我写了以下php代码:
if ($insert_client_stmt = $mysqli->prepare("INSERT INTO client (username, email, password) VALUES (?, ?, ?)")) {
$insert_client_stmt->bind_param('ssss', $username, $email, $password);
// Execute the prepared query.
if (! $insert_client_stmt->execute()) {
header('Location: ../error.php?err=Registration failure: INSERT');
}
}
$client_id = $mysqli->insert_id;
if ($insert_client_details_stmt = $mysqli->prepare("INSERT INTO client_addr (addr_id, client_id, client_name, contact_name) VALUES (?, ?, ?, ?)")) {
$insert_client_details_stmt->bind_param('ssss', $client_id, $_POST['company'], $_POST['contact']);
// Execute the prepared query.
if (! $insert_client_details_stmt->execute()) {
header('Location: ../error.php?err=Registration failure: INSERT');
}
}
header('Location: ./register_success.php');
和经过该代码后，我的浏览器将我重定向到页面register_success.php，但在我的数据库中，我只有新客户端，client_addr中的条目缺失。是什么导致的呢?我需要重新设置我自己吗?谢谢!
你的代码有两个问题。
首先，你以最糟糕的方式处理错误。因此，使您无法获得有用的错误消息，这可能会帮助您。为了在mysqli中正确处理错误，你必须摆脱所有这些无用的if，并告诉mysqli在发生错误时抛出异常。
第二个错误本身很简单:您将三个变量传递给第二个查询，而在INSERT查询中有四个字段。您必须从查询
中取出addr_id和相应的占位符所以，代码应该是。
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$stmt = $mysqli->prepare("INSERT INTO client (username, email, password) VALUES (?,?,?)");
$stmt->bind_param('sss', $username, $email, $password);
$stmt->execute();
$client_id = $mysqli->insert_id;
$stmt = $mysqli->prepare("INSERT INTO client_addr (client_id, client_name, contact_name) VALUES (?,?,?)");
$stmt->bind_param('sss', $client_id, $_POST['company'], $_POST['contact']);
$stmt->execute();
header('Location: ./register_success.php');