如何使用两个表中的选择子句，连接它们并显示其中一个结果？

Question

我从一个表中选择数据，但我还需要使用另一个表的选择条款：

sql＆amp; PHP：

  $condition = "
AND CONCAT_WS('-    ', products_description.products_id, products_attributes.products_attributes_id) = '".$product_number."' 
OR products_description.products_id = '".$product_number."'";

    $show_products_query_string = "
SELECT 
    products_description.products_id AS id,
    products_description.products_name AS product_name, 
    products.products_image AS product_image 
FROM 
    products_description 
    LEFT JOIN products ON products_description.products_id = products.products_id  
    LEFT JOIN warehouse_products ON products_description.products_id = warehouse_products.id" . $filter.$condition;

在这个查询中，我只需要显示我选择的内容：id，name，image_name。

提前谢谢！