SQL从同一个表中选择两个最大行并与第三个表连接

Question

混合中的两个表格:

项目

| item_id | user_id | data |
----------------------------
| 10       | 100    | A    |
| 11       | 100    | C    |
| 12       | 101    | E    |
| 13       | 101    | G    |

item_detail

| id | item_id | ignore | detail1 | detail2 | detail3 | detail4 |
-----------------------------------------------------------------
| 1  | 10      |   0    |   h1    |   h2    |  h3     |  h4     |
| 2  | 10      |   0    |   g1    |   g2    |  g3     |  g4     |
| 3  | 10      |   1    |   f1    |   f2    |  f3     |  f4     |
| 4  | 11      |   0    |   e1    |   e2    |  e3     |  e4     |
| 5  | 11      |   0    |   d1    |   d2    |  d3     |  d4     |
| 6  | 11      |   1    |   c1    |   c2    |  c3     |  c4     |
| 7  | 12      |   0    |   b1    |   b2    |  b3     |  b4     |
| 8  | 13      |   0    |   a1    |   a2    |  a3     |  a4     |

我需要通过item_id找到item_detail的MAX id中的detail1,detail2和item_id的item_detail的MAXNON IGNORED ID的detail3,detail4,并显示项目行.
预期结果:

| item_id | user_id | data | detail1 | detail2 | detail3 | detail4 |
--------------------------------------------------------------------
| 10      | 100     | A    |   f1    | f2      | g3      | g4      |       
| 11      | 100     | C    |   c1    | c2      | d3      | d4      |
| 12      | 101     | E    |   b1    | b2      | b3      | b4      |
| 13      | 101     | G    |   a1    | a2      | a3      | a4      |

这是SQLFiddle与这些数据集:http://sqlfiddle.com/#!2/52839
任何接受者？您的意见非常感谢.
谢谢!

Answer 1

怎么样...

SELECT i.* 
     , x.detail last_detail
     , y.detail last_non_ignored_detail
  FROM items i
  LEFT
  JOIN 
     ( SELECT a.* 
         FROM item_detail a 
         JOIN 
            ( SELECT item_id
                  , MAX(id) max_id 
               FROM item_detail 
              GROUP 
                 BY item_id
            ) b 
           ON b.item_id = a.item_id 
          AND b.max_id = a.id
     ) x
    ON x.item_id = i.item_id
  LEFT
  JOIN  
     ( SELECT a.* 
         FROM item_detail a 
         JOIN 
            ( SELECT item_id
                  , MAX(id) max_id 
               FROM item_detail
              WHERE ignored = 0
              GROUP 
                 BY item_id
            ) b 
           ON b.item_id = a.item_id 
          AND b.max_id = a.id
     ) y
    ON y.item_id = i.item_id;

... 或类似的东西.

哦,傻我......

SELECT i.*
     , MAX(d.detail) x
     , MAX(CASE WHEN d.ignored = 0 THEN d.detail END) y
  FROM items i 
  JOIN item_detail d 
    ON d.item_id = i.item_id 
 GROUP  
    BY item_id;

请注意,IGNORE是一个保留字,所以我改变了它.(我冒昧地改变了戈登的答案!)