在MySQL中用name替换select result_id

Question

所以我确实有2个这样的表：

____________________________________________________
|categories_id|categories_name|categories_description|

和

__________________________________________
|categories_id|img_link|parent_id|

和其他一些不相关的列  我希望以这种方式展示：

id|categories_name|parent_id|img_link

这是我到目前为止所提出的......

SELECT  '' as 'id', c.categories_id, cd.categories_name as 'Name', c.parent_id as 'Parent Category', concat('http://www.url.com/images/',c.categories_image) as 'Image URL'
FROM categories c, categories_description cd
WHERE c.categories_id=cd.categories_id
GROUP BY c.categories_id;

这让我有了这样的事情：

id|categories_name|parent_id|img_link

我试图以某种方式离开加入，但没有成功。

SELECT c.categories_id, '' AS 'id', cd.categories_name AS 'Name', c.parent_id AS 'Parent Category', concat('http://www.url.com/images/',c.categories_image) AS 'Image URL'
FROM categories_description cd
LEFT JOIN categories c ON c.parent_id = cd.categories_name
WHERE c.categories_id=cd.categories_id AND language_id=1
GROUP BY c.categories_id;

这将在“父类别”列中返回零。问题是，如何将此parent_id替换为categories_name parent_id=categories_id ...

Answer 1

在第二个查询中，您将加入c.parent_id = cd.category_name而不是c.parent_id = cd.parent_id。