基于MySQL具有相同行值的特定列对列进行排名

Question

我想创建一些东西来对通过查询最接近用户的距离进行排名。 现在唯一的问题是我不确定如何为MySQL实现它。 我正在考虑像按Oracle中的实现对Rank分区那样的东西。 现在这是我的查询：

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

并添加了我应该得到的排名：

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，我将得到：

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

我在这里看到了@curRank之类的东西（ MySQL中的Rank函数 ），但是不确定如何根据当前查询实现它。

我尝试将GROUP BY用于common_name列，但我想这绝对不是正确的方法。 希望有人能帮忙。

Answer 1

该查询在MySQL可以很好地进行排名：

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小提琴

Answer 2

这是在您的描述中获得最终结果集的解决方案：

  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

这是SQL Fiddle中的代码

我假设已经计算出距离表，因此在上面的查询中，只要提到distance_table ，都可以用将距离结果集作为输出的查询来代替。

这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

这是SQL Fiddle的代码让我知道它是否解决了您的问题。