PLSQL SUBSTR function ignore the trailing zero - plsql

Question

select TO_NUMBER (SUBSTR(10.31, INSTR (10.31, '.') + 1)) from dual
Above query returns 31 as the output. But below query returns 3 as the output.
select TO_NUMBER (SUBSTR(10.30, INSTR (10.30, '.') + 1)) from dual
How could I get the 30 as the output instead of the 3?

Answer 1

As it seems (from comments) that you are starting with a numeric value that you want to turn into words, you should begin by splitting it into dollars and cents.
If you really need to use substr etc, then you could start with a known format, such as to_char(amount,'fm9990.00'), so it will be a string with exactly two decimal places. However, if you have the numeric value it would be easier to convert it into the desired units using arithmetic functions. Whole dollars are trunc(amount) and cents are 100 * mod(amount,1).
Another issue is that the 'Jsp' date format approach can't handle zeroes. If you are using Oracle 12.2 or later there is a workaround using the default on conversion error clause:
create table demo
( amount number(6,2) );
insert into demo values (10.3);
insert into demo values (.25);
insert into demo values (25);
select amount
, nvl(to_char(to_date(trunc(amount) default null on conversion error,'J'),'Jsp'),'Zero') as dollars
, nvl(to_char(to_date(100 * mod(amount,1) default null on conversion error,'J'),'Jsp'),'Zero') as cents
from demo;
AMOUNT DOLLARS CENTS
-------- ------------ -------------
10.30 Ten Thirty
25.00 Twenty-Five Zero
0.25 Zero Twenty-Five
In 12.1 you could get around it using an inline function (maybe not a bad idea even in later versions, to simplify the rest of the query):
with
function to_words(num number) return varchar2 as
begin
return
case num
when 0 then 'Zero'
else to_char(to_date(num,'J'),'Jsp')
end;
end;
select amount
, to_words(trunc(amount)) as dollars
, to_words(100 * mod(amount,1)) as cents
from demo;
For values greater than 5373484 (the Julian representation of date '9999-12-31'), you can use this from Ask Tom: Spell the number (converted here to a WITH clause, but you can create it as a standalone function):
with function spell_number
( p_number in number )
return varchar2
as
-- Tom Kyte, 2001:
-- https://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1407603857650
l_num varchar2(50) := trunc(p_number);
l_return varchar2(4000);
type myarray is table of varchar2(15);
l_str myarray :=
myarray
( ''
, ' thousand '
, ' million '
, ' billion '
, ' trillion '
, ' quadrillion '
, ' quintillion '
, ' sextillion '
, ' septillion '
, ' octillion '
, ' nonillion '
, ' decillion '
, ' undecillion '
, ' duodecillion ');
begin
for i in 1 .. l_str.count loop
exit when l_num is null;
if substr(l_num, length(l_num) -2, 3) <> 0 then
l_return := to_char(to_date(substr(l_num, length(l_num) - 2, 3), 'J'), 'Jsp') || l_str(i) || l_return;
end if;
l_num := substr(l_num, 1, length(l_num) - 3);
end loop;
return l_return;
end spell_number;
select amount
, spell_number(trunc(amount)) as dollars
, spell_number(100 * mod(amount,1)) as cents
from demo
/

Answer 2

I am actually surprised that your current query is even running without error, given that Oracle's SUBSTR function is supposed to operate on strings, not numbers. That being said, if you properly use your current query with strings, then it works:
SELECT TO_NUMBER(SUBSTR('10.30', INSTR ('10.30', '.') + 1)) FROM dual; -- returns 30
A more compact (though not necessarily more performant) way of doing this might be to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR('10.30', '[0-9]+$') FROM dual;
This would retain only digits appearing after the decimal point, in the case that a decimal point be present. Otherwise, it would just return all numbers for inputs which have no decimal component.