Question

我需要拆分消息：

 500 Oracle Parkway.Redwood Shores.*.=13

现在我有一个针对Substr1 / 2/4

的解决方案

  SELECT '500 Oracle Parkway.Redwood Shores.*.=13' string1,
  REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','.[^.]+') 
  "SUBSTR1" ,
  replace(REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[$.]+
  [^.]+'),'.',null) "SUBSTR2" ,
  REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[$.]+.[$.]+[^.]') 
  "SUBSTR3" ,
  REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[^=]+$') 
  "SUBSTR4" 
  FROM DUAL;

但是，Substr3包含＆＃39; =＆＃39;。我想至少拥有＆＃39;。*。＆＃39;或者＆＃39; *＆＃39;

你能不能给我一个提示如何＆＃34;排除＆＃34; regexp中的任何字符（例如＆＃39; =＆＃39;）

非常感谢任何帮助！

谢谢

已解决，请参阅SUBSTR3.1

      SELECT
     '500 Oracle Parkway.Redwood Shores.*.=13' string1,
      REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','.[^.]+') 
      "SUBSTR1" ,
      replace(REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[$.]+
      [^.]+'),'.',null) "SUBSTR2" ,
      REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[$.]+.[$.]+
      [^.]') "SUBSTR3" ,
      REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[^.]+',1,3) 
      "SUBSTR3.1" ,
      REGEXP_SUBSTR('500 Oracle Parkway.Redwood Shores.*.=13','[^=]+$') 
      "SUBSTR4" 
      FROM DUAL;

Answer 1

非常尊重Alex Poole，如果缺少列表中的某个元素，格式'[^.]+'的正则表达式将失败。它将默默地返回不正确的数据。请改用此表格。注意我从第一个示例中删除了城市。尝试一下，你可能会感到惊讶：

with t (str) as (
  select '500 Oracle Parkway..*.=13' from dual union 
  select 'One Microsoft Way.Redmond.Washington.=27' from dual
)
select str,
  regexp_substr(str, '(.*?)(\.|$)', 1, 1, NULL, 1) as substr1,
  regexp_substr(str, '(.*?)(\.|$)', 1, 2, NULL, 1) as substr2,
  regexp_substr(str, '(.*?)(\.|$)', 1, 3, NULL, 1) as substr3,
  ltrim(regexp_substr(str, '(.*?)(\.|$)', 1, 4, NULL, 1), '=') as substr4
from t;

有关详细信息，请参阅此处：Split comma separated values to columns in Oracle

Answer 2

看起来您正在尝试根据句点对源字符串进行标记，然后它们（可能）从第四个标记中删除前导等号。您用于＆＃39; substring3.1＆＃39;的解决方案可以用于所有这些：

with t (str) as (
  select '500 Oracle Parkway.Redwood Shores.*.=13' from dual
  union all select 'One Microsoft Way.Redmond.Washington.=27' from dual
)
select str,
  regexp_substr(str, '[^.]+', 1, 1) as substr1,
  regexp_substr(str, '[^.]+', 1, 2) as substr2,
  regexp_substr(str, '[^.]+', 1, 3) as substr3,
  ltrim(regexp_substr(str, '[^.]+', 1, 4), '=') as substr4
from t;

STR                                      SUBSTR1              SUBSTR2              SUBSTR3    SUBSTR4
---------------------------------------- -------------------- -------------------- ---------- -------
500 Oracle Parkway.Redwood Shores.*.=13  500 Oracle Parkway   Redwood Shores       *          13     
One Microsoft Way.Redmond.Washington.=27 One Microsoft Way    Redmond              Washington 27

Oracle使用regexp_substr拆分消息

2 个答案: