How to get rightmost 10 places of a string in oracle

Question

Hi,

i am trying to fetch an id from an oracle table. Its something like "TN0001234567890345". What i want is to sort the values according to the right most 10 positions. I am using oracle 11g. Is there any funcrion to cut the rightmost 10 places in oracle ?

thanks in advance

tismon

Answer 1

You can use SUBSTR function as:

select substr('TN0001234567890345',-10) from dual;

Output:

4567890345

Answer 2

Calling substr(id,-10,10) will return the rightmost 10 characters of id.