MySQL：特定类别的独特产品计数

我有三张桌子：

• 产品

• 类别

• 类别产品

  （用于“多对多”关系）

“类别产品”表示例：

+-------------+------------+
| category_id | product_id |
+-------------+------------+
|           3 |          9 |
|           3 |         28 |
|           3 |        100 |
|         ... |        ... |
|           7 |         13 |
|           7 |         21 |
|           7 |         81 |
|           7 |        100 |
|         ... |        ... |
|          25 |         22 |
|          25 |         28 |
|          25 |        100 |
+-------------+------------+

---

我需要找到这个类别中独特产品的数量。在我的例子中，我必须从

COUNT

行中排除，其中

product\u id=28

和100。如何做到这一点？

我认为这应该是可行的：

SELECT
    c.*,
    COALESCE( cnts.distinct_product_count, 0 ) AS distinct_product_count
FROM
    categories AS c
    LEFT OUTER JOIN
    (
        SELECT
            category_id,
            COUNT( DISTINCT product_id ) AS distinct_product_count
        FROM
            category_product
        WHERE
            product_id NOT IN ( 28, 100 )
        GROUP BY
            category_id

    ) AS cnts ON
        cnts.category_id = c.category_id

---

使用一个不同的计数

架构（MySQL v5.7）

---

查询#1

SELECT COUNT(DISTINCT product_id)  FROM category_product
WHERE `category_id` IN (3,7,25);

| COUNT(DISTINCT product_id) |
| -------------------------- |
| 7                          |

---

---

您可以使用

不存在

排除属于一个以上类别的产品：

SELECT cp.category_id, COUNT(*) counter
FROM category_product cp
WHERE NOT EXISTS (SELECT 1 FROM category_product WHERE product_id = cp.product_id AND category_id <> cp.category_id)
GROUP BY cp.category_id

作为在中使用运算符

的子查询：

SELECT category_id, COUNT(*) counter
FROM category_product
WHERE product_id IN (
  SELECT product_id
  FROM category_product
  GROUP BY product_id
  HAVING COUNT(*) = 1
)
GROUP BY category_id

请参阅。

结果:

为什么一个类别会有非唯一的产品？
category\u product
是否定义了复合键？因为存在这样一个条件：一个产品可以属于不同的类别。复合密钥已定义。谢谢！这就是我需要的！
SELECT product_id
FROM category_product
GROUP BY product_id
HAVING COUNT(*) = 1

SELECT category_id, COUNT(*) counter
FROM category_product
WHERE product_id IN (
  SELECT product_id
  FROM category_product
  GROUP BY product_id
  HAVING COUNT(*) = 1
)
GROUP BY category_id

category_id | counter
----------: | ------:
          3 |       1
          7 |       3
         25 |       1