[sql] Extract number from string with Oracle function

I need to create an Oracle DB function that takes a string as parameter. The string contains letters and numbers. I need to extract all the numbers from this string. For example, if I have a string like RO1234, I need to be able to use a function, say extract_number('RO1234'), and the result would be 1234.

To be even more precise, this is the kind of SQL query which this function would be used in.

SELECT DISTINCT column_name, extract_number(column_name) 
FROM table_name
WHERE extract_number(column_name) = 1234;

QUESTION: How do I add a function like that to my Oracle database, in order to be able to use it like in the example above, using any of Oracle SQL Developer or SQLTools client applications?

This question is related to sql oracle oracle11g oracle-sqldeveloper sqltools

The answer is


This works for me, I only need first numbers in string:

TO_NUMBER(regexp_substr(h.HIST_OBSE, '\.*[[:digit:]]+\.*[[:digit:]]*'))

the field had the following string: "(43 Paginas) REGLAS DE PARTICIPACION".

result field: 43


You can use regular expressions for extracting the number from string. Lets check it. Suppose this is the string mixing text and numbers 'stack12345overflow569'. This one should work:

select regexp_replace('stack12345overflow569', '[[:alpha:]]|_') as numbers from dual;

which will return "12345569".

also you can use this one:

select regexp_replace('stack12345overflow569', '[^0-9]', '') as numbers,
       regexp_replace('Stack12345OverFlow569', '[^a-z and ^A-Z]', '') as characters
from dual

which will return "12345569" for numbers and "StackOverFlow" for characters.


If you are looking for 1st Number with decimal as string has correct decimal places, you may try regexp_substr function like this:

regexp_substr('stack12.345overflow', '\.*[[:digit:]]+\.*[[:digit:]]*')

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