I was trying to reverse a number in PL/SQL. It's working fine, but when my number contains any 0, the output is unexpected. For example:
1234 output 4321
1000 output 1
1203 ouput 3021
10001 output 1
DECLARE
r number(9);
num number(9):=&p_num;
BEGIN
WHILE num>=1 LOOP
IF mod(num,10)=0 THEN -- extracting last digit of a number .
r:=r||0; --if end digit is 0 then concat r with 0
ELSE
r:=r||mod(num,10);--if mod is not 0 then it would be last digit.
END IF;
num:=trunc(num/10);--Removing last digit from number
END LOOP;
dbms_output.put_line(r);
END;
Try changing the type of the variable "r" to varchar2.
Since it's declared as number, leading zeros are discarded.
'Reversing a number' is fundamentally a string operation, not a numerical one. Numerically, the reverse of 10, 100, 1000, etc are all 1 - the leading zeroes in the result don't count. And the operation is not, therefore, invertible; all numbers with the same leading (significant) digits and with zero or more trailing zeroes produce the same output when reversed.
So, you need to revise your code to generate a character string, not a number.
You can't preserve leading zeros with numbers; you must use strings (varchar2). Try something like this to see:
DECLARE
r varchar2(9);
num varchars(9):=&p_num;
BEGIN
r := REVERSE(num);
dbms_output.put_line(r);
END;
I'm not sure what's going wrong in your code Vineet but perhaps this will work.
DECLARE
newStr varchar2(9) = "";
numStr varchar2(9) := to_char(&p_num);
i number;
BEGIN
i = length(numStr);
WHILE i>0 LOOP
newStr := newStr || substr(numStr, i, i + 1);
i = i - 1;
END LOOP;
dbms_output.put_line(r);
END;
Edit: Or as gabe correctly points out, just use the REVERSE function.
The problem is you're dealing with a NUMBER value. When you reverse 1000, you get 0001, which when output unformatted is 1.
What you really need is something akin to:
CREATE OR REPLACE FUNCTION rev(p_num NUMBER)
RETURN VARCHAR2 IS
v_chr VARCHAR2(50);
BEGIN
v_chr := p_num;
IF LENGTH(v_chr) > 1 THEN
RETURN SUBSTR(v_chr, -1, 1)||rev(SUBSTR(v_chr, 1, LENGTH(v_chr)-1));
END IF;
RETURN v_chr;
EXCEPTION
WHEN OTHERS THEN
RETURN 'Bad Input';
END;
/
Function created
SQL> SELECT rev(100000) FROM dual;
REV(100000)
--------------------------------------------------------------------------------
000001
SQL>
Related
following issue:
I get a String as such "512, 986, 571, 665" transferred as parameter, basically a set of 3 digit numbers and these need to be returned sorted from highest to lowest.
I have only found answers pertaining such an issue when the String is found in a table and not given as parameter to a function
Here's one option: split input string into rows (that's what subquery in lines #4 - 6 does), and then aggregate them back using listagg with appropriate order by clause (line #3).
SQL> with test (col) as
2 (select '512, 986, 571, 665' from dual)
3 select listagg(val, ', ') within group (order by val) result
4 from (select to_number(trim(regexp_substr(col, '[^,]+', 1, level))) val
5 from test
6 connect by level <= regexp_count(col, ',') + 1
7 );
RESULT
--------------------------------------------------------------------------------
512, 571, 665, 986
SQL>
A programmatic approach would be to use each element in the list as the index of an associative array (so integers only, or it'll break), which will have the effect of sorting it, then loop through the resulting array constructing a new list:
create or replace function sort_number_list
( p_list varchar2 )
return varchar2
as
type aa is table of number index by pls_integer;
sort_tab aa;
sorted_list varchar2(4000);
begin
for i in 1..regexp_count(p_list, ',') +1 loop
sort_tab(regexp_substr(p_list,'[^,]+', 1, i)) := i;
end loop;
for i in indices of sort_tab loop
sorted_list := sorted_list || ', ' || i;
end loop;
return ltrim(sorted_list,', ');
end sort_number_list;
I've used the indices of syntax added in 21c which makes looping around an associative array less verbose. Alternatively you can use the first and next collection methods.
create or replace TYPE t_numlist IS table OF varchar2(32700);
create or replace function sort_str_of_num(p_string in varchar2,p_delimiter in char) return
varchar2 as
v_strnum varchar2(32767):=p_string;
v_delimiter char(1):=p_delimiter;
l_num t_numlist:=t_numlist();
incr integer :=1;
b_bool boolean:= true;
begin
v_strnum := trim(replace(v_strnum,v_delimiter,' '));
loop
l_num.extend;
l_num(incr):= regexp_substr(v_strnum,'[0-9]{1,}');
v_strnum := trim(SUBSTR(v_strnum,instr(v_strnum,' ')+1));
if (instr(v_strnum,' ')+1)=1 then
l_num.extend;
l_num(incr+1):=regexp_substr(v_strnum,'[0-9]{1,}');
exit;
end if;
incr:=incr+1;
end loop;
for i in (select column_value from table(l_num) order by to_number(column_value))loop
if b_bool then v_strnum:=i.column_value;
else v_strnum:=i.column_value||', '||v_strnum;
end if;
b_bool:=false;
end loop;
return v_strnum;
exception
when others then dbms_output.put_line('wrong entry');
end;
This will sort any list of numbers in a string and numbers can have any digit count. Also u can enter any character to be the delimiter between numbers (delimiter must be 1 character) and space character in number list string is ignored.
One way would be to use the ability of xmltable to parse a comma-separated list into XML elements, then use its getstringval() method to convert each element into a string, then finally re-aggregate the strings using listagg:
create or replace function sort_number_list
( p_list varchar2 )
return varchar2
as
sorted_list varchar2(4000);
begin
select listagg(x.column_value.getstringval(), ', ') within group (order by to_number(x.column_value.getstringval()))
into sorted_list
from xmltable(p_list) x;
return sorted_list;
end sort_number_list;
Test:
select sort_number_list('97,3,-123,4')
from dual;
SORT_NUMBER_LIST('97,3,-123,4')
------------------------------------------------------------
-123, 3, 4, 97
I am converting number into word with own code using length function.
Here is the code:
CREATE OR REPLACE FUNCTION num_to_words (NUM IN NUMBER)
RETURN VARCHAR2
IS
v_length NUMBER;
v_words VARCHAR2(100);
word_1 VARCHAR2(100);
BEGIN
word_1 := 'HUNDRED';
v_length := LENGTH(NUM);
IF v_length = 3
THEN v_words := SUBSTR(TO_CHAR(TO_DATE(NUM,'J'),'JSP'),1,3) ||' '|| word_1;
END IF;
RETURN(v_words);
END;
Problem is when I enter "100", it successfully converts into "ONE HUNDRED".
How can I implement the code for converting "101" into "ONE HUNDRED ONE".
The JSP conversion will do that for you; you don't need to extract the first digit, or supply the 'hundred' text yourself:
IF v_length <= 3 THEN
v_words := TO_CHAR(TO_DATE(NUM,'J'),'JSP');
END IF;
If you pass in 101 the result is ONE HUNDRED ONE. And because I changed = 3 to <= 3 it will work for any 1, 2 or 3-digit value, so passing in 11 returns ELEVEN.
For longer values you might be looking for something like this, which breaks the number down into groups and converts each group onto words plus the corresponding scale (e.g. 'million'). If you want non-standard grouping then it's a bit more involved, but there's a recent example for lakh here.
I am trying to create a way to find the exponent of a number (in this case the base is 4 and the exponent 2 so the answer should be 16) using a procedure without using the POW Function or any built in functions to find the exponent. Eventually I would like to take input numbers from the user.
set serveroutput on;
CREATE OR REPLACE PROCEDURE Exponent(base number, exponent number) as
answer number;
BEGIN
base := 4;
exponent := 2;
LOOP
IF exponent > 1 THEN
answer := base * base;
END IF;
END LOOP;
dbms_output.put_line('Answer is: ' || answer);
END;
/
Error(7,25): PLS-00103: "expression 'BASE' cannot be used as an assignment target" and "expression 'EXPONENT' cannot be used as an assignment target"
Any ideas on how to solve the error and/or better ways of getting the exponent without using built-in functions like POW?
In your procedure base and exponent are input parameters and can't be changed. You've got a couple of options:
1) copy the parameters to variables internal to the procedure and manipulate those internal values, or
2) change the parameters to be input/output parameters so you can change them.
Examples:
1)
CREATE OR REPLACE PROCEDURE Exponent(pin_base number, pin_exponent number) as
base number := pin_base;
exponent number := pin_exponent;
answer number;
BEGIN
base := 4;
exponent := 2;
LOOP
IF exponent > 1 THEN
answer := base * base;
END IF;
END LOOP;
dbms_output.put_line('Answer is: ' || answer);
END;
2)
CREATE OR REPLACE PROCEDURE Exponent(base IN OUT number,
exponent IN OUT number) as
answer number;
BEGIN
base := 4;
exponent := 2;
LOOP
IF exponent > 1 THEN
answer := base * base;
END IF;
END LOOP;
dbms_output.put_line('Answer is: ' || answer);
END;
The best thing is that whatever Oracle provides as inbuilt functionality that serves the purpose in a best possible. (Almost all the times better then customized codes) Try to use EXP function. I have tried to make customized code per my understanding. Hope this helps.
