Magento 사전 검색 문제

Question

Magento의 고급 검색 app \ code \ core \ Mage \ CatalogSearch \ Model \ Mysql4을 수정하여 일부 맞춤 조건을 사용했습니다. 특정 속성 대신 검색을 사용하고 싶었 기 때문입니다. IN

하지만 아무것도 반환하지 않습니다.

`SELECT DISTINCT `e`.*, `price_index`.`price`, `price_index`.`tax_class_id`, `price_index`.`final_price`, IF(`price_index`.`tier_price`, LEAST(`price_index`.`min_price`, `price_index`.`tier_price`), `price_index`.`min_price`) AS `minimal_price`, `price_index`.`min_price`, `price_index`.`max_price`, `price_index`.`tier_price`, `cat_index`.`position` AS `cat_index_position` FROM `catalog_product_entity` AS `e` 
INNER JOIN `catalog_product_index_price` AS `price_index` ON price_index.entity_id = e.entity_id AND price_index.website_id = '1' AND price_index.customer_group_id = 0 
INNER JOIN `catalog_category_product_index` AS `cat_index` ON cat_index.product_id=e.entity_id AND cat_index.store_id='3' AND cat_index.visibility IN(3, 4) AND cat_index.category_id='2' 
INNER JOIN `catalog_product_index_eav` AS `ast_types` ON e.entity_id=ast_types.entity_id AND ast_types.attribute_id=150 AND ast_types.store_id=3 
INNER JOIN `catalog_product_index_eav` AS `ast_condom_size` ON e.entity_id=ast_condom_size.entity_id AND ast_condom_size.attribute_id=151 AND ast_condom_size.store_id=3 WHERE (( ast_types.value= 131 AND ast_types.value= 135)) AND (ast_condom_size.`value` IN('141', '140'))` 

새로운 수정 된 쿼리 : 무슨 일에

SELECT DISTINCT `e`.*, `price_index`.`price`, `price_index`.`tax_class_id`, `price_index`.`final_price`, IF(`price_index`.`tier_price`, LEAST(`price_index`.`min_price`, `price_index`.`tier_price`), `price_index`.`min_price`) AS `minimal_price`, `price_index`.`min_price`, `price_index`.`max_price`, `price_index`.`tier_price`, `cat_index`.`position` AS `cat_index_position` FROM `catalog_product_entity` AS `e` 
INNER JOIN `catalog_product_index_price` AS `price_index` ON price_index.entity_id = e.entity_id AND price_index.website_id = '1' AND price_index.customer_group_id = 0 
INNER JOIN `catalog_category_product_index` AS `cat_index` ON cat_index.product_id=e.entity_id AND cat_index.store_id='3' AND cat_index.visibility IN(3, 4) AND cat_index.category_id='2' 
INNER JOIN `catalog_product_index_eav` AS `ast_types` ON e.entity_id=ast_types.entity_id AND ast_types.attribute_id=150 AND ast_types.store_id=3 
INNER JOIN `catalog_product_index_eav` AS `ast_condom_size` ON e.entity_id=ast_condom_size.entity_id AND ast_condom_size.attribute_id=151 AND ast_condom_size.store_id=3 WHERE (( ast_types.`value`= 135 OR ast_types.`value`= 132 OR ast_types.`value`= 131 )) AND (ast_condom_size.`value` IN('141', '140')) GROUP BY sku HAVING COUNT(*)=3 

어떤 생각

여기

if(in_array($value,$cSizeArray) && isset($_GET['condom_size'])){ 
    $select->where("{$tableAlias}.`value` IN(?)", $value); 
}elseif(in_array($value,$lSizeArray) && isset($_GET['lubepack'])){ 
    $select->where("{$tableAlias}.`value` IN(?)", $value); 
}else{ 
    if($tableAlias == 'ast_types' || $tableAlias== 'ast_lubetype'){ 
     $condition = "("; 
     foreach($value as $v){ 
      $condition.=" {$tableAlias}.`value`= $v AND"; 
     } 
     $condition = substr($condition,0,-3);   
     $condition .= ") ";  
     $select->where($condition); 
    }else{ 
     $select->where("{$tableAlias}.`value` IN(?)", $value); 
    } 
} 

내가 echo $select를 사용하여 가지고 쿼리는 다음과 같습니다 내가 수정 한 쿼리입니다 접근 방식이 잘못되었습니다 ...

Answer 1

WHERE (( ast_types.value= 131 AND ast_types.value= 135)) 

동시에 131과 135 값이 필요합니다. 이것은 결코 사실이 아니므로 결과 세트는 비어 있습니다.