Hello talkphp members,

I am very very new to php and mysql, and i have a problem.

Problem: I am buiding a simple embedd youtube-link site (school project) and i cant get my "add to favorites"-php to throw any values into my database. In my opinion the php below should work, i cant find any errors and i am completely blinded by now - why wont this work?

Additional things you need to know:

• this lies in a while loop that grabs the $Id.
• $Current_Id is from sessions id.

There was a point in time where this worked. please help me.

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print "<div class='video_post_bottom'>";//BOTTOM
print "<div class='favorit'><form action='Main4.php' method='post' class='favorites'><input type='image' src='Images/Favorites.png' name='FAV' value=''/><input type='hidden' name='FAV_Id' value='$Id'></form></div>"; //"Favorites" - BUTTON

//ADD TO FAVORITES
if (isset($_POST['FAV']))
{
print"<p>isset _POSTFAV</p>";
$Video_Id = $_POST['FAV_Id'];

$query7 ="SELECT * FROM Favorites WHERE '$Video_Id' = Favorites.Video_Id AND '$Current_Id' = User_Id";
$result_C = mysql_query($query7);
print"<p>SELECT * FROM FAVORITES</p>";

while ($row7 = mysql_fetch_array($result_C)) {
$favorite = $row7['Video_Id'];
print"<p>favorite = row7Video_Id</p>";
}

If(!$favorite)//not favorite yet?
{
$query = "INSERT INTO Favorites (User_Id, Video_Id) VALUES ('$Current_Id', '$Video_Id ')";
print"<p>INSERT INTO Favorites</p>";

$result_C = mysql_query($query);
print "<p> $Title has been added to your favorites</p>";

unset($_POST['FAV']);
}
else if($favorite && $Video_Id == $Id) //if voted has value
{
print "<p> THIS IS ALREADY IN YOUR FAVOURITES </p>";
}
}

print"<div class='comments'><a href='../Comment/Comment.php?V_Id=$Id'><img src='Images/Comments.png'></a></div>";//COMMENTS
print "</div>";