Supplied argument is not a valid MySQL result resource

Help I get the error message:

Warning: Supplied argument is not a valid MySQL result resource ... in line x

The line that gives this error is :

$rs = mysql_fetch_array($myquery)

The confusing thing is that on my PC (offline) all the pages work when I upload onto my hosting site I get this error.

Can anybody assist, any help would be greatly appreciated?

Thanks.

===============
<?
$db_handle = mysql_connect("xxx","xxx","xxx");
mysql_select_db("fixtures",$db_handle);

$sqlVar="SELECT * FROM teamFixtures ";
{
$sqlVar.="WHERE teamFixtures.TeamName LIKE
'%".$_GET['TeamID']."'";

}

$myquery = mysql_query($sqlVar,$db_handle);
$rs = mysql_fetch_array($myquery)
?>

use

mysql_error();

to see where the problem is, check your tables structure on the remote server, maybe there is something not like on youl local machine

Thanks malbera,

Being a newbie to mysql & php .....

Can you provide me with a little but more detail regarding :

mysql_error();

to see where the problem is, check your tables structure on the remote server, maybe there is something not like on youl local machine

Many thanks again.

php.net
this about mysql_error();

and why do you need this

$sqlVar="SELECT * FROM teamFixtures "; 
{ 
$sqlVar.="WHERE teamFixtures.TeamName LIKE '%".$_GET['TeamID']"'"; 

}

How to use mysql_error:

see below:
<?
$db_handle = mysql_connect("xxx","xxx","xxx") or die(mysql_error());
mysql_select_db("fixtures",$db_handle) or die(mysql_error());

$sqlVar="SELECT * FROM teamFixtures ";
{
$sqlVar.="WHERE teamFixtures.TeamName LIKE
'%".$_GET['TeamID']."'";

}

$myquery = mysql_query($sqlVar,$db_handle) or die(mysql_error());
$rs = mysql_fetch_array($myquery)
?>

Now if any of the commands fail mysql will return an error stating what the problem is

HTH

Many thanks to malbera & trooper I have now sussed it out with your advice.

Thanks again.