爲什麼這個MySQL查詢產生錯誤的行號？

Question

給出下面的查詢形式的

SET @num :=0, @current_shop_id := NULL, @current_product_id := NULL; 

SELECT * FROM (

SELECT products.shop_id, products.product_id, @num := IF(@current_shop_id=shops.shop_id, IF(@current_product_id=products.product_id,@num,@num+1),0) AS row_number, @current_shop_id := shops.shop_id AS shop_dummy, @current_product_id := products.product_id AS product_dummy 

    FROM 
    favorites fav1 INNER JOIN 
    products ON 
    fav1.product_id=products.product_id AND 
    fav1.current=1 AND 
    fav1.closeted=1 AND 
    fav1.user_id=30 INNER JOIN 

    shops ON 
    shops.shop_id = products.shop_id 

    ORDER BY shops.shop ASC, products.product_id DESC 

) AS rowed_results WHERE rowed_results.row_number>=0 AND rowed_results.row_number<(20) AND shop_id=130 

我期待ROW_NUMBER值

+---------+------------+------------+------------+---------------+ 
| shop_id | product_id | row_number | shop_dummy | product_dummy | 
+---------+------------+------------+------------+---------------+ 
|  130 | 1153746 |   0 |  130 |  1153746 | 
|  130 | 1153736 |   1 |  130 |  1153736 | 
|  130 | 1139944 |   2 |  130 |  1139944 | 
|  130 | 1098296 |   3 |  130 |  1098296 | 
|  130 | 1017455 |   4 |  130 |  1017455 | 
|  130 |  551953 |   5 |  130 |  551953 | 
|  130 |  551914 |   6 |  130 |  551914 | 
+---------+------------+------------+------------+---------------+ 

（即，與給定shop_id值相關聯的所有獨特的product_id值獲得一個唯一的行號，從0開始）。相反，我越來越

+---------+------------+------------+------------+---------------+ 
| shop_id | product_id | row_number | shop_dummy | product_dummy | 
+---------+------------+------------+------------+---------------+ 
|  130 | 1153746 |   1 |  130 |  1153746 | 
|  130 | 1153736 |   0 |  130 |  1153736 | 
|  130 | 1139944 |   0 |  130 |  1139944 | 
|  130 | 1098296 |   0 |  130 |  1098296 | 
|  130 | 1017455 |   0 |  130 |  1017455 | 
|  130 |  551953 |   1 |  130 |  551953 | 
|  130 |  551914 |   0 |  130 |  551914 | 
+---------+------------+------------+------------+---------------+ 

我做錯了什麼？

編輯：

邁克爾Berkowski的解決方案（使用三個查詢，一個用於檢索數據，一個用於添加行號，一個由行限制）的作品（語法略有變化）：

SET @num :=0, @current_shop_id := NULL, @current_product_id := NULL; 

#limit by row 

SELECT * FROM (

    #add row 

    SELECT limit_query.*, @num := IF(@current_shop_id=shop_id, IF(@current_product_id=product_id,@num,@num+1),0) AS row_number, @current_shop_id := shop_id AS shop_dummy, @current_product_id := product_id AS product_dummy FROM (

     #retrieve data 

     SELECT row_query.* FROM (

      SELECT products.shop_id, products.product_id 

      FROM 
      favorites fav1 INNER JOIN 
      products ON 
      fav1.product_id=products.product_id AND 
      fav1.current=1 AND 
      fav1.closeted=1 AND 
      fav1.user_id=30 INNER JOIN 

      shops ON 
      shops.shop_id = products.shop_id 

     ) AS row_query ORDER BY shop_id ASC, product_id DESC 

    ) AS limit_query 

) AS rowed_results WHERE rowed_results.row_number>=0 AND rowed_results.row_number<(20) AND shop_id=130; 

純作爲教育興趣的問題，我想知道爲什麼不可能同時添加行號和行號限制，如下面的（非功能性）示例

SET @num :=0, @current_shop_id := NULL, @current_product_id := NULL; 

#limit by row 

SELECT rowed_results.*, @num := IF(@current_shop_id=shop_id, IF(@current_product_id=product_id,@num,@num+1),0) AS row_number, @current_shop_id := shop_id AS shop_dummy, @current_product_id := product_id AS product_dummy FROM (

    #retrieve data 

    SELECT row_query.* FROM (

     SELECT products.shop_id, products.product_id 

     FROM 
     favorites fav1 INNER JOIN 
     products ON 
     fav1.product_id=products.product_id AND 
     fav1.current=1 AND 
     fav1.closeted=1 AND 
     fav1.user_id=30 INNER JOIN 

     shops ON 
     shops.shop_id = products.shop_id 

    ) AS row_query ORDER BY shop_id ASC, product_id DESC 

) AS rowed_results WHERE row_number>=0 AND row_number<(20) AND shop_id=130; 

Answer 1

要使行增加，它應該與將@num計算移動到外部查詢中一樣簡單。在這種情況下，@current_product_id不應該需要，因爲@num可以直接增加。

SET @num :=0, @current_shop_id := NULL, @current_product_id := NULL; 

SELECT 
    *, 
    /* Perform the row increment in the outer query so it acts on the final rowset */ 
    @num := @num+1 AS row_number 
FROM (
    SELECT products.shop_id, products.product_id, @current_shop_id := shops.shop_id AS shop_dummy, @current_product_id := products.product_id AS product_dummy 

    FROM 
    favorites fav1 INNER JOIN 
    products ON 
    fav1.product_id=products.product_id AND 
    fav1.current=1 AND 
    fav1.closeted=1 AND 
    fav1.user_id=30 INNER JOIN 

    shops ON 
    shops.shop_id = products.shop_id 

    ORDER BY shops.shop ASC, products.product_id DESC 

) AS rowed_results 
WHERE 
    rowed_results.row_number>=0 
    AND rowed_results.row_number<(20) 
    AND shop_id=130 

Answer 2

您的邏輯檢查商店ID和產品ID。如果它們相同，則計數器遞增，否則重置爲零。雖然您的結果集具有所有相同的商店ID，但它具有不同的產品ID。