Question

我有一个表product_category，其中包含字段id，name，parent_id和level。

例如， Agriculture 类别包含 id = 75 ， level = 1 和parent_id = NULL。子类别玉米，小麦， Rye 等具有 level = 2 和 {{ 1}}

在我的网站中，我想显示顶级类别，并且每个类别下方只显示5个子类别。但是进行查询以检索它们比我想象的更难。

如果我执行以下查询：

parent_id = 75

我检索所有顶级类别和子类别，但是有数千个子类别，因此当我只想要每个子类别中的前5个时，它会非常昂贵。

如果我执行以下操作：

SELECT a.name, a.parent_id FROM product_category a
WHERE (
    SELECT b.level
    FROM product_category b
    WHERE b.id = a.parent_id
    LIMIT 1
) = 1

仅检索5个子类别，而不是每个顶级类别的5个子类别。

然后我想到了以下方式：

SELECT a.name, a.parent_id FROM product_category a
WHERE (
    SELECT b.level
    FROM product_category b
    WHERE b.id = a.parent_id
    LIMIT 1
) = 1
LIMIT 5

它看起来非常脏和硬编码，但它是我现在能想到的唯一方式。还有其他解决办法吗？

谢谢！

Answer 1

以下是每个根类别最多返回两个子类别的示例：

select  parent.name as Category
,       child.name as SubCategory
from    (
        select  name
        ,       parent_id
        ,       @rn := if(@cur = parent_id, @rn+1, 1) as rn
        ,       @cur := parent_id
        from    product_category pc
        join    (select @rn := 0, @cur := '') i
        where   level = 2
        order by
                parent_id
        ,       id
        ) as child
join    product_category as parent
on      child.parent_id = parent.id
where   child.rn < 3

Live example at SQL Fiddle.

Answer 2

此解决方案按字母顺序排列子结果......

SELECT * FROM product_category;
+-----+---------------------+-----------+-------+
| id  | name                | parent_id | level |
+-----+---------------------+-----------+-------+
|  75 | Agriculture         |      NULL |     1 |
|  76 | Corn                |        75 |     2 |
|  77 | Wheat               |        75 |     2 |
|  78 | Rye                 |        75 |     2 |
|  85 | Vehicles            |      NULL |     1 |
|  86 | Cars                |        85 |     1 |
|  87 | Planes              |        85 |     1 |
|  88 | Trains              |        85 |     1 |
|  95 | Painters            |      NULL |     1 |
|  96 | Surrealists         |        95 |     2 |
|  97 | Impressionists      |        95 |     2 |
|  98 | Post-Impressionists |        95 |     2 |
|  99 | Max Ernst           |        96 |     3 |
| 100 | Claude Monet        |        97 |     3 |
| 101 | Gauguin             |        98 |     3 |
| 102 | Van Gogh            |        98 |     3 |
+-----+---------------------+-----------+-------+

SELECT a.*
  FROM
     ( SELECT x.*
            , y.name subcategory
         FROM product_category x
         JOIN product_category y
           ON y.parent_id = x.id
        WHERE x.parent_id IS NULL
     ) a
  JOIN
     ( SELECT x.*
            , y.name subcategory
         FROM product_category x
         JOIN product_category y
           ON y.parent_id = x.id
        WHERE x.parent_id IS NULL
     ) b
    ON b.id = a.id
   AND b.subcategory <= a.subcategory
 GROUP
    BY a.id,a.subcategory
HAVING COUNT(*) <= 2;
+----+-------------+-----------+-------+---------------------+
| id | name        | parent_id | level | subcategory         |
+----+-------------+-----------+-------+---------------------+
| 75 | Agriculture |      NULL |     1 | Corn                |
| 75 | Agriculture |      NULL |     1 | Rye                 |
| 85 | Vehicles    |      NULL |     1 | Cars                |
| 85 | Vehicles    |      NULL |     1 | Planes              |
| 95 | Painters    |      NULL |     1 | Impressionists      |
| 95 | Painters    |      NULL |     1 | Post-Impressionists |
+----+-------------+-----------+-------+---------------------+

每个类别选择5个子类别

2 个答案: