Ajax＆Google reCaptcha

Question

我正在嘗試將reCaptcha添加到我在我的網站上的回執表單中。我已經跟隨了一個關於如何去做的視頻教程，但是我正努力使它適應我的表單，它使用ajax調用一個PHP文件，而實際上並沒有提交表單。我已經嘗試了幾個在以前的問題中提出的建議，但沒有一個能夠得到預期的結果，而是在註冊頁面顯示「我不喜歡機器人」。如果你能想到任何的話，一些提示/建議將會很好。

<div class="g-recaptcha" data-sitekey="6LcXMg0UAAAAABmlDlOGa6onxqqzERZ483XOJbFm"></div> 

的Javascript

function Register(){ 
     var Forename = $("#txtForename").val(); 
     var Surname = $("#txtSurname").val(); 
     var Password = $("#txtPassword").val(); 
     var PasswordR = $("#txtPasswordR").val(); 
     var response = $("#g-recaptcha").val(); 
      $.post('functions/php/fncregister.php', {Forename: Forename, Surname: Surname, Password: Password, PasswordR: PasswordR, response: response}, function(data) { 
       var returnValue = JSON.parse(data); 
       if (returnValue['data'] == 0){ 
        $('#mdlInfo').html('<p>Your account has been created under the username: <strong><span id="spnUsername">'+returnValue['username']+'</span></strong>. You <strong>must</strong> remember this as you will require it to log into your account.</p><p>Your account has also been added to a moderation que. <strong>You must wait until a member of staff activates your account!</strong></p>'); 
        $("#mdlRegister").modal("show"); 
       } 
       else if (returnValue['data'] == 1){ 
        $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">Passwords did not match!</p>'); 
       } 
       else if (returnValue['data'] == 3){ 
        $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">An error occured when adding your details to the Database!</p>'); 
       } 
       else if (returnValue['data'] == 4){ 
        $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">I don\'t like Robots!</p>'); 
       } 
      }); 
    } 

PHP

<?php 
//Retrieves variables from Javascript. 
$Forename = $_POST["Forename"]; 
$Surname = $_POST["Surname"]; 
$Password = $_POST["Password"]; 
$PasswordR = $_POST["PasswordR"]; 

//reCaptcha 
$Url = "https://www.google.com/recaptcha/api/siteverify"; 
$SecretKey = "---KEY---"; 
$Response = file_get_contents($Url."?secret=".$SecretKey."&response=".$_POST['response']."remoteip=".$_SERVER['REMOTE_ADDR']); 
$Robot = json_decode($response); 

$data = 0; 

if(isset($Robot->success) AND $Robot->success==true){ 
    //OTHER CODE 
} 

else{ 
    //This code always runs (though this is only meant to happen if reCaptcha detects a robot. 
    $data = 4; 
     echo json_encode(["data"=>"$data"]); 
?> 

Answer 1

不太清楚我是如何得到它的工作，但我做到了。

首先，我在Javascript中添加了一個新的變量「Response」，並使用文檔中列出的函數來檢索當用戶證明他們不是機器人時返回的密鑰值。我加入這個變量到我的AJAX調用了，所以它傳遞到PHP文件，像這樣：

function Register(){ 
    var Forename = $("#txtForename").val(); 
    var Surname = $("#txtSurname").val(); 
    var Password = $("#txtPassword").val(); 
    var PasswordR = $("#txtPasswordR").val(); 
    var Response = grecaptcha.getResponse(); 
     $.post('functions/php/fncregister.php', {Forename: Forename, Surname: Surname, Password: Password, PasswordR: PasswordR, Response: Response}, function(data) { 
      var returnValue = JSON.parse(data); 
      if (returnValue['data'] == 0){ 
       $('#mdlInfo').html('<p>Your account has been created under the username: <strong><span id="spnUsername">'+returnValue['username']+'</span></strong>. You <strong>must</strong> remember this as you will require it to log into your account.</p><p>Your account has also been added to a moderation que. <strong>You must wait until a member of staff activates your account!</strong></p>'); 
       $("#mdlRegister").modal("show"); 
      } 
      else if (returnValue['data'] == 1){ 
       $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">Passwords did not match!</p>'); 
      } 
      else if (returnValue['data'] == 3){ 
       $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">An error occured when adding your details to the Database!</p>'); 
      } 
      else if (returnValue['data'] == 4){ 
       $('#divError').html('<p class="text-center text-danger bg-danger" id="pUPInc">I don\'t like Robots!</p>'); 
      } 
     }); 
} 

在我的PHP文件，我刪除從URL後，用戶的IP地址，因爲它根據文件是可選（不知道這樣做的好處）。谷歌隨後會返回有關請求的信息，如果成功，則該項目的「成功」在我的代碼是正確的，因此進行到生成的帳戶：

$Url = "https://www.google.com/recaptcha/api/siteverify"; 
$SecretKey = "---KEY---"; 
$Response = file_get_contents($Url."?secret=".$SecretKey."&response=".$_POST['Response']); 
$Robot = json_decode($Response); 

$data = 0; 

if(isset($Robot->success) AND $Robot->success==true){ 

我沒有到底使用POST，但使用取而代之。我相信有一些安全方面的好處，因爲它會隱藏祕密密鑰，所以我會盡快研究它。

感謝@WEBjuju和我的隊友「Bridge Troll」的幫助。沒有他們中的任何一個，我都做不到。

Answer 2

這裏，看你的變量的外殼是很重要的：

$Response = file_get_contents($Url."?secret=".$SecretKey."&response=".$_POST['response']."remoteip=".$_SERVER['REMOTE_ADDR']); 
// because variables are case sensitive... 
$Robot = json_decode($Response); // it is $Response, not $response 

如果不解決這個問題，請用這個p的輸出更新你的問題roduces：

$Response = file_get_contents($Url."?secret=".$SecretKey."&response=".$_POST['response']."remoteip=".$_SERVER['REMOTE_ADDR']); 
die('Response file: <pre>'.$Response); 
$Robot = json_decode($Response); 

也嘗試一定要進行urlencode（）您要發送的增值經銷商，以谷歌：

$Response = file_get_contents($Url."?secret=".urlencode($SecretKey)."&response=".urlencode($_POST['response'])."remoteip=".urlencode($_SERVER['REMOTE_ADDR'])); 

，但到目前爲止，你必須

$Robot = json_decode($Response); // and NOT $response 

這裏是截圖顯示如何從ajax調用中得到輸出，在你的php處理中你死（）說ajax調用：

最佳解決方案

Review this guide on how to install Google reCAPTCHA with PHP