如何从 Oracle SQL 中的 https 地址解析数字？答案

【问题标题】：How parse numbers from https arddress in Oracle SQL?如何从 Oracle SQL 中的 https 地址解析数字？
【发布时间】：2016-03-14 08:19:01
【问题描述】：

我想从 https-address 中提取两个字符串集。我正在使用 Oracle 11g，这是我的数据和我的解决方案：

with mcte as (
select 'https://id.adsf.fi/dfad2/services/org/Organizations/9884752-6/2542595-3/Corlsdf-asdf/Corlsdf/MOB50202282' as addr from dual
union all 
select 'https://id.asdf.fi/dfad2/services/org/Organizations/2435213-2/Corlsdf-asdf/Corlsdf/KVY239229' as addr from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0' as addr from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0' as addr  from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0/' as addr from dual
)
select  addr, 
        REGEXP_SUBSTR(addr,  '(Organizations\/)(\d+)', 1, 1, null, 2) AS num1,
        REGEXP_SUBSTR(addr,  '(Organizations\/)(\d+)', 1, 1, null, 3) AS num2
from mcte
;

这不会产生正确的集合，对于 NUM1 和 NUM2 应该是这样的

+----------------------------------------------------------------------------------------------------------+-----------+-----------+
|                                                    id                                                    |   num1    |   num2    |
+----------------------------------------------------------------------------------------------------------+-----------+-----------+
| https://id.adsf.fi/dfad2/services/org/Organizations/9884752-6/2542595-3/Corlsdf-asdf/Corlsdf/MOB50202282 | 9884752-6 | 2542595-3 |
| https://id.asdf.fi/dfad2/services/org/Organizations/2435213-2/Corlsdf-asdf/Corlsdf/KVY239229             | 2435213-2 | NULL      |
| https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0                                            | 6356334-0 | NULL      |
| https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0                                  | 6356334-0 | 6356324-0 |
| https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0/                                 | 6356334-0 | 6356324-0 |
+----------------------------------------------------------------------------------------------------------+-----------+-----------+

我怎样才能做到这一点？

【问题讨论】：

标签： sql regex oracle oracle11g

【解决方案1】：

你的表达不正确/不完整：

您只查找数字 (\d)，而不是查找减号 ([0-9-])。
他们缺少您要查找的第二个号码。

应该是这样的：

REGEXP_SUBSTR(addr, '(Organizations\/)([0-9-]+)', 1, 1, null, 2) AS num1,
REGEXP_SUBSTR(addr, '(Organizations\/)([0-9-]+)\/([0-9-]+)', 1, 1, null, 3) AS num2

【讨论】：

【解决方案2】：

尝试以下方法：

REGEXP_SUBSTR(addr,  '(Organizations\/)([^\/]+)', 1, 1, null, 2) AS num1,
REGEXP_SUBSTR(addr,  '(Organizations\/)([^\/]+\/)([0-9\-]+)', 1, 1, null, 3) AS num2

首先是：跳过Organizations/ 并在符号不是/ 时获取所有内容。

第二个是：跳过Organizations/..../，然后取连续数字和-。

这是小提琴http://sqlfiddle.com/#!4/9eecb7d/12905

【讨论】：

【解决方案3】：

REGEXP_SUBSTR 的第四个参数是出现参数。这应该可以让您到达您需要去的地方：

with mcte as (
select 'https://id.adsf.fi/dfad2/services/org/Organizations/9884752-6/2542595-3/Corlsdf-asdf/Corlsdf/MOB50202282' as addr from dual
union all 
select 'https://id.asdf.fi/dfad2/services/org/Organizations/2435213-2/Corlsdf-asdf/Corlsdf/KVY239229' as addr from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0' as addr from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0' as addr  from dual
union all
select 'https://id.asdf.fi/dfad2/services/org/Organizations/6356334-0/6356324-0/' as addr from dual
)
select  REGEXP_SUBSTR(addr, '/([[:digit:]\-])+', 1, 1) AS p1,
        REGEXP_SUBSTR(addr, '/([[:digit:]\-])+', 1, 2) AS p2
from mcte;

结果：

p1          p2
/9884752-6  /2542595-3
/2435213-2  
/6356334-0  
/6356334-0  /6356324-0
/6356334-0  /6356324-0

【讨论】：