MySQL基于ENUM值进行选择

Question

我在MySQL数据库中有一个供应商目录表,它有一个相关的供应商映射表.此表针对单个供应商映射以下标准:

1. 服务.
2. 特产
3. 标准

仅用于背景信息,每个都具有存储在3个单独对应表中的值.可以由任何供应商提供的所有服务都列在服务表中,映射表捕获由单个供应商提供的服务.供应商可以提供多种服务,因此他们可以为他们提供的每项服务制作映射.同样适用于专业和标准.

我遇到困难的地方是搜索查询,用户可以在上述任何一个或全部三个标准中查询供应商列表.例如,他们可以搜索具有服务a,专业b和标准c的供应商.他们无法在服务,专业或标准上搜索多个值

我的映射表如下所示:

id int(11) unsigned NOT NULL 
supplier_id int(11) unsigned NOT NULL
entity_type enum('KEY_SERVICE','STANDARD','SPECIALITY') NOT NULL
entity_id int(11) NOT NULL 

其中entity_type映射用于指示要映射的实体类型,entity_id用于指示单个实体.

我的查询如下:

SELECT DISTINCT supplier_mappings.supplier_id, suppliers.company_name
FROM supplier_mappings
JOIN suppliers ON suppliers.id = supplier_mappings.supplier_id
WHERE (supplier_mappings.entity_type = 'KEY_SERVICE' AND supplier_mappings.entity_id =    '55')
AND (supplier_mappings.entity_type = 'SPECIALITY' AND supplier_mappings.entity_id = '218')
AND (supplier_mappings.entity_type = 'STANDARD' AND supplier_mappings.entity_id = '15'); 

这应该返回所有具有id为55的密钥服务的供应商,id为218的专业和id为15的标准.但是它只返回一个空的结果集,即使我知道至少有一个供应商具有这些定义.它似乎与复合的AND子句有关,但无法弄清楚是什么.

如果有人有任何想法,会很感激吗？

Answer 1

如果您查看查询的条件：

(supplier_mappings.entity_type = 'KEY_SERVICE' AND sellers_mappings.entity_id = '55') AND (supplier_mappings.entity_type = '特殊' AND sellers_mappings.entity_id = '218') AND (supplier_mappings.entity_type = '标准' AND sellers_mappings.entity_id = ' 15')

例如：supplier_mappings.entity_type = 'KEY_SERVICE'并且supplier_mappings.entity_type ='SPECIALITY'将始终返回FALSE值。

您必须为每个条件使用单独的表，如下所示

SELECT DISTINCT suppliers.ID, suppliers.company_name
           FROM suppliers,
                supplier_mappings mapA,
                supplier_mappings mapB,
                supplier_mappings mapC
          WHERE suppliers.ID = mapA.supplier_id
            AND suppliers.ID = mapB.supplier_id
            AND suppliers.ID = mapC.supplier_id
            AND mapA.entity_type = 'KEY_SERVICE'  AND mapA.entity_id = '55'
            AND mapB.entity_type = 'SPECIALITY'   AND mapB.entity_id = '218'
            AND mapC.entity_type = 'STANDARD'     AND mapC.entity_id = '15';