Question

目前，我正在尝试接收使用网站上联系表格的用户的ip地址，并将其与其他信息一起发送到firebase。对于HTML和PHP，我使用了本教程中的确切代码：http://www.danielpinero.com/how-to-add-ip-address-html-form。

这是firebase代码：

    // Initialize Firebase (ADD YOUR OWN DATA)
var config = {
    apiKey: "XXXXXXXXXXXXXXXXXXXXXXXXX",
    authDomain: "XXXXXXXXXXXXXXXXXXXXXXXX",
    databaseURL: "XXXXXXXXXXXXXXXXXXXX",
    projectId: "XXXXXXXXXXXXXXXXX",
    storageBucket: "XXXXXXXXXXXXXXXXX",
    messagingSenderId: "XXXXXXXXXXXX"
  };
  firebase.initializeApp(config);

// Reference messages collection
var messagesRef = firebase.database().ref('messages');

// Listen for form submit
el = document.getElementById('contactForm')
if(el){
  el.addEventListener('submit', submitForm);
}

// Submit form
function submitForm(e){
  e.preventDefault();

  // Get values
  var name = getInputVal('name');
  var email = getInputVal('email');
  var message = getInputVal('msg');
  var ip = getInputVal('ip')

  // Save message
  saveMessage(name, email, message, ip);

  // Clear form
  document.getElementById('contactForm').reset();
}

// Function to get get form values
function getInputVal(id){
  return document.getElementById(id).value;
}

// Save message to firebase
function saveMessage(name, email, message, ip){
  var newMessageRef = messagesRef.push();
  newMessageRef.set({
    name: name,
    email:email,
    message:message,
    ip:ip
  });
}

这是我在表单上单击“提交”时从firebase收到的输出：

相反，我希望显示用户的实际IP。我是PHP的新手。如果有人可以帮助我，那就太好了。

编辑（index.php）：

<form id="contactForm" method="post" class="cta">
                <div class="col gtr-uniform gtr-50">
                    <div class="col-8 col-12-xsmall inputs"><input type="text" name="name" id="name" placeholder="Name" /></div>
                    <div class="col-8 col-12-xsmall inputs"><input type="email" name="email" id="email" placeholder="Email Address" /></div>
                    <div class="col-8 col-12-xsmall inputs"><textarea id="msg" rows="4" cols="50" placeholder="Message"></textarea></div>
                    <input id="ip" type="hidden" name="ip" value="<?=$ip;?>">
                    <div class="col-4 col-12-xsmall"><input type="submit" value="Get Started" class="fit primary" /></div>
                </div>
            </form>

Answer 1

第一个问题是您在var ip = getInputVal('ip')

附近缺少分号

第二个问题是将此<input id="ip" type="hidden" name="ip" value="<?=$ip;?>">更改为<input id="ip" type="hidden" name="ip" value="<?php echo $ip;?>"> 使用它来获得更好的客户端IP结果：

<?php
function get_client_ip() {
$ipaddress = '';
if (isset($_SERVER['HTTP_CLIENT_IP']))
    $ipaddress = $_SERVER['HTTP_CLIENT_IP'];
else if(isset($_SERVER['HTTP_X_FORWARDED_FOR']))
    $ipaddress = $_SERVER['HTTP_X_FORWARDED_FOR'];
else if(isset($_SERVER['HTTP_X_FORWARDED']))
    $ipaddress = $_SERVER['HTTP_X_FORWARDED'];
else if(isset($_SERVER['HTTP_FORWARDED_FOR']))
    $ipaddress = $_SERVER['HTTP_FORWARDED_FOR'];
else if(isset($_SERVER['HTTP_FORWARDED']))
    $ipaddress = $_SERVER['HTTP_FORWARDED'];
else if(isset($_SERVER['REMOTE_ADDR']))
    $ipaddress = $_SERVER['REMOTE_ADDR'];
else
    $ipaddress = 'UNKNOWN';
return $ipaddress;
} 
$ip = get_client_ip();
?>

现在，对于js变量，您可以通过多种方式填充它：

<script>var ip = "<?php echo $ip;?>"; </script>

或者您可以将其填充为隐藏输入的值，并通过js获取该值：

<script>var ip = document.getElementById('inputID').value; </script>

如何使用PHP从HTML表单获取IP地址？

1 个答案: