Question

我是oracle 11g的新手，不知道这样做的最好方法。感谢您的任何提示。

T1
id_set    id_
1         100 
1         101
1         102
1         103
2         100
2         101
2         102
2         108

T2
id_    tag    name
100    A       a
101    B1      b
102    B2      c
103    B3      d
108    B8      e

这里的规则是检查id_set，检查它是否包含tag ='A'和tag = B *，选择B之后具有最高整数的那个。所以样本输出将是：

id_set_    name    
1          d
2          e

我有B15，达到这个目标的最佳方法是什么？谢谢！ PS。 id增加时标签号不会增加

Answer 1

SQL Fiddle

Oracle 11g R2架构设置：

CREATE TABLE T1 (id_set, id_) AS
          SELECT 1, 100 FROM DUAL 
UNION ALL SELECT 1, 101 FROM DUAL
UNION ALL SELECT 1, 102 FROM DUAL
UNION ALL SELECT 1, 103 FROM DUAL
UNION ALL SELECT 2, 100 FROM DUAL
UNION ALL SELECT 2, 101 FROM DUAL
UNION ALL SELECT 2, 102 FROM DUAL
UNION ALL SELECT 2, 108 FROM DUAL;

CREATE TABLE T2 (id_, tag, name) AS
          SELECT 100,    'A',       'a' FROM DUAL
UNION ALL SELECT 101,    'B1',      'b' FROM DUAL
UNION ALL SELECT 102,    'B2',      'c' FROM DUAL
UNION ALL SELECT 103,    'B3',      'd' FROM DUAL
UNION ALL SELECT 108,    'B8',      'e' FROM DUAL;

查询1 ：

WITH max_tags AS (
  SELECT id_set, MAX( tag ) AS max_tag
  FROM   T1
         INNER JOIN
         T2
         ON (T1.id_ = T2.id_)
  GROUP  BY id_set
  HAVING MIN( tag ) = 'A'
  AND    MAX( tag ) LIKE 'B%'
)
SELECT id_set, name
FROM   max_tags m
       INNER JOIN
       T2 t
       ON (m.max_tag = t.tag)

<强> Results ：

| ID_SET | NAME |
|--------|------|
|      1 |    d |
|      2 |    e |

在oracle / sql中获取给定集合中的最高值

1 个答案: