點擊Like按鈕後，Facebook Like按鈕在牆上顯示變量

我正在使用Facebook like按鈕。它的工作原理，顯示我的網站的域名和圖像。但我想在牆上顯示變量$submission。我怎樣才能做到這一點？點擊Like按鈕後，Facebook Like按鈕在牆上顯示變量

的Like按鈕：

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>

來源

2011-09-02 John

我想你可能會得太多這

使每個 '提交' 有自己的URL（指數。 php？submitid = X在你的情況下聽起來正確）
在此URL上有描述特定提交的開放圖形元標記
將Like按鈕（在您希望的任何頁面上）指向該提交專用URL
Facebook將爲元數據刮取特定於提交的URL並使用在最終的飼料故事

來源

2011-09-02 04:30:06 Igy

點擊Like按鈕後，Facebook Like按鈕在牆上顯示變量

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