How to with oracle regexp_replace parse string

Question

I have a string: "ABC-30 (40D)", the result is 30, my example:

regexp_replace(f, '^[^-]+-(\d+)*')

don't work...

without regexp it works, but I'm afraid this :D substr(substr(f, - instr(reverse(f), '-') + 1), 1, instr(substr(f, - instr(reverse(f), '-') + 1), ' ')-1), — ExcepOra
– ExcepOra, Commented Oct 7, 2017 at 11:34

Kaushik Nayak · Accepted Answer · 2017-10-07 11:40:49Z

2

Use REGEXP_SUBSTR with this pattern.

SELECT regexp_substr('ABC-30 (40D)', '^[^-]+-(\d+)[^0-9]+.*$' , 1 ,1 , 'i', 1  ) FROM DUAL;

answered Oct 7, 2017 at 11:40

Kaushik Nayak

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2 Comments

Ok, thx, pls advise, how to with regexp_like in where clause filter only digits in this example result is same: 30?

REGEXP_LIKE(f, '^[^-]+-(\d+)[^0-9]+.*$') this is work