How to get rightmost 10 places of a string in oracle

You can use SUBSTR function as:

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

Output:

4567890345

codaddict's solution works if your string is known to be at least as long as the length it is to be trimmed to. However, if you could have shorter strings (e.g. trimming to last 10 characters and one of the strings to trim is 'abc') this returns null which is likely not what you want.

Thus, here's the slightly modified version that will take rightmost 10 characters regardless of length as long as they are present:

select substr(colName, -least(length(colName), 10)) from tableName;

Another way of doing it though more tedious. Use the REVERSE and SUBSTR functions as indicated below:

SELECT REVERSE(SUBSTR(REVERSE('TN0001234567890345'), 1, 10)) FROM DUAL;

The first REVERSE function will return the string 5430987654321000NT.

The SUBSTR function will read our new string 5430987654321000NT from the first character to the tenth character which will return 5430987654.

The last REVERSE function will return our original string minus the first 8 characters i.e. 4567890345