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I use this regex for extract email:

regexp_replace(xxx, '.*=([A-Za-z0-9._%-]*@[A-Za-z0-9._%-]*\.[A-Za-z]{2,4}).*','\1')
regexp_like(xxx,'=+[A-Za-z0-9._%-]+@[A-Za-z0-9._%-]+\.[A-Za-z]{2,4}')

I can extract string like this:

select regexp_replace('/AAA-SSS-CCC?User=testmail@mail.com?Id=12323424','.*=([A-Za-z0-9._%-]*@[A-Za-z0-9._%-]*\.[A-Za-z]{2,4}).*','\1')

Output: testmail@mail.com

But I can't extract string like this:

select regexp_replace('/AAA/SSS/CCC/testmail@mail.com?Id=12323424','.*=([A-Za-z0-9._%-]*@[A-Za-z0-9._%-]*\.[A-Za-z]{2,4}).*','\1')

Output: /AAA/SSS/CCC/testmail@mail.com?Id=12323424

How can I do that?

Thanks.

onur
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1 Answers1

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Try This. I have modified the pattern to be matched before @ appears. It matches one or more occurrence of any character other than @, which matches / . 

SELECT REGEXP_REPLACE (
          '/AAA/SSS/CCC/testmail@mail.com?Id=12323424',
          '^([^@]+)@[A-Za-z0-9._%-]*\.[A-Za-z]{2,4}.*',
          '\1')
  FROM DUAL;
Kaushik Nayak
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