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sqlregexoracle-databaseregexp-substr

oracle get all matched occurrences from a column


I have a table, which has 2 columns: ID & JOB_Description(Text). I would like to write an oracle SQL to extract all substrings in the Description column which match a regular pattern.

However, I have learnt how to extract matched substrings from a string with below SQL, but I have no idea to apply below SQL on all data in one go on the aforementioned table(column:JOB_Description).

SQL to get all matched occurrences from a string:

 SELECT REGEXP_SUBSTR(JOB_Description, '(ABC|DE)([[:digit:]]){5}', 1, LEVEL) AS substr
   FROM (
         select 'Please help to repair ABC12345, DE22222' as JOB_Description 
          from DUAL)
  CONNECT BY LEVEL <= REGEXP_COUNT(JOB_Description, '(ABC|DE)([[:digit:]]){5}');

Solution

  • you can try this query out.

     with test as(
      select 'ABC12345, DE22222' as JOB_Description from DUAL union
      select 'Please help to repair ABC12345, DE22222' as JOB_Description from DUAL 
    )
    SELECT REGEXP_SUBSTR(JOB_Description, '(ABC|DE)([[:digit:]]){5}', 1, LEVEL) AS substr
    FROM test
    CONNECT BY LEVEL <= REGEXP_COUNT(JOB_Description, '(ABC|DE)([[:digit:]]){5}')
      AND PRIOR JOB_Description = JOB_Description
      AND PRIOR DBMS_RANDOM.VALUE IS NOT NULL
    

    Result:

    ABC12345
    DE22222
    ABC12345
    DE22222
    

    A good explanation of the last two lines can be found here