php - MySQL 插入在一次提交中发生两次

Question

我有这个 PHP 文件，它处理使用 mysql 进行注册的用户输入...我有一个问题，它使用户输入被输入两次... 因此，这只是注册表单中的一项输入。下面是我的注册表单的大约一半(最有用的部分)...

  if ($_SERVER["REQUEST_METHOD"] == "POST") {

    require("db-settings.php");

    // Security

    if (empty($_POST['name'])) {
      echo "Sorry, fullname input was empty, please retry if you like.";
      die();
    } else {
      $fullname = $_POST['name'];
    }
    if (empty($_POST['email'])) {
      echo "Sorry, email input was emty, please retry if you like.";
      die();
    } else { 
      $email = $_POST['email'];
    }
    if (empty($_POST['password'])) {
      echo "Sorry, password was empty, please retry if you like.";
      die();
    } else {
      $password = $_POST['password'];

      // If password variable is success to set, let's encrypt it now!
      $password = password_hash($password, PASSWORD_DEFAULT)."\n";
    }

  // Log users IP and store in variable
    $ip = $_SERVER["REMOTE_ADDR"];

  // Create connection
  $conn = new mysqli($servername, $username, $db_password, $dbname);
  // Check connection
  if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
  }

  $sql = "INSERT INTO `table-ex` (fullname, email, password, ip) VALUES ('$fullname', '$email', '$password', '$ip')";
  $stmt = $conn->prepare($sql);
  //$stmt->bind_param('sss', $fullname, $email, $password, $ip);
  $stmt->execute();

  if ($conn->query($sql) === TRUE) {
      echo "New user was created successfully, please wait for activation...";
  } else {
    echo "Error: " . $sql . "<br>" . $conn->error;
  }

  $conn->close();

所以，这一切都在这里。我还将在下面的 html 代码中给出整个表单部分...

  <form action="signup.php" method="post">
    <h1>Sign up</h1><br/>

    <span class="input"></span>
  <input type="text" name="name" placeholder="Full name" title="Format: Xx[space]Xx (e.g. John Doe)" autofocus autocomplete="off" required pattern="^\w+\s\w+$" />
    <span class="input"></span>
    <input type="email" name="email" placeholder="Email address" required />
    <span id="passwordMeter"></span>
    <input type="password" name="password" id="password" placeholder="Password" title="Password min 10 characters. At least one UPPERCASE and one lowercase letter" required pattern="(?=^.{10,}$)(?=.*[a-z])(?=.*[A-Z])(?!.*\s).*$"/>

    <button type="submit" value="Sign Up" title="Submit form" class="icon-arrow-right"><span>Sign up</span></button>
  </form>

所以，代码中一定有什么东西让它输入两次......另外，我如何重置ID号？因为每次我创建一个新用户，并且这种情况发生(每次)，然后我只需删除用户，它仍然算作它们仍然存在。

Answer 1

就是因为这条线。您不需要添加 if else 语句。

  if ($conn->query($sql) === TRUE) {
          echo "New user was created successfully, please wait for activation...";
      }

只需这样做-

    $sql = "INSERT INTO `table-ex` (fullname, email, password, ip) VALUES ('$fullname', '$email', '$password', '$ip')";
  $stmt = $conn->prepare($sql);
  //$stmt->bind_param('sss', $fullname, $email, $password, $ip);
//Set the variables here for $fullname, $email, $password and $ip
 if($stmt->execute())

 {
      echo "New user was created successfully, please wait for activation...";
} 
  else { echo "There was a problem";}

  $stmt->close();
  $conn->close();

更新

对于 id 部分，我假设您正在使用自动增量，但我建议您手动插入它们而不是依赖它。我建议您使用唯一的 key 派生函数并对它们进行编码(如果您希望它们是纯文本并将它们用作 ID)。

如果您想跟踪其中有多少条目，您可以随时使用 mysqli_num_rows() 计算行数。