Question

我有这个PHP文件处理用户使用mysql注册的用户输入...我有一个问题，它使用户输入被输入两次... 所以，这只是注册表单中的一个输入。以下是我注册表格的一半（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，代码中必须有一些东西让它进入两次......另外，如何重置id号？因为每次我创建一个新用户，这都发生了（这是每次）然后我只是删除用户，它仍然计算好像它们仍然存在。

Answer 1

您同时使用了execute()和query()，因此执行了两次。

首先，它在$stmt->execute();处插入了一行。然后它在$conn->query($sql)处插入了另一行。

$stmt->execute();

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

您应该只$stmt->execute();：

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

注意：

更好的做法是坚持使用预先准备好的语句并使用execute()来提高安全性，而不是使用$conn->query($sql)。有关PDO's query vs execute的差异的更多信息。

Answer 2

这是因为这条线。你不需要发表if else声明。

  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();

<强>更新

对于id部分，我假设您正在使用auto increment，但我建议您手动插入它们而不是依赖它。我建议你使用一个唯一的密钥派生函数并对它们进行编码（如果你希望它们是纯文本并将它们用作ID）。

如果您想跟踪其中的条目数，您可以随时使用mysqli_num_rows()计算行数。

Answer 3

 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);


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

MySQL Insert在一次提交时出现两次

3 个答案: