因为....

Question

我一直在使用FB SDK，以便我可以让用户通过他们的FB帐户登录，但是在这个阶段我希望他们的用户数据存储在我的数据库中，甚至在关注SO之后的多个解决方案之后它也没有。似乎工作。此外，数据是从变量中打印出来的，从而证明这没有任何问题，说实话，我认为我的SQL INSERT语句有问题，但我似乎无法看到它。

我的主页代码如下：

 <?php
// Path to PHP-SDK
include('config.php');
require '......./src/facebook.php ';
$facebook = new Facebook(array(
'appId'  => '',
'secret' => '',
 )); 

  /* $facebook = new Facebook(array(
 'appId' => AppInfo::appID(),
  'secret' => AppInfo::appSecret(),
  'sharedSession' => true,
 'trustForwarded' => true,
   )); */

// See if there is a user from a cookie
## connect mysql server
$mysqli = new mysqli(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE);
# check connection
if ($mysqli->connect_errno) {
    echo "<p>MySQL error no {$mysqli->connect_errno} : {$mysqli->connect_error}</p>";
    exit();}
$user = $facebook->getUser();

if ($user) {
 try {
  // Proceed knowing you have a logged in user who's authenticated.
  $user_profile = $facebook->api('/me');
  // $fbid = $user_profile['id'];                 // To Get Facebook ID
  $fbuname = $mysqli->real_escape_string($user_profile['name']);  // To Get Facebook Username
    $fbfname = $mysqli->real_escape_string($user_profile['first_name']); // To Get Facebook full name
    $fblname = $mysqli->real_escape_string($user_profile['last_name']); // To Get Facebook full name        
    $fbemail = $mysqli->real_escape_string($user_profile['email']);    // To Get Facebook email ID
         } catch (FacebookApiException $e) {
         echo '<pre>'.htmlspecialchars(print_r($e, true)).'</pre>';
     $user = null;
  }
  }

 ?>
  <!DOCTYPE html>
<html xmlns:fb="http://www.facebook.com/2008/fbml">
<head>
    <title>Facebook Application Local Development</title>
    <link type="text/css" rel="stylesheet" href="css/reset.css">
    <link type="text/css" rel="stylesheet" href="css/main.css">
</head>
<body>
<div id="login">
    <?php if (!$user) { ?>
      <fb:login-button></fb:login-button>
    <?php } ?>
</div>
<div id="cont">
<!--redirect to local welcome page?!!-->
     <?php if ($user): 


    // $result = $mysqli->query("SELECT username from users WHERE username = '$fbuname' LIMIT 1");
  // to check if its getting anything in result
 //print "<pre>";
//print_r($result);
//print "</pre>"; 


 $result = $mysqli->query("SELECT username from users WHERE username = '{$fbuname}' LIMIT 1");
 print "<pre>";
 print_r($result);
 print "</pre>"; 
if ($result->num_rows == 0){
// to check if it enters a loop
echo "true";
echo "<br/>" ; 
$sql= mysqli_query($db, "INSERT INTO users (username,first_name,last_name,email) 
VALUES ('{$user_profile['name']}', '{$user_profile['first_name']}', '{$user_profile['last_name']}', '{$user_profile['email']}')");
//$sql= mysqli_query($db,$sql);
if ( $sql === false ){
 //handled the  SQL error here
 //echo "ERROR";
 die($mysqli->error);
 }  
}
echo "<br/>" ; 


 echo $fbuname; echo $fbfname; echo $fblname; echo $fbemail; 
     //header("location: welcome.php"); 
    //<!--img src="https://graph.facebook.com/ <//?php echo $user; picture"-->
    //<!--p Hello // echo $user_profile['name']; !</p-->


     else: 
    echo "You are not Connected.";
     endif ?>

Answer 1

为什么你的代码不能正常工作

您正在混合程序和面向对象的样式。另外，$ db不包含数据库连接的凭据，因为它是空的。

解决方案

替换

$sql= mysqli_query($db, "INSERT INTO users (username,first_name,last_name,email) 
VALUES ('{$user_profile['name']}', '{$user_profile['first_name']}', '{$user_profile['last_name']}', '{$user_profile['email']}')");

带

$sql= $mysqli->query("INSERT INTO users (username,first_name,last_name,email) 
VALUES ('{$user_profile['name']}', '{$user_profile['first_name']}', '{$user_profile['last_name']}', '{$user_profile['email']}')");

Answer 2

因为....

无效

$ db未声明连接值。

将值声明如下......

$db = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE) or die("Error " . mysqli_error($db);

，然后才能执行查询。

$sql= mysqli_query($db, "INSERT INTO users (username,first_name,last_name,email) 
    VALUES ('{$user_profile['name']}', '{$user_profile['first_name']}', '{$user_profile['last_name']}', '{$user_profile['email']}')");

Facebook PHP SDK-在MySql数据库中存储用户数据无法正常工作

2 个答案:

为什么你的代码不能正常工作

解决方案

因为....