Pl / SQL-oracle 9i-手动透视

|| 我们有一个包含三列的表:
Customer_name, Age_range, Number_of_people.
1              1-5         10
1              5-10        15
我们需要将不同年龄范围的所有人数作为单个查询的行返回。如果我们搜索客户#1,则查询应仅返回一行:
Header- Age Range (1-5)             Age Range (5-10)
        10                             15
我们需要将所有结果都放在一行中。当我查询客户1时,结果应该是单个行组中的人数乘以
age_range
。 解决这个问题的最佳方法是什么?     
已邀请:
        您需要手动执行数据透视:
  SELECT   SUM(CASE WHEN age_range = \'5-10\' 
                    THEN number_of_people 
                    ELSE NULL END) AS nop5,
           SUM(CASE WHEN age_range = \'10-15\' 
                    THEN number_of_people 
                    ELSE NULL END) AS nop10
    FROM   customers
   WHERE   customer_name = 1;
    
        在9i之后增加了使用LISTGAGG,COLLECT或其他功能的10g和11g的简单解决方案,但我相信以下内容将在9i中起作用。 来源(http://www.williamrobertson.net/documents/one-row.html) 您只需要将deptno替换为customer_name,将ename替换为Number_of_people
SELECT deptno, 
       LTRIM(SYS_CONNECT_BY_PATH(ename,\',\'))
FROM   ( SELECT deptno,
                ename,
                ROW_NUMBER() OVER (PARTITION BY deptno ORDER BY ename) -1 AS seq 
         FROM   emp ) 
WHERE  connect_by_isleaf = 1
CONNECT BY seq = PRIOR seq +1 AND deptno = PRIOR deptno 
START WITH seq = 1;


    DEPTNO CONCATENATED
---------- --------------------------------------------------
        10 CLARK,KING,MILLER
        20 ADAMS,FORD,JONES,SCOTT,SMITH
        30 ALLEN,BLAKE,JAMES,MARTIN,TURNER,WARD

3 rows selected.
    
        这将创建一个存储的“ 5”,表示您可以随时访问它。
CREATE OR REPLACE FUNCTION number_of_people(p_customer_name VARCHAR2)
RETURN VARCHAR2
IS
  v_number_of_people NUMBER;
  v_result VARCHAR2(500);

  CURSOR c1
  IS
    SELECT Number_of_people FROM the_table WHERE Customer_name = p_customer_name;
BEGIN
  OPEN c1;
  LOOP
    FETCH c1 INTO v_number_of_people;
    EXIT WHEN c1%NOTFOUND;

    v_result := v_result || v_number_of_people || \' \' || CHR(13);
  END;
END;
要运行它,请使用:
SELECT number_of_people(1) INTO dual;
希望对您有所帮助,请让我知道是否有任何错误,我自己没有对该函数进行测试。     
        做就是了
select Number_of_people
from table
where Customer_name = 1
我们缺少一些细节吗?     

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