拆分正则表达式并循环

Question

我需要在我认为我写的连接内循环。 我正在发布代码。

    select listagg(request_num,',') within group (order by request_num) as request_num,segmentation_name from (
select MST.REQUEST_NUM,seg_dtls.SEGMENT_NAME,LAST_UPDATED_date,seg_dtls.segmentation_name from 
(select  * from rp_sr_master ) Mst,
(select  SUBSTR(ANSWER,1,INSTR (ANSWER, '~', 1)-1) AS SM_ID,sr_id from rp_sR_details 
WHERE Q_ID in (SELECT Q_ID FROM RP_QUESTIONS WHERE field_id='LM_LRE_Q6')
    ) Dtls, (select SM_ID, SQL_STATEMENT, CREATION_DATE, UPDATED_DATE, SEGMENT_NAME,segmentation_name ,TOTAL_COUNT
from rp_sEGMENT_master ) seg_dtls
where Dtls.SM_ID=seg_dtls.SM_ID
and Dtls.sr_id=Mst.sr_id)
group by segmentation_name;

我在这里面临的问题如下，

(select  SUBSTR(ANSWER,1,INSTR (ANSWER, '~', 1)-1) AS SM_ID,sr_id from rp_sR_details 
    WHERE Q_ID in (SELECT Q_ID FROM RP_QUESTIONS WHERE field_id='LM_LRE_Q6')
        )

在上面的代码中，答案是这样的：

2603~NG non IaaS IT Professional^2600~NG non IaaS Senior IT^2598~NG data profiling SENIOR IT professional^2595~Nigeria data profiling IT professiona

它只选择第一个数字 2603，其他数字将被忽略。

有什么方法可以循环遍历“ANSWER”中的所有数字。 我正在寻找想法。

谢谢。

Answer 1

一个想法是使用一种将逗号分隔的字符串拆分为行的方法，
您可以在以下答案中找到该方法的示例：

Splitting comma separated values in Oracle

How can I use regex to split a string, using a string as a delimiter?

以上解决方案使用regexp_substr函数。
如果您深入了解 Oracle 的 REGEXP_SUBSTR function 的详细信息，您会发现那里有可选的 position 参数。

此参数可以与此答案中显示的解决方案结合使用：
SQL to generate a list of numbers from 1 to 100
（即SELECT LEVEL n FROM DUAL CONNECT BY LEVEL <= 100）通过以下方式：

with xx as (
select '2603~NG non IaaS  IT Professional^2600~NG non IaaS Senior '
       || 'IT^2598~NG data profiling SENIOR IT professional^2595~Nigeria '
       || 'data profiling IT professiona' as answer
from dual
)
select LEVEL AS n, regexp_substr( answer, '\d+',  1, level) as nbr
from xx
connect by level <= 6
;

---

上述查询产生以下结果：

N |NBR  |
--|-----|
1 |2603 |
2 |2600 |
3 |2598 |
4 |2595 |
5 |     |
6 |     |

我们需要从结果集中消除空值，可以使用简单的条件IS NOT NULL

with xx as (
select '2603~NG non IaaS  IT Professional^2600~NG non IaaS Senior '
       || 'IT^2598~NG data profiling SENIOR IT professional^2595~Nigeria '
       || 'data profiling IT professiona' as answer
from dual
)
select LEVEL AS n, regexp_substr( answer, '\d+',  1, level) as nbr
from xx
connect by regexp_substr( answer, '\d+',  1, level) IS NOT NULL
;

N |NBR  |
--|-----|
1 |2603 |
2 |2600 |
3 |2598 |
4 |2595 |

---

上面的查询非常适合单条记录，但是当我们尝试解析 2 行或更多行时会感到困惑。幸运的是，SO 上有另一个答案有助于解决这个问题：

Is there any alternative for OUTER APPLY in Oracle?

---

--  source data
WITH xx as (
select 1 AS id,
       '2603~NG non IaaS  IT Professional^2600~NG non IaaS Senior '
       || 'IT^2598~NG data profiling SENIOR IT professional^2595~Nigeria '
       || 'data profiling IT professiona' as answer
from dual
UNION ALL
select 2 AS id,
       '11111~NG non IaaS  IT Professional^22222~NG non IaaS Senior '
       || 'IT^2598~NG data 33333 profiling SENIOR IT professional^44~Nigeria '
       || 'data profiling 5 IT professiona 66' as answer
from dual
)
-- end of source data


SELECT t.ID, t1.n, t1.nbr
FROM xx t
CROSS JOIN LATERAL (
        select LEVEL AS n, regexp_substr( t.answer, '\d+',  1, level) as nbr
        from dual
        connect by regexp_substr( t.answer, '\d+',  1, level) IS NOT NULL
) t1;

上述查询从两条记录中解析数字，然后以以下形式显示：

ID |N |NBR   |
---|--|------|
1  |1 |2603  |
1  |2 |2600  |
1  |3 |2598  |
1  |4 |2595  |
2  |1 |11111 |
2  |2 |22222 |
2  |3 |2598  |
2  |4 |33333 |
2  |5 |44    |
2  |6 |5     |
2  |7 |66    |

---

我相信您会设法将这个简单的“解析”查询合并到您的主查询中。