JS / PHP表单到mysql

Question

我试图从js表单中使用php到mysql数据库获取数据，以便将新用户注册到我的大学项目网站。 我很新，我试图弄清楚发生了什么，但无法让它发挥作用。 一切似乎都在工作，但数据不会被发送到数据库。

<?php

// initializing variables
$username = "";
$password = "";
$title = "";
$firstname = "";
$lastname = "";
$gender = "";
$address = "";
$postcode = "";
$email = "";
$telephone = "";
$errors = array(); 
$_SESSION['success'] = "";

// connect to the database
$db = mysqli_connect('localhost', 'root', '', 'cmetmarketplace');

// REGISTER USER
if (isset($_POST['reg_user'])) {
  // receive all input values from the form
  $username = mysqli_real_escape_string($db, $_POST['username']);
  $password = mysqli_real_escape_string($db, $_POST['password']);
  $password2 = mysqli_real_escape_string($db, $_POST['password2']);
  $title = mysqli_real_escape_string($db, $_POST['title']);
  $firstname = mysqli_real_escape_string($db, $_POST['firstname']);
  $lastname = mysqli_real_escape_string($db, $_POST['lastname']);
  $gender = mysqli_real_escape_string($db, $_POST['gender']);
  $address = mysqli_real_escape_string($db, $_POST['address']);
  $postcode = mysqli_real_escape_string($db, $_POST['postcode']);
  $email = mysqli_real_escape_string($db, $_POST['email']);
  $telephone = mysqli_real_escape_string($db, $_POST['telephone']);



  // form validation: ensure that the form is correctly filled ...
  // by adding (array_push()) corresponding error unto $errors array
  if (empty($username)) { array_push($errors, "Username is required"); }
  if (empty($password)) { array_push($errors, "Password is required"); }
  if (empty($title)) { array_push($errors, "Title is required"); }
  if (empty($firstname)) { array_push($errors, "First name is required"); }
  if (empty($lastname)) { array_push($errors, "Last name is required"); }
  if (empty($gender)) { array_push($errors, "Gender is required"); }
  if (empty($address)) { array_push($errors, "Address is required"); }
  if (empty($postcode)) { array_push($errors, "Postcode is required"); }
  if (empty($email)) { array_push($errors, "Email is required"); }
  if (empty($telephone)) { array_push($errors, "Telephone number is required"); }
  if ($password != $password2) {
    array_push($errors, "The two passwords do not match");
  }


  // Finally, register user if there are no errors in the form
  if (count($errors) == 0) {
    $password = md5($password_1);//encrypt the password before saving in the database

    $query = "INSERT INTO users (username, password, title, first_name, last_name, gender, address, postcode, email, telephone) 
              VALUES('$username', '$password', '$title', '$firstname', '$lastname', '$gender', '$address', '$postcode', '$email', '$telephone')";
    mysqli_query($db, $query);
  }
}

Answer 1

如果没有错误，我会建议做类似的事情：

if( mysqli_query($db, $query) {
    //SUCCESS
} else {
    printf("Error: %s\n", $mysqli->error);
}

这应该有助于调试问题。

不能说现在有什么问题，因为我们必须看到数组的打印等。

否则，请检查错误是什么。