CREATE OR REPLACE
FUNCTION EXP_DUMMY(
BASE_IN IN NUMBER,
EXPO_IN IN NUMBER)
RETURN PLS_INTEGER
AS
lv PLS_INTEGER:=1;
BEGIN
FOR I IN
(SELECT base_in COL1 FROM DUAL CONNECT BY level < expo_in+1
)
LOOP
lv:=lv*i.col1;
END LOOP;
RETURN
CASE
WHEN EXPO_IN = 0 THEN
1
ELSE
lv
END;
END;
SELECT EXP_DUMMY(2,4) FROM DUAL;
I want to make a procedure called OE that will have text output based on the number that I define.
For example, inputting the number 6 will give the following output:
odd
even
odd
even
odd
even
= even steven!
and inputting the number 5 will give the following output:
odd
even
odd
even
odd
= you oddball!
I'm completely new at this and have been struggling to get the odd number to load correctly (for some reason, it gets stuck in an infinite loop). Any help would be appreciated! Here is what I got so far:
CREATE OR REPLACE procedure oe
(p_n IN number)
AS
v_n number;
v_on number;
BEGIN
v_n := p_n;
v_on := p_n;
IF v_n>0 THEN LOOP
dbms_output.put_line('odd');
v_n := v_n-1;
dbms_output.put_line('even');
v_n := v_n-1;
If v_n=0 then
exit;
if v_on mod 2 > 0 then dbms_output.put_line('=' || ' you oddball!');
exit;
else
dbms_output.put_line('=' || ' even steven!');
exit;
end if;
end if;
end loop;
end if;
END;
/
You are not using exit conditions properly hence your code is going in infinite loop. You simplify your logic as below. Let me know it it works for you.
You may add few validations to make sure you get proper input parameters such as p_n > 0 and other.
CREATE OR REPLACE procedure oe
(p_n IN number)
AS
begin
for i in 1..p_n
loop
if mod(i,2)=1 then dbms_output.put_line('odd');
else dbms_output.put_line('even');
end if;
end loop;
if mod(p_n,2)=1 then dbms_output.put_line('= you oddball!');
else dbms_output.put_line('= even steven!');
end if;
end;
hemalp108 has already answered this, but I just wanted to add that you don't even need the if/else logic that fills the procedure (except perhaps for handling values less than 1, which I'll leave as an exercise), because we have case:
create or replace procedure oe
( p_n in number )
as
begin
for i in 1 .. p_n loop
dbms_output.put_line(case mod(i,2) when 1 then 'odd' else 'even' end);
end loop;
dbms_output.put_line(case mod(p_n,2) when 1 then '= you oddball!' else '= even steven!' end);
end;
(You may also notice how laying your code out neatly is half the way towards debugging it.)
There is this code for generatig unique shortname from a table MMSTREPHDR .
I have shortnames kv,kv1,kv2,kv3 already in MMSTREPHDR . But on passing parameter kv it gives me kv1 and not kv4(since it's in LOOP) . Can't figure out what's wrong ?
FUNCTION FUN_GENERATE_SNAME (p_name VARCHAR2)
RETURN VARCHAR2
IS
vl_sname VARCHAR2 (15);
n_cnt NUMBER := 1;
vl_sub NUMBER;
CURSOR c1 (vl_sname VARCHAR2)
IS
SELECT a.repsname, a.repcode
FROM MMSTREPHDR a
WHERE TRIM (UPPER (a.repsname)) = TRIM (UPPER (vl_sname));
BEGIN
vl_sname := TRIM (SUBSTR (p_name, 1, 15));
FOR i IN c1 (vl_sname)
LOOP
vl_sub := LENGTH (TO_CHAR (n_cnt));
vl_sname := SUBSTR (vl_sname, 1, (15 - vl_sub)) || n_cnt;
n_cnt := n_cnt + 1;
END LOOP;
RETURN vl_sname;
EXCEPTION
WHEN OTHERS
THEN
RETURN vl_sname;
END fun_generate_sname;
Your starting parameter is 'kv'. This is what you pass to the cursor. Consequently your cursor will select one row , the row where MMSTREPHD.repsname = 'kv'.
So your loop logic will be executed once. So cnt = . Hencevl_sname` becomes 'kv1', which is the value you get when the loop exits cleanly.
The cleanest way to fix this would be to admit that MMSTREPHD.repsname is a smart key, consisting of two elements: a subsystem name and a report number. Splitting the column into two columns would make it a cinch to find the next report number for a given sub-system. You can even retain the composite value as a virtual column (11g or later), or maintain it with triggers which sucks a bit.
Otherwise:
select concat(p_name
, trim(to_char(max(to_number(nvl(replace(repsname,p_name),'0')))+1)) )
into vl_sname
from MMSTREPHD
where repsname like p_name||'%'
Caveat - I haven't tested this (yet) so ths brackets may not pair up correctly.
You are concatenating the in below statement as
vl_sname := SUBSTR (vl_sname, 1, (15 - vl_sub)) || n_cnt;
here n_cnt is given initial value to it as 1 in code above thats why its fetching you value as KV1.You should keep it null and after statement should increment it by 1 in loop. Hope will be usefull
try this function
function FUN