与组Oracle查询通过并相对于密钥

排序顺序我有在具有键值对存储如下与组Oracle查询通过并相对于密钥

KEY  VALUE  SortOrder REGN NO 
--------------------------------------- 
KEY1 VALUE1  1   123 
KEY2 VALUE2  2   123 
KEY1 VALUE3  3   123 
KEY1 VALUE4  1   456 
KEY1 VALUE5  3   456 
KEY1 VALUE6  2   456 
KEY2 VALUE7  1   678 
KEY2 VALUE8  3   678 
Key 2 VALUE9  2   678

对于每个键我已排序顺序针对REGN NO Oracle数据库值。每个REGN NO允许最多三个条目，但它可以是任何键。我需要通过将特定键的REGN NO分组来获得结果，并且应该按排序顺序排序。对于如对于KEY1结果看起来像

REG NO OPTION1  OPTION 2 OPTION 3 
------------------------------------------ 
123  VALUE 1  VALUE 3 
456  VALUE 4  VALUE 6  VALUE 5

我怎样才能获取该使用Oracle的SQL语句？

来源

2013-10-31 Jijil PK

试试这个：

CREATE TABLE key_value (
    KEY VARCHAR2(20), 
    VALUE VARCHAR2(20), 
    SortOrder NUMBER, 
    REGN_NO NUMBER 
); 

INSERT INTO key_value VALUES ('KEY1', 'VALUE1', 1, 123); 
INSERT INTO key_value VALUES ('KEY2', 'VALUE2', 2, 123); 
INSERT INTO key_value VALUES ('KEY1', 'VALUE3', 3, 123); 
INSERT INTO key_value VALUES ('KEY1', 'VALUE4', 1, 456); 
INSERT INTO key_value VALUES ('KEY1', 'VALUE5', 3, 456); 
INSERT INTO key_value VALUES ('KEY1', 'VALUE6', 2, 456); 
INSERT INTO key_value VALUES ('KEY2', 'VALUE7', 1, 678); 
INSERT INTO key_value VALUES ('KEY2', 'VALUE8', 3, 678); 
INSERT INTO key_value VALUES ('KEY2', 'VALUE9', 2, 678); 

SELECT key, regn_no, option1, option2, option3 
    FROM (
    SELECT 
     key, 
     regn_no, 
     sortorder, 
     value AS option1, 
     LEAD(value, 1) OVER (PARTITION BY key, regn_no ORDER BY sortorder) AS option2, 
     LEAD(value, 2) OVER (PARTITION BY key, regn_no ORDER BY sortorder) AS option3 
     FROM key_value 
) kv 
WHERE 
    sortorder = (SELECT MIN(sortorder) FROM key_value WHERE key = kv.key AND regn_no = kv.regn_no) 
;

输出：

KEY  REGN_NO OPTION1  OPTION2  OPTION3  
------ ---------- ----------- ------------ ------------- 
KEY1   123 VALUE1  VALUE3      
KEY1   456 VALUE4  VALUE6  VALUE5  
KEY2   123 VALUE2         
KEY2   678 VALUE7  VALUE9  VALUE8

来源

2013-10-31 00:47:33

我用你的DML和DDL IN的SQL Server基本得到相同的输出我的第二个查询。感谢dml – realnumber3012

感谢它为我工作 –

RY这个解决方案）

SELECT REGNO, KEY, MAX(CASE WHEN SortOrder = 1 THEN VALUE ELSE '' END) AS OPTION1, 
    MAX(CASE WHEN SortOrder = 2 THEN VALUE ELSE '' END) AS OPTION2, 
    MAX(CASE WHEN SortOrder = 3 THEN VALUE ELSE '' END) AS OPTION3 
    FROM SomeTable 
    GROUP BY REGNO, KEY 




    SELECT REGNO, KEY, MAX(CASE WHEN num= 1 THEN VALUE ELSE '' END) AS OPTION1, 
      MAX(CASE WHEN num= 2 THEN VALUE ELSE '' END) AS OPTION2, 
      MAX(CASE WHEN num= 3 THEN VALUE ELSE '' END) AS OPTION3 
    FROM 
    (
    SELECT ROW_NUMBER() OVER(PARTITION BY REGNO, KEY ORDER BY SortOrder) AS num, * FROM SomeTable 
    ) T GROUP BY REGNO, KEY

来源

2013-10-31 00:36:24 realnumber3012

与组Oracle查询通过并相对于密钥

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