Question

我想创建一些能够通过查询对距离最近的距离进行排名的东西。现在唯一的问题是我不确定如何为MySQL实现它。我想在Oracle中实现Rank分区之类的东西。现在这是我的问题：

SELECT  p.idproduct,
p.common_name,
ROUND(
SQRT(
    POW(69.1 * (s.store_lat - 4.946966), 2) +
    POW(69.1 * (114.960770 - s.store_long) * COS(s.store_lat / 57.3), 2)),2) AS distance
FROM    product p
        INNER JOIN branches b
            ON b.idproduct = p.idproduct
        INNER JOIN store s
            ON b.idstore = s.idstore
        INNER JOIN
        (   SELECT DISTINCT p.common_name
            FROM    shopping_list_content s
                    INNER JOIN product p
                        ON s.iditem = p.idproduct
            WHERE   s.idlist =64
        ) s
            ON s.common_name = p.common_name

现在它的结果如下：

idproduct | common_name | distance
1         | item 1      |   0
1         | item 1      |   1
2         | item 2      |   3
2         | item 2      |   1
3         | item 3      |   2
3         | item 3      |   0

并添加了我应该获得的排名：

idproduct | common_name | distance | rank
1         | item 1      |   0      | 1
1         | item 1      |   1      | 2
2         | item 2      |   3      | 2
2         | item 2      |   1      | 1
3         | item 3      |   2      | 2
3         | item 3      |   0      | 1

最后通过嵌套选择我会得到：

idproduct | common_name | distance | rank
1         | item 1      |   0      | 1
2         | item 2      |   1      | 1
3         | item 3      |   0      | 1

我在这里见过类似@curRank的内容（Rank function in MySQL），但不确定如何根据我当前的查询实现它。

我尝试使用GROUP BY作为common_name列，但我想这完全不是正确的方法。希望有人可以提供帮助。

Answer 1

此查询在MySQL中可以正常排名：

SELECT TAB1.idproduct,TAB1.common_name,TAB1.distance,
(TAB1.RN - TAB2.MN) + 1 RANK FROM
(SELECT T1.*,@ROWNUM := @ROWNUM + 1 RN FROM
(SELECT * FROM (SELECT  p.idproduct,
p.common_name,
ROUND(
SQRT(
POW(69.1 * (s.store_lat - 4.946966), 2) +
POW(69.1 * (114.960770 - s.store_long) * COS(s.store_lat / 57.3), 2)),2) AS distance
FROM    product p
    INNER JOIN branches b
        ON b.idproduct = p.idproduct
    INNER JOIN store s
        ON b.idstore = s.idstore
    INNER JOIN
    (   SELECT DISTINCT p.common_name
        FROM    shopping_list_content s
                INNER JOIN product p
                    ON s.iditem = p.idproduct
        WHERE   s.idlist =64
    ) s
        ON s.common_name = p.common_name)TABLE1 
ORDER BY idproduct,common_name,distance)T1,
(SELECT @ROWNUM := 0) RN)TAB1
INNER JOIN
(SELECT T2.*,MIN(RN) MN FROM
(SELECT T1.*,@ROWNUM := @ROWNUM + 1 RN FROM
(SELECT * FROM (SELECT  p.idproduct,
p.common_name,
ROUND(
SQRT(
POW(69.1 * (s.store_lat - 4.946966), 2) +
POW(69.1 * (114.960770 - s.store_long) * COS(s.store_lat / 57.3), 2)),2) AS distance
FROM    product p
    INNER JOIN branches b
        ON b.idproduct = p.idproduct
    INNER JOIN store s
        ON b.idstore = s.idstore
    INNER JOIN
    (   SELECT DISTINCT p.common_name
        FROM    shopping_list_content s
                INNER JOIN product p
                    ON s.iditem = p.idproduct
        WHERE   s.idlist =64
    ) s
        ON s.common_name = p.common_name)TABLE1 
ORDER BY idproduct,common_name,distance)T1,
(SELECT @ROWNUM := 0) RN)T2
GROUP BY idproduct,common_name)TAB2
ON TAB1.idproduct = TAB2.idproduct AND
TAB1.common_name = TAB2.common_name;

SQL Fiddle

Answer 2

这是一个解决方案，可以在您的描述中实现最终结果集：

  SELECT a.idproduct, a.common_name, a.distance FROM
  (
      SELECT (@rownumber1:= @rownumber1 + 1) AS rn, dt.*
      FROM distance_table dt,(SELECT @rownumber1:= 0) nums
      ORDER BY common_name, distance 
  ) a  
  JOIN
  ( 
     SELECT MIN(rn) AS minRn, common_name FROM  
     (
         SELECT (@rownumber:= @rownumber + 1) AS rn, dt.*
         FROM distance_table dt,(SELECT @rownumber:= 0) nums
         ORDER BY common_name, distance 
     ) c 
     GROUP BY common_name
   ) b
   ON a.common_name = b.common_name
   AND a.rn = b.minRn

以下是SQL Fiddle

的代码

我假设已经计算了距离表，所以在上面的查询中，只要提到distance_table，就可以用查询替换它，它将距离结果集作为输出。

以下是每个组中排名的查询：

  SELECT a.idproduct, a.common_name, a.distance, (a.rn - b.minRn + 1) AS rank FROM
  (
      SELECT (@rownumber1:= @rownumber1 + 1) AS rn, dt.*
      FROM distance_table dt,(SELECT @rownumber1:= 0) nums
      ORDER BY common_name, distance 
  ) a  
  JOIN
  ( 
     SELECT MIN(rn) AS minRn, common_name FROM  
     (
         SELECT (@rownumber:= @rownumber + 1) AS rn, dt.*
         FROM distance_table dt,(SELECT @rownumber:= 0) nums
         ORDER BY common_name, distance 
     ) c 
     GROUP BY common_name
   ) b
   ON a.common_name = b.common_name

以下是SQL Fiddle的代码如果它能解决您的问题，请告诉我。

基于具有相同MySQL值的特定列的Rank列

2 个答案:

SQL Fiddle