REGEXP\u子类

到目前为止，正则表达式是我的弱点。 我正在尝试分解以下字符串

Node 51 Path 1 Route 4
Node 51A Path 12 Route 3
Node 5 Path 12 Route 2
Node 7B Path 1 Route 1

我需要的是节点，节点的字母，路径和路线

提取节点的字母时遇到问题。节点的字母是单个非数字字符，始终紧跟在节点编号之后，不带空格

对于第2行和第4行

Node 51A Path 12 Route 3 - Nodes letter is A
Node 5 Path 12 Route 2 - Nodes letter is NULL 
Node 7B Path 1 Route 1- Nodes letter is B

到目前为止,

with gen as (
    select 'Node 51 Path 1 Route 4' x from dual union all 
    select 'Node 51A Path 12 Route 3' x from dual union all 
    select 'Node 5 Path 12 Route 2' x from dual union all 
    select 'Node 7B Path 1 Route 1' x from dual
) 
select  x , 
        regexp_substr(x, '(\d+)',1,1) as Node , 
        regexp_substr(x, '(\d+)',1,2) as Path , 
        regexp_substr(x, '(\d+)',1,3) as Route
from    gen  

X                        NODE   PATH   ROUTE
------------------------ ------ ------ -------
Node 51 Path 1 Route 4   51     1      4
Node 51A Path 12 Route 3 51     12     3
Node 5 Path 12 Route 2   5      12     2
Node 7B Path 1 Route 1   7      1      1

---

Oracle 10gR2。

因此您提取了整个节点并删除了所有数字。美好的谢谢

with gen as (
    select 'Node 51 Path 1 Route 4' x from dual union all 
    select 'Node 51A Path 12 Route 3' x from dual union all 
    select 'Node 5 Path 12 Route 2' x from dual union all 
    select 'Node 7B Path 1 Route 1' x from dual
) 
select  x , 
        regexp_substr(x, '\d+') as Node, 
        regexp_replace(regexp_substr(x, '\d+\S*'),'\d+') as NodeLetter , 
        regexp_substr(x, '\d+',1,2) as Path , 
        regexp_substr(x, '\d+',1,3) as Route
from    gen