ส่งค่า $_GET มาจาก คนละที่ แล้วแสดงผล ในหน้า select_asset.php ส่งค่ามาได้ แต่ไม่ดึงข้อมูลจาก sql มาแสดง

$sql="SELECT * FROM tb_company WHERE year='2556' ";
$query=mysql_query($sql) or die (mysql_error());
//echo $sql;
while($result=mysql_fetch_assoc($query)){
<a href="select_asset.php?select=<?=$result["code_company"];?>&&year=<?=$result["year"];?>&&company=<?=$result["company_name"];?>"><?=strtoupper($result["code_company"]);?></a>
}

<?
if($objquery){
      echo"Updated.";
      $sql_back="SELECT * FROM tb_asset ";
      $query_back=mysql_query($sql_back) or die (mysql_error());
	  while($result_back=mysql_fetch_assoc($query_back)){
			header( "location:select_asset.php?select=$result_back[code_company]&year=$result_back[year]&company=$result_back[company_name]" );   
	  exit(0);  
	  }
	}else{
		echo"Not Update ";
	}
 }
 ?>

<?
//$sql="SELECT * FROM tb_asset WHERE owner='$_GET[select]' AND year='$_GET[year]'  ";
//$sql="SELECT * FROM tb_asset WHERE owner='$_GET[select]' AND year='$_GET[year]' AND company='$_GET[company_name]' ";
$sql= "SELECT tb_company.*,tb_asset.* FROM tb_company,tb_asset  WHERE owner='$_GET[select]' AND year='$_GET[year]' AND company='$_GET[company]' ";
$query = mysql_query($sql) or die ("Error Query = ".$sql." "); 
echo $sql;
$no =1;
while($result=mysql_fetch_assoc($query)){
?>

Error Query = SELECT tb_company.*,tb_asset.* FROM tb_company,tb_asset WHERE owner='tva' AND year='2556' AND company='[TVA] Toscana Valley Archiscape Co,.Ltd' 

Error

SQL query: Documentation

SELECT tb_company. * , tb_asset. *
FROM tb_company, tb_asset
WHERE owner = 'tva'
AND year = '2556'
AND company = '[TVA] Toscana Valley Archiscape Co,.Ltd'
LIMIT 0 , 30

MySQL said: Documentation
#1052 - Column 'year' in where clause is ambiguous