列值在一排

Question

我具有以下表

id count hour age range 
    ------------------------------------- 
    0  5  10  61  10-200 
    1  6  20  61  10-200 
    2  7  15  61  10-200 
    5  9  5  61  201-300 
    7  10  25  61  201-300 
    0  5  10  62  10-20 
    1  6  20  62  10-20 
    2  7  15  62  10-20 
    5  9  5  62  21-30 
    1  8  6  62  21-30 
    7  10  25  62  21-30 
    10  15  30  62  31-40 

我需要选择我尝试以下查询列范围 的不同值

Select distinct range as interval from table name where age = 62; 

其结果是在一列中，如下所示：

interval 
---------- 
10-20 
21-30 
31-41 

如何得到如下结果？

10-20, 21-30, 31-40 

编辑： 我现在尝试以下查询：

select sys_connect_by_path(range,',') interval 
from 
    (select distinct NVL(range,'0') range , ROW_NUMBER() OVER (ORDER BY RANGE) rn 

from table_name where age = 62) 

where connect_by_isleaf = 1 CONNECT BY rn = PRIOR rn+1 start with rn = 1; 

这是给我的输出：

家伙plz帮助我，让我所需的输出。

Answer 1

如果你是在11.2，而不是仅仅11.1，你可以，如果你使用的是Oracle的早期版本使用LISTAGG聚合函数

SELECT listagg(interval, ',') 
     WITHIN GROUP(ORDER BY interval) 
    FROM (SELECT DISTINCT range AS interval 
      FROM table_name 
     WHERE age = 62) 

，您可以使用其他Oracle string aggregation techniques之一蒂姆·霍尔的页面上。此前11.2，我个人的偏好是创建一个user-defined aggregate function，这样你就可以

SELECT string_agg(interval) 
    FROM (SELECT DISTINCT range AS interval 
       FROM table_name 
      WHERE age = 62) 

如果你不希望创建一个功能，但是，你可以使用ROW_NUMBER and SYS_CONNECT_BY_PATH approach尽管这往往得到位难以遵循

with x as (
    SELECT DISTINCT range AS interval 
      FROM table_name 
     WHERE age = 62) 
select ltrim(max(sys_connect_by_path(interval, ',')) 
       keep (dense_rank last order by curr), 
       ',') range 
    from (select interval, 
       row_number() over (order by interval) as curr, 
       row_number() over (order by interval) -1 as prev 
      from x) 
connect by prev = PRIOR curr 
    start with curr = 1