Привет * прошу ответа на тупой вопрос: не вижу, где ошибка. Код (Text): $result = mysql_query("SELECT name AS vName FROM model WHERE name LIKE '%. $modelName .%'") or die("Invalid query: " . mysql_error()); $res = mysql_query("SELECT name AS mName FROM vendor WHERE name LIKE '%. $vendorName .%'") or die("Invalid query: " . mysql_error()); $fetch = mysql_fetch_array($result, MYSQL_ASSOC); $row = mysql_fetch_array($res, MYSQL_ASSOC); $vendor = $fetch['vName']; $model = $row['mName']; $vendorModel = $vendor . " " . $model; эхом vendorModel не выводится. в упор не вижу, что не так. затык. помогите, пжт
Попробуй так: PHP: <? $result = mysql_query("SELECT name AS vName FROM model WHERE name LIKE '%$modelName%'") or die("Invalid query: " . mysql_error()); $res = mysql_query("SELECT name AS mName FROM vendor WHERE name LIKE '%$vendorName%'") or die("Invalid query: " . mysql_error()); $fetch = mysql_fetch_array($result, MYSQL_ASSOC); $row = mysql_fetch_array($res, MYSQL_ASSOC); $vendor = $fetch['vName']; $model = $row['mName']; $vendorModel = $vendor . " " . $model;