PHP- Kodu:
Parse error: syntax error, unexpected ':' in C:AppServwwwyou.php on line 10
böle bi hata verio belkide ben yanlış yapıyorum
orjinali budur.
PHP- Kodu:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<?
if ($_POST[git] == 'ok'){
header("Location: http://rd.cache.l.google.com/get_video?video_id=$_POST[video_id] ");
}
echo '<form method="POST" action="youtube.php">'."\n";
echo ' <input type="text" name="video_id" size="22">'."\n";
echo ' <input type="hidden" name="git" size="22" value="ok">'."\n";
echo ' <input type="submit" value="Gönder" name="B1">'."\n";
echo '</form>'."\n";
?>
</body>
</html>
Değiştirdiğim...
PHP- Kodu:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<?
if ($_POST[git] == 'ok') echo "<script language=\"JavaScript\">window.location.href = '$url';</script>";
http://rd.cache.l.google.com/get_video?video_id=$_POST[video_id] ");
}
echo '<form method="POST" action="you.php">'."\n";
echo ' <input type="text" name="video_id" size="22">'."\n";
echo ' <input type="hidden" name="git" size="22" value="ok">'."\n";
echo ' <input type="submit" value="Gönder" name="B1">'."\n";
echo '</form>'."\n";
?>
</body>
</html>