Mysql：在组中选择唯一值？

共有3栏-类别、材质和品牌。老实说，它不是来自单个表的列，而是来自使用联接的8个其他表的查询结果

    |   category  |   material   | brand          |
    ----------------------------------------------|
    |  engines    |  product 1   | abb            | -> unique for category "engines"
    |  engines    |  product 2   | wika           | -> unique for category "engines"
    |  engines    |  product 3   | allen-bradley  | -> unique for category "engines"
    |  engines    |  product 5   | wika           |
    |  engines    |  product 6   | e+h            | -> unique for category "engines"
    |  drives     |  product 7   | abb            | -> unique for category "drives"
    |  drives     |  product 8   | wika           | -> unique for category "drives"
    |  drives     |  product 9   | allen-bradley  | -> unique for category "drives"
    |  drives     |  product 10  | e+h            | -> unique for category "drives"
    |  drives     |  product 11  | e+h            | 

因此，我需要这样的smt：

    |   category  |   material   | brand          | concat(category, brand) |
    ----------------------------------------------|-------------------------|
    |  engines    |  product *   | abb            | engines/abb             |
    |  engines    |  product *   | wika           | engines/wika            |
    |  engines    |  product *   | allen-bradley  | engines/allen-brandley  |
    |  engines    |  product *   | e+h            | engines/e+h             |
    |  drives     |  product *   | abb            | drives/abb              |
    |  drives     |  product *   | wika           | drives/wika             |  
    |  drives     |  product *   | allen-bradley  | drives/allen-bradley    |
    |  drives     |  product *   | e+h            | drives/e+h              |

如果我使用GROUPBY语句GROUPBY CONCAT，每300个结果的查询时间超过10秒，这并不能让我高兴

任何人都知道如何获得唯一的值​​团队内部

UPD：

尝试：

你可以这样试试

选择子字符串new1,0，charindex'/'，new1作为类别“product*” 作为材料，子字符串new1，charindex'/'，new1+1，lennew1作为品牌， 从表中选择不同类别+'/'+品牌作为新1 临时工

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发布你的问题，首先，也许它只需要一个小的更正或类似的，如果没有，那么我们将有一个更好的了解你的问题有愚蠢的选择。。。左连接。。。左连接。。。左连接。。。这是第一个表格的结果。从第一个表中获取第二个表的主要用途；为什么要将驱动器->e+h排除在最终结果之外？你真的想在“材料”列中“产品”旁边的*吗？我应该得到唯一的类别URL，每个产品引用一个品牌和无限数量的类别。产品的标题一点也不重要。您应该尽量避免在联接中使用复杂的条件，尤其是在列上使用函数：在CONCAT'taxonomy/term/'上，th0.tid=url\u alias.sourceHm，我首先尝试了您的解决方案。它执行约10秒。

SELECT *, CONCAT(url_alias.alias, '/', LOWER(brand_alias.field_brand_path_value)) as real_url 
FROM `taxonomy_term_hierarchy` as th0
LEFT JOIN `taxonomy_term_hierarchy` as th1 ON th0.tid = th1.parent
LEFT JOIN `taxonomy_term_hierarchy` as th2 ON th1.tid = th2.parent
LEFT JOIN `taxonomy_term_hierarchy` as th3 ON th2.tid = th3.parent
LEFT JOIN `taxonomy_term_hierarchy` as th4 ON th3.tid = th4.parent

LEFT JOIN field_data_field_cat_reference as cat_reference ON    
cat_reference.field_cat_reference_tid IN (th0.tid, th1.tid, th2.tid, th3.tid, th4.tid)
LEFT JOIN node n ON cat_reference.entity_id = n.nid
LEFT JOIN field_data_field_brand_reference as brand_reference ON n.nid = brand_reference.entity_id
LEFT JOIN taxonomy_term_data as td_brand ON brand_reference.field_brand_reference_tid = td_brand.tid
LEFT JOIN field_data_field_brand_path as brand_alias ON td_brand.tid = brand_alias.entity_id
LEFT JOIN url_alias ON CONCAT('taxonomy/term/', th0.tid) = url_alias.source
WHERE 1
GROUP BY CONCAT(url_alias.alias, '/', LOWER(brand_alias.field_brand_path_value))

SELECT   a.category, 
         CONCAT(SUBSTRING_INDEX(a.material, ' ', 1), ' *') AS material,
         brand,
         CONCAT(a.category, '/', a.brand) AS concattedval
FROM     (
             [Your Sub-Select Query]
         ) a
GROUP BY a.category, 
         material, 
         a.brand