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[SOLVED] Resource ID #3????


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I get "Resource ID #3" for the result of a query, which has me completely baffled.

 

Any ideas?

 

The code that generates this answer-

 

<?php 
$con=mysql_connect('localhost','root','');
mysql_select_db('LolStrat',$con);
$Champion=$_GET['Champion'];
echo $Champion;
$Tag=mysql_query("SELECT Tag FROM champions WHERE Name='$Champion'");
echo "<br>";
echo $Tag;
?>

 

The first part works, it fetches the champion name from the URL and prints that. Then I want the query to find the tag of that champion, easily enough explained by the query. I thought $Tag would therefore be "The ChronoKeeper", which is what the database has for that record, only it prints "Resource id #3" in those exact words...

 

Ideas?

 

Thanks.

 

Edit- After a bit more googling, I got some examples and tried those, to no success.

 

<?php 
$con=mysql_connect('localhost','root','');
mysql_select_db('LolStrat',$con);
$Champion=$_GET['Champion'];
echo $Champion;
$queryone=mysql_query("SELECT Tag FROM champions WHERE Name='$Champion'");
$Tag=mysql_fetch_row($queryone); 
echo "<br>";
echo $Tag;
?>

 

This generates the word "Array" for $Tag, when it should be "The ChronoKeeper".

I also tried Mysql_fetch_object and got a fatal error, could not convert to string.

 

Really confused...

 

2nd Edit-

 

<?php 
$con=mysql_connect('localhost','root','');
mysql_select_db('LolStrat',$con);
$Champion=$_GET['Champion'];
echo $Champion;
$queryone=mysql_query("SELECT Tag FROM champions WHERE Name='$Champion'");
$query_row=mysql_fetch_array($queryone);
echo($query_row[Tag]); 
?>

 

This code generates

 

Zilean
Notice: Use of undefined constant Tag - assumed 'Tag' in C:\wamp\www\LolStrategy\ViewChampion.php on line 30
The ChronoKeeper
  

 

Now If i could just get rid of that notice, i'm done!

 

Any ideas at all?

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