如何通过PHP将数据插入嵌套表中

Question

我认为这与PDO有关。

这是我的病人表

patientid | name | age | email | address

这是我的言论表

patientid | remarksid | date | description

我想将数据插入患者和备注表中，这些表将同步这两个表的患者。

问题是我不知道该如何查询。这是我要做的，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));
    }

我在耐心Info自动插入中做了患者。请！非常感谢您的帮助！

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));
}

您应该写这样的东西。请检查列的名称（dateberId或attery_id？）