点击Like按钮后，Facebook Like按钮在墙上显示变量

我正在使用Facebook like按钮。它的工作原理，显示我的网站的域名和图像。但我想在墙上显示变量$submission。我怎样才能做到这一点？点击Like按钮后，Facebook Like按钮在墙上显示变量

的Like按钮：

echo '<div class="like2">'; 

echo '<script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script><fb:like href="http://www.domain.com/directory/directory/fblike.php?submissionid='.$submissionid.'&submission='.$submission.'&uid='.$uid.'" send="true" layout="button_count" width="150" show_faces="false" font="arial"></fb:like>'; 

echo '</div>';

在www.domain.com/directory/directory/fblike.php：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 

<div class="hslogo"><a href="http://www.domain.com/directory/"><img src="images/image.png" alt="Name" border="0"/></a></div> 

<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 



<?php 
session_start(); 

$submissionid = $_GET['submissionid']; 
$uid = $_GET['uid']; 

$submission = $_GET['submission']; 




?> 

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title><?php echo urldecode($submission); ?></title> 

<meta property="og:title" content="FB Like Page Test"/> 

</head> 
<body> 


<?php 



mysql_connect("server", "username", "password") or die(mysql_error()); 
mysql_select_db("database") or die(mysql_error()); 


echo $submission; 

$q = "INSERT INTO fblikes VALUES (NULL, '$submissionid', '$uid', NULL)"; 

$r = mysql_query($q); 

if($r) //voting done 
    { 

// echo "Success!"; 
    } 
elseif(!$r) //voting failed 
    { 
// echo "Failed!"; 
    } 



header("Location: index.php?submission=".urlencode($submission)."&submissionid=".$submissionid"&uid=".$uid."); 


?> 

</body> 
</html>

来源

2011-09-02 John

我想你可能会得太多这

使每个 '提交' 有自己的URL（指数。 php？submitid = X在你的情况下听起来正确）
在此URL上有描述特定提交的开放图形元标记
将Like按钮（在您希望的任何页面上）指向该提交专用URL
Facebook将为元数据刮取特定于提交的URL并使用在最终的饲料故事

来源

2011-09-02 04:30:06 Igy

点击Like按钮后，Facebook Like按钮在墙上显示变量

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