Php 连接具有不同列值的行

我正在做一个可以使用过滤器的搜索引擎

我的SQL看起来像这样

SELECT l.location_id, og.*,f.*
        FROM object_types ot
        LEFT JOIN object_group_types ogt ON ogt.object_type_id = ot.object_type_id
        LEFT JOIN object_groups og ON og.object_group_type_id = ogt.object_group_type_id
        LEFT JOIN object_groups2filters og2f ON og2f.object_group_id = og.object_group_id
        LEFT JOIN filters f ON f.filter_id = og2f.filter_id
        LEFT JOIN locations l ON l.location_id = og.location_id
        WHERE ot.object_type_key = 'TYPE_TENNIS';

现在根据用户过滤器输入，我想为它选择正确的位置。 但因为我用左连接所有东西，所以我得到了不同行上的所有过滤器项，见图

所以我想选择一个位置，在这个位置上，一个位置id既有filter\u key filter\u GRASS又有filter\u PARKING

如果我使用“AND f.filter\u key='filter\u PARKING'和f.filter\u key='filter\u GRASS'”，它将不起作用，因为过滤器值位于不同的行上

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任何人都有线索可以选择位置id同时具有两个筛选键的位置吗？

您需要使用

分组依据
和
具有
来筛选出位置

尝试：

您可以使用
中的
和条件聚合来尝试下面的方法

    SELECT l.location_id 
    FROM object_types ot
    LEFT JOIN object_group_types ogt ON ogt.object_type_id = ot.object_type_id
    LEFT JOIN object_groups og ON og.object_group_type_id = ogt.object_group_type_id
    LEFT JOIN object_groups2filters og2f ON og2f.object_group_id = og.object_group_id
    LEFT JOIN filters f ON f.filter_id = og2f.filter_id
    LEFT JOIN locations l ON l.location_id = og.location_id
    WHERE ot.object_type_key = 'TYPE_TENNIS'
   and f.filter_key in ('FILTER_PARKING','FILTER_GRAS')
    GROUP BY l.location_id 
    HAVING SUM(case when f.filter_key = 'FILTER_PARKING' then 1 else 0 end)>=1 AND 
           SUM(case whenf.filter_key = 'FILTER_GRAS' then 1 else 0 end)>=1

使用f.filter_键（‘filter_PARKING’、‘filter_GRASS’）谢谢，伙计，这就成功了！非常感谢。这确实奏效了。有一个真正的锻炼计划well@Tycho乐意帮忙：）
    SELECT l.location_id 
    FROM object_types ot
    LEFT JOIN object_group_types ogt ON ogt.object_type_id = ot.object_type_id
    LEFT JOIN object_groups og ON og.object_group_type_id = ogt.object_group_type_id
    LEFT JOIN object_groups2filters og2f ON og2f.object_group_id = og.object_group_id
    LEFT JOIN filters f ON f.filter_id = og2f.filter_id
    LEFT JOIN locations l ON l.location_id = og.location_id
    WHERE ot.object_type_key = 'TYPE_TENNIS'
   and f.filter_key in ('FILTER_PARKING','FILTER_GRAS')
    GROUP BY l.location_id 
    HAVING SUM(case when f.filter_key = 'FILTER_PARKING' then 1 else 0 end)>=1 AND 
           SUM(case whenf.filter_key = 'FILTER_GRAS' then 1 else 0 end)>=1