我有一個RSS源,由於某種原因沒有顯示我的PHP變量。當返回料它回來與原標籤例如標題將時間表$ yesterdayd,而不是時間表 - 週二4月10日RSS源不顯示PHP變量
<?PHP
include("../config.php");
#// Timetable Clearup Variabls
$yesterday = strtotime('yesterday');
$yesterdow = date('l',$yesterday);
$order = "SELECT * FROM timetable WHERE day = '$yesterdow' ORDER BY time";
$result = mysql_query($order);
$yesterdayd = date('F jS, Y', time()-86400);
//SET XML HEADER
header('Content-type: text/xml');
//CONSTRUCT RSS FEED HEADERS
$output = '<rss version="2.0">';
$output .= '<channel>';
$output .= '<title>Timetable - $yesterdayd </title>';
$output .= '<description>Timetable.</description>';
$output .= '<link>http://site.com/</link>';
### $output .= '<copyright>Your copyright details</copyright>';
//BODY OF RSS FEED
$output .= '<item>';
$output .= '<title>Timetable for $yesterdayd</title>';
$output .= '<description>" . htmlspecialchars($row['username']) . "</td><td>" . htmlspecialchars($row['time']) . "</description>';
$output .= '<link>Link to Item</link>';
$output .= '<pubDate>Date Published</pubDate>';
$output .= '</item> ';
//CLOSE RSS FEED
$output .= '</channel>';
$output .= '</rss>';
//SEND COMPLETE RSS FEED TO BROWSER
echo($output);
?>
任何想法?
'$輸出= '說明> 「'...'。」';。' - 在這裏你有一個引號不匹配(''''''')... –
glglgl
2012-04-11 19:51:44