MySQL if条件然后左连接

我在页面底部有一个工作选择。正如你所看到的，它有53行，在我看来，这太多了。有人告诉我MySQL中有一个“如果”条件，但我无法使它工作。工作选择联合4选择，因为在每个选择中，我需要连接另一个表。是否可以根据表的内容连接表

SELECT *
FROM (
    SELECT m.`identificator` , ml.id AS link_id, ml.parent, ml.type, 
            ml.destination, ml.disabled, ml.order, mld.`name` AS link_name, 
            mld.`alt` , mld.`title` , cpd.`slug` 
    FROM  `wf_menu` m
    LEFT JOIN  `wf_menu_link` ml ON m.`id` = ml.`menu_id` 
    LEFT JOIN  `wf_menu_link_desc` mld ON ml.`id` = mld.`link_id` 
    LEFT JOIN  `wf_cms_post_desc` cpd ON ml.destination = cpd.post_id
    WHERE mld.`lang_id` =1
    AND mld.`lang_id` = cpd.`lang_id` 
    AND (ml.`type` =  'page'
    OR ml.`type` =  'article')

    UNION

    SELECT m.`identificator` , ml.id AS link_id, ml.parent, ml.type, 
            ml.destination, ml.disabled, ml.order, mld.`name` AS link_name, 
            mld.`alt` , mld.`title` , cpd.`slug` 
    FROM  `wf_menu` m
    LEFT JOIN  `wf_menu_link` ml ON m.`id` = ml.`menu_id` 
    LEFT JOIN  `wf_menu_link_desc` mld ON ml.`id` = mld.`link_id` 
    LEFT JOIN  `wf_cms_category_desc` cpd ON ml.destination = cpd.category_id
    WHERE mld.`lang_id` =1
    AND mld.`lang_id` = cpd.`lang_id` 
    AND ml.`type` =  'cmscat'

    UNION

    SELECT m.`identificator` , ml.id AS link_id, ml.parent, ml.type, 
            ml.destination, ml.disabled, ml.order, mld.`name` AS link_name, 
            mld.`alt` , mld.`title` , spd.`slug` 
    FROM  `wf_menu` m
    LEFT JOIN  `wf_menu_link` ml ON m.`id` = ml.`menu_id` 
    LEFT JOIN  `wf_menu_link_desc` mld ON ml.`id` = mld.`link_id` 
    LEFT JOIN  `wf_shop_category_desc` spd ON ml.destination = spd.category_id
    WHERE mld.`lang_id` =1
    AND mld.`lang_id` = spd.`lang_id` 
    AND ml.`type` =  'shopcat'

    UNION

    SELECT m.`identificator` , ml.id AS link_id, ml.parent, ml.type, 
            ml.destination, ml.disabled, ml.order, mld.`name` AS link_name, 
            mld.`alt` , mld.`title` , 'link' as slug 
    FROM  `wf_menu` m
    LEFT JOIN  `wf_menu_link` ml ON m.`id` = ml.`menu_id` 
    LEFT JOIN  `wf_menu_link_desc` mld ON ml.`id` = mld.`link_id` 
    WHERE mld.`lang_id` =1
    AND ml.`type` =  'link'

) a
ORDER BY `order` DESC

---

我认为您需要使用动态SQL。我的意思是，将您的终端SQL构建成一个临时变量并执行它：

set @query = (SELECT ...);
PREPARE st FROM @query;
EXECUTE st;

您的第一个案例可能是：

SET @query= (SELECT CONCAT(GROUP_CONCAT(CONCAT('SELECT m.`identificator` , ml.id AS link_id, ml.parent, ml.type, 
            ml.destination, ml.disabled, ml.order, mld.`name` AS link_name, 
            mld.`alt` , mld.`title` , cpd.`slug` FROM  `wf_menu` m LEFT JOIN  `wf_menu_link` ml ON m.`id` = ml.`menu_id` ',
            CASE WHEN TTYPES.type='article' THEN 
                CONCAT('LEFT JOIN  `wf_menu_link_desc` mld ON ml.`id` = mld.`link_id` LEFT JOIN  `wf_cms_post_desc` cpd ON ml.destination = cpd.post_id WHERE mld.`lang_id` =1 AND mld.`lang_id` = cpd.`lang_id` AND (ml.`type` =  \'page\' OR ml.`type` =  \'',TTYPES.type,'\')')

            END )
SEPARATOR ' UNION '), ' SELECT NULL FROM wf_menu_link WHERE 1=2') FROM (SELECT DISTINCT wf_menu_link.`type` FROM wf_menu_link) TTYPES);

（您可以完成重新标记案例）

运行该查询将使用生成的查询填充

@query

变量。生成的查询应该与您在问题中发布的查询相同。然后您可以运行它，正如您在上面的回答中所看到的

---

这是一个示例/想法，您必须完成它和/或检查它在您的模型中是否有效

您可能只需执行一个查询，其中包含每个左连接，然后使用IF语句对生成的“slug”列进行排序。但是，这完全取决于您想要输出什么，以及其他表上是否有多个匹配行

大概是这样的：-

SELECT m.identificator , ml.id AS link_id, ml.parent, ml.type, 
        ml.destination, ml.disabled, ml.order, mld.name AS link_name, 
        mld.alt , mld.title , cpd.slug 
FROM  wf_menu m
LEFT JOIN  wf_menu_link ml ON m.id = ml.menu_id 
LEFT JOIN  wf_menu_link_desc mld ON ml.id = mld.link_id 
LEFT JOIN  wf_cms_post_desc cpd ON ml.destination = cpd.post_id AND (ml.type = 'page' OR ml.type = 'article')
LEFT JOIN  wf_cms_category_desc cpd ON ml.destination = cpd.category_id AND ml.type =  'cmscat' AND mld.lang_id = cpd.lang_id 
LEFT JOIN  wf_shop_category_desc spd ON ml.destination = spd.category_id AND ml.type =  'shopcat' AND mld.lang_id = spd.lang_id 
WHERE mld.lang_id =1
AND mld.lang_id = cpd.lang_id 

---

另一方面，我建议将其创建为一个过程，因为它只需要通过网络进行一次调用，并且您可以在该过程中执行所需的所有逻辑。我是否应该理解，您正在使用不同案例的并集来用备选方案的结果补充一个空案例？在这种情况下，选择备选方案ml.type的条件是什么？如果ml.type=页面/文章，我将加入cms\u post\u desc，如果cmscat cms\u category\u desc，如果shopcat shop\u category\u desc，如果ml.type=链接，这意味着目标中有直接链接，因此不需要留下任何链接