ORACLE:SQL REGEXP_SUBSTR that returns the column value after last backslash(/)

Question

Gurwinder Singh · Accepted Answer · 2017-03-21 18:00:39Z

1

You don't need regular expressions for this. You can simply using substr and instr which are likely to perform faster:

select
  substr(col, instr(col, '/', -1) + 1)
from t;

If you must use regexp_substr (for some reason) then use:

select regexp_substr(col, '[^/]+$') from t;

answered Mar 21, 2017 at 17:54

Gurwinder Singh

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Comments

Oto Shavadze · Accepted Answer · 2017-03-21 17:58:44Z

0

If you need with REGEXP_SUBSTR also, then:

SELECT REGEXP_SUBSTR ('https://test/test/test/test/getTest/1234' , '[^/]+$' )  from dual

answered Mar 21, 2017 at 17:58

Oto Shavadze

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