Mysql剖析项目并选择匹配的配置文件

Question

是否有有效的算法来确定项目配置文件并选择与查询匹配的配置文件？

例如

TABLE MEN 
id name 
1 man1 
2 man2 
3 man3 


TABLE PROPERTIES 
id name 
1 health_points 
2 strenght 
3 speed 


TABLE MEN_PROPERTIES 
id man_id property_id property_counter 
1 1  1   1000 
2 1  2   100 
3 1  3   50 
4 2  1   100 
5 2  2   200 
6 2  3   100 
7 3  1   100 
8 3  2   10 
9 3  3   5 

这意味着

man1 { 
    health_point:1000, 
    strenght:100 
    speed:50 
} 

man2 { 
    health_point:100, 
    strenght:200 
    speed:100 
} 

man3 { 
    health_point:100, 
    strenght:10 
    speed:5 
} 

比方说，我在man_1工作，我们直观地了解它的轮廓与man_3轮廓相匹配。我希望mysql将man_3作为与man_1配置文件匹配的配置文件返回。

达到结果的最佳方法是什么？

Answer 1

SELECT x.* 
FROM  
     (
      SELECT a.ID, 
        a.Name, 
        MAX(IF(c.Name = 'health_points', b.property_counter, NULL)) health_points, 
        MAX(IF(c.Name = 'strenght', b.property_counter, NULL)) strenght, 
        MAX(IF(c.Name = 'speed', b.property_counter, NULL)) speed 
      FROM Men a 
        INNER JOIN Men_Properties b 
         ON a.ID = b.man_ID 
        INNER JOIN Properties c 
         ON b.Property_ID = c.ID 
      WHERE a.ID <> 1 
      GROUP BY a.ID, a.Name 
     ) x 
     CROSS JOIN 
     (
      SELECT a.ID, 
        a.Name, 
        MAX(IF(c.Name = 'health_points', b.property_counter, NULL)) health_points, 
        MAX(IF(c.Name = 'strenght', b.property_counter, NULL)) strenght, 
        MAX(IF(c.Name = 'speed', b.property_counter, NULL)) speed 
      FROM Men a 
        INNER JOIN Men_Properties b 
         ON a.ID = b.man_ID 
        INNER JOIN Properties c 
         ON b.Property_ID = c.ID 
      WHERE a.ID = 1 
      GROUP BY a.ID, a.Name 
     ) y 
WHERE (x.health_points * 1.0/y.health_points) = (x.strenght * 1.0/y.strenght) AND 
     (x.strenght * 1.0/y.strenght) = (x.speed * 1.0/y.speed) 

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