正则表达式解析至少20个字符的文本，字符串中不带空格

Question

i need to parse a code from hand inputed string field in SQL query for Oracle DB which can look something lik this:

"i would ?! like / to make * bb a8 001/XYZ/0002/65432178 thank you very much friends"

Number of words totally differs each row only thing that is the same are the spaces between the words. Somewhere in the middle (but can be also very beginning or very end) of the string is CODE with variable length (always at least 20 chars) that i need to parse out - code is always without spaces and divided from the rest of the random text by spaces. I need to parse out just the code cutting all the words. So only way how to identify the code in my opinion is that it sould be at least 20 char sequence without space. Can you recommed regex to do this kind of thing? Thank you very much

so i expect to get string like this "001/XYZ/0002/65432178"

Answer 1

You can just look for a sequence of 20+ instances of anything except a space:

select regexp_substr(
  'i would ?! like / to make * bb a8 001/XYZ/0002/65432178 thank you very much friends',
  '[^ ]{20,}') as result
from dual;

RESULT               
---------------------
001/XYZ/0002/65432178

[^ ] is a pattern that excludes spaces; {20,} means that has to be repeated a minimum of 20 times, with no maximum.

If you want to exclude any whitespace - in case, for instance, there's a tab instead of a space immediately before or after the part you want - you can use a character class instead:

regexp_substr(<your string>, '[^[:space:]]{20,}')

As @MTO points out, these will match the first 20-character string within the value, and it's feasible your user-inputted text could contain long non-code values that you don't really want to see. It would be better if you could match on an expected pattern for the code.

Answer 2

If you will never have any words longer than 20 characters then you can naively use:

SELECT REGEXP_SUBSTR( value, '\S{20,}' ) AS code,
       value
FROM   data d;

However, if you have words longer than 20 characters such as:

CREATE TABLE data ( value ) AS
SELECT 'long words like floxinoxinihilipilification and antidisestablishmentarianism with your code 001/XYZ/0002/65432178' FROM DUAL UNION ALL
SELECT 'i would ?! like / to make * bb a8 001/XYZ/0002/65432178 thank you very much friends' FROM DUAL;

Then the above code outputs:

CODE                        | VALUE                                                                                                            
:-------------------------- | :----------------------------------------------------------------------------------------------------------------
floxinoxinihilipilification | long words like floxinoxinihilipilification and antidisestablishmentarianism with your code 001/XYZ/0002/65432178
001/XYZ/0002/65432178       | i would ?! like / to make * bb a8 001/XYZ/0002/65432178 thank you very much friends

Instead you could try to do something like returning the word with more than 20 characters which also has the greatest number of / characters:

SELECT ( SELECT MAX( REGEXP_SUBSTR( d.value, '\S{20,}', 1, LEVEL ) ) KEEP ( DENSE_RANK LAST ORDER BY REGEXP_COUNT( REGEXP_SUBSTR( d.value, '\S{20,}', 1, LEVEL ), '/' ) )
         FROM   DUAL
         CONNECT BY LEVEL <= REGEXP_COUNT( d.value, '\S{20}' )
       ) AS code,
       value
FROM   data d;

Which outputs:

CODE                  | VALUE                                                                                                            
:-------------------- | :----------------------------------------------------------------------------------------------------------------
001/XYZ/0002/65432178 | long words like floxinoxinihilipilification and antidisestablishmentarianism with your code 001/XYZ/0002/65432178
001/XYZ/0002/65432178 | i would ?! like / to make * bb a8 001/XYZ/0002/65432178 thank you very much friends

db<>fiddle here