21

有两个 SQL 表：

Parents:
+--+---------+
|id|   text  |
+--+---------+
| 1|  Blah   |
| 2|  Blah2  |
| 3|  Blah3  |
+--+---------+

Childs
+--+------+-------+
|id|parent|feature|
+--+------+-------+
| 1|   1  |  123  |
| 2|   1  |   35  |
| 3|   2  |   15  |
+--+------+-------+

我想用单个查询从父母表中选择每一行，并为孩子表中的每一行选择关系“父”-“id”值和最大“特征”列值。在此示例中，结果应为：

+----+------+----+--------+---------+
|p.id|p.text|c.id|c.parent|c.feature|
+----+------+----+--------+---------+
|  1 | Blah |  1 |    1   |    123  |
|  2 | Blah2|  3 |    2   |    15   |
|  3 | Blah3|null|   null |   null  |
+----+------+----+--------+---------+

其中 p = 父表和 c = 子表

我尝试 LEFT OUTER JOIN 和 GROUP BY 但 MSSQL Express 告诉我使用 GROUP BY 的查询需要在每个非分组字段上使用聚合函数。而且我不想将它们全部分组，而是选择顶行（使用自定义排序）。

我完全没有想法......

4

5 回答 5

18 
select p.id, p.text, c.id, c.parent, c.feature
from Parents p
left join (select c1.id, c1.parent, c1.feature
             from Childs c1
             join (select p1.id, max(c2.feature) maxFeature
                     from Parents p1
                left join Childs c2 on p1.id = c2.parent
            group by p1.id) cf on c1.parent = cf.id 
                              and c1.feature = cf.maxFeature) c
on p.id = c.parent
于 2009-10-13T22:00:21.987 回答
11

使用 CTE (SQL Server 2005+)：

WITH max_feature AS (
   SELECT c.id,
          c.parent,
          MAX(c.feature) 'feature'
     FROM CHILD c
 GROUP BY c.id, c.parent)
   SELECT p.id,
          p.text,
          mf.id,
          mf.parent,
          mf.feature
     FROM PARENT p
LEFT JOIN max_feature mf ON mf.parent = p.id

非 CTE 等价物：

   SELECT p.id,
          p.text,
          mf.id,
          mf.parent,
          mf.feature
     FROM PARENT p
LEFT JOIN (SELECT c.id,
                  c.parent,
                  MAX(c.feature) 'feature'
             FROM CHILD c
         GROUP BY c.id, c.parent) mf ON mf.parent = p.id

您的问题缺少处理决胜局的详细信息（当 2+CHILD.id值具有相同的特征值时）。Agent_9191 的答案使用TOP 1，但这将采用返回的第一个而不一定是您想要的。

于 2009-10-14T03:03:12.107 回答
5

这应该有效：

SELECT p.id, p.text, c.id, c.parent,c.feature
FROM parent p
 LEFT OUTER JOIN (SELECT TOP 1 child.id,
                               child.parent,
                               MAX(child.feature)
                  FROM child
                  WHERE child.parent = p.id
                  GROUP BY child.id, child.parent
                  ) c ON p.id = c.parent
于 2009-10-13T22:03:55.670 回答
4

manji 的查询不处理最大功能的决胜局。这是我测试过的方法：

;WITH WithClause AS (SELECT p.id, p.text, 
        (SELECT TOP 1 c.id from childs c 
            where c.parent = p.id order by c.feature desc) 
        AS BestChildID
    FROM Parents p) 
SELECT WithClause.id, WithClause.text, c.id, c.parent, c.feature
FROM WithClause 
LEFT JOIN childs c on WithClause.BestChildID = c.id
于 2014-06-16T21:05:33.123 回答
3

如果您需要通过封闭嵌套选择连接不同于 MAX 列和组中描述的任何列，则可以使用 APPLY 函数。这是一个最简单的解决方案。您也可以使用 WITH 运算符。但这看起来更难。

SELECT p.id, p.text, CHILD_ROW.ANY_COLLUMN
FROM parent p
OUTER APPLY (SELECT TOP 1 child.ANY_COLLUMN
                  FROM child
                  WHERE child.parent = p.id
                  ORDER BY child.feature DESC 
                  ) CHILD_ROW
于 2017-09-11T15:02:03.810 回答