regexp_substr 当标点符号在字符串中间时忽略 (oracle)

Question

我正在查看文本中的大字符串。文中很多时候都有。或者特别是？？作为分隔符。

所以我想从字符串 'Delivery Note ?Patient was at home. Delivered at home'

是在家交付但不是送货单？患者在家这是我目前得到的。下面的代码在其他方面运行良好。

已添加：考虑到下面的评论，这里使用了正则表达式，因为实际数据源中有许多排列，例如在家中分娩、在救护车中分娩、在车辆中分娩或出生救护车，在家生等等等等

任何建议表示赞赏-

with txt as (select 1 as ID,
            'Delivery Note ?Patient was at home. Delivered at home' as note_text from dual)
    select txt.*,
regexp_substr(NOTE_TEXT, '(born|birth|sv(b|d)|deliver(ed|y|ing)|vbac)[^.|?|,|(|)|;|:|-].{1,24}(vehicle|bathroom|ambulance|home|[^a-z]car([^a-z|]|$))',1,1,'i') 
    as results
    
    from txt;

Answer 1

看起来这就是您需要的：

with txt as (select 1 as ID,
            'Delivery Note ?Patient was at home. Delivered at home' as note_text from dual)
    select txt.*,
regexp_substr(NOTE_TEXT, '(born|birth|sv(b|d)|deliver(ed|y|ing)|vbac)[^.?,)(;:-]{1,24}(vehicle|bathroom|ambulance|home|[^a-z]car([^a-z|]|$))',1,1,'i') 
    as results    
    from txt;

如您所见，我在 [^...] 中删除了 |，因为您只需要指定不带 or 的排除字符，并在其后删除 .。

结果：

        ID NOTE_TEXT                                             RESULTS
---------- ----------------------------------------------------- --------------------
         1 Delivery Note ?Patient was at home. Delivered at home Delivered at home

regexp_substr 当标点符号在字符串中间时忽略 (oracle)

regexp_substr ignore when punctuation in middle of string (oracle)

oracle

regexp-substr