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Oracle 11g REGEXP_SUBSTR question

Posted on 2011-02-16
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Last Modified: 2012-08-13
I have a field (varchar2 type) in oracle table with sample contents:
f1
---
33345
223
230.0
234 567
abc345
34902,345
2
Country of
3345-45

I want an O/P:
33345
223
230
234
NULL -- For abc345
34902
2
NULL -- Country of
3345


-----
So I want to extract the first numeric part of the string (of any length) . If the string does not start with number (0-9), return null. I tried with REGEXP_SUBSTR('my string', ....) --- need some help.
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Question by:toooki
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5 Comments
 
LVL 22

Expert Comment

by:Ivo Stoykov
ID: 34913882
Hi this the code below. table in mine select is named t

HTH

Ivo Stoykov
select f1, case when regexp_like(f1, '[a-zA-Z]') then null else f1 end case from t t

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LVL 74

Accepted Solution

by:
sdstuber earned 1800 total points
ID: 34915416
regexp_substr(f1,'^[0-9]+')
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Author Comment

by:toooki
ID: 34919629
Thank you all. I am checking.
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LVL 74

Expert Comment

by:sdstuber
ID: 34921329
 34913882 doesn't produce the requested results

here's the test case I used


WITH yourtable
     AS (SELECT '33345' f1 FROM DUAL
         UNION ALL
         SELECT '223' FROM DUAL
         UNION ALL
         SELECT '230.0' FROM DUAL
         UNION ALL
         SELECT '234 567' FROM DUAL
         UNION ALL
         SELECT 'abc345' FROM DUAL
         UNION ALL
         SELECT '34902,345' FROM DUAL
         UNION ALL
         SELECT '2' FROM DUAL
         UNION ALL
         SELECT 'Country of' FROM DUAL
         UNION ALL
         SELECT '3345-45' FROM DUAL)
SELECT f1,
       CASE WHEN REGEXP_LIKE(f1, '[a-zA-Z]') THEN NULL ELSE f1 END bad,
       REGEXP_SUBSTR(f1, '^[0-9]+') good
  FROM yourtable
0
 

Author Comment

by:toooki
ID: 34922363
Thanks a lot. regexp_substr(f1,'^[0-9]+')  worked perfectly.
Thanks everyone.
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