INSERT INTO remarks (patientid, description) VALUES (:patient_id, :remarks)"; 
    $query_params = array(':remarks' => $_POST['remarks'],':patient_id'=>$patient_id); 

    $q = $dbname->prepare($query); 
    $q->execute($query_params); 

}catch(PDOException $ex){ 
     $response["success"] = 0; 
     $response["message"] = $ex ; 

     die(json_encode($response)); 
} 

你应该写这样的东西。请检查列名（patientid或patient_id？）