foreach

$car_query="	SELECT

				`MOD_ID`,

				`TEX_TEXT` AS `MOD_CDS_TEXT`,

				`MOD_PCON_START`,

				`MOD_PCON_END`

			FROM

	           `MODELS`

			INNER JOIN `COUNTRY_DESIGNATIONS` ON `CDS_ID` = `MOD_CDS_ID`

			INNER JOIN `DES_TEXTS` ON `TEX_ID` = `CDS_TEX_ID`

			WHERE

				`MOD_MFA_ID` = '$car_id' AND

				`CDS_LNG_ID` = 16

			ORDER BY

				`MOD_CDS_TEXT`";

	

$cars_select = mysql_query($car_query);



$car_mod_output = array();



foreach ($cars_select->rows as $car => $carslist) 

		{

			$model_id = $carslist['MOD_ID'];

			$car_text=$carslist['MOD_CDS_TEXT'];

			$car_year_s=substr($carslist['MOD_PCON_START'],0,4);

			if (strlen($carslist['MOD_PCON_END'])<1)

				{

					//$car_year_e=date("Y");

					$car_year_e=" ";

				}

			else

				{

					$car_year_e=substr($carslist['MOD_PCON_END'],0,4);

				}

			$year = $car_year_s." - ".$car_year_e;



			$car_mod_output[]=array('model' => $model_id, 'text' => $car_text, 'year' => $year);

		}

ошибка PHP Warning: Invalid argument supplied for foreach()

что тут не так?