我有一個網站,並在其上可以發佈YouTube視頻鏈接,當你這樣做時,它從它(11個字符)獲取ID並將其放入數據庫。然後,您可以在一個頁面上查看視頻,並獲得YouTube標題和作者http://gdata.youtube.com/feeds/api/videos/ID,並將其放到帶有嵌入代碼的頁面上。我想知道如何不讓他們發佈私人視頻鏈接,就像我可以做某些檢查一樣。這是我迄今爲止張貼鏈接:不允許別人發佈私人YouTube視頻鏈接
if(isset($_POST['video'])){
$error = array();
if(filter_var($_POST['videourl'], FILTER_VALIDATE_URL) !== false){
if(strpos($_POST['videourl'],'youtube.com')){
preg_match('/[\\?\\&]v=([^\\?\\&]+)/',$_POST['videourl'],$video_embed);
$video_embed = $video_embed[1];
}elseif(strpos($_POST['videourl'],'youtu.be')){
$video_embed = substr(parse_url($_POST['videourl'], PHP_URL_PATH), 1);
}else{
$error[] = 'Invalid link';
}
}else{
$error[] = 'Invalid link';
}
$video_exist = mysql_num_rows(mysql_query("SELECT interest_vid FROM interest_videos WHERE interest_vid = '$video_embed'"));
$interest_exist = mysql_query("SELECT name FROM interests WHERE name = '".$_POST['interest_for_video']."'");
if(!empty($_POST['interest_for_video']) && mysql_num_rows($interest_exist) != 0){
$interest = strtolower(mysql_real_escape_string(strip_tags($_POST['interest_for_video'])));
$interest_id = mysql_result(mysql_query("SELECT id FROM interests WHERE name = '$interest'"), 0);
}else{
$error[] = 'Must specify an interest. ';
}
if(empty($error)){
if($video_exist == 0){
$result2 = mysql_query(" INSERT INTO interest_videos (user_id,interest_id,interest_vid) VALUES ('".$_SESSION['id']."','$interest_id','$video_embed')") or die(mysql_error());
if(!$result2){
die('Could not delete from database: '.mysql_error());
}else{
//$error_message = '<a href="#" onclick="toggle2(\'deletewebsite\', this); return false;"><div id="deletewebsite" class="success">Video Created</div></a>';
header("Location: /interest/video.php?interest=".$interest_id."&video=".$video_embed."");
}
}else{
$error_message = '<a href="#" onclick="toggle2(\'deletewebsite\', this); return false;"><div id="deletewebsite" class="error">That video already exists</div></a>';
}
}else{
$error_message = '<a href="#" onclick="toggle2(\'deletewebsite\', this); return false;"><div id="deletewebsite" class="error">';
foreach($error as $key => $values){
$error_message.= "$values";
}
$error_message.="</div></a>";
}
}
有沒有辦法,我可以把在那裏,使他們不能發佈私人視頻支票?