Mysql剖析項目並選擇匹配的配置文件

Question

是否有有效的算法來確定項目配置文件並選擇與查詢匹配的配置文件？

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 

這意味着

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

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

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

比方說，我在man_1工作，我們直觀地瞭解它的輪廓與man_3輪廓相匹配。我希望mysql將man_3作爲與man_1配置文件匹配的配置文件返回。

達到結果的最佳方法是什麼？

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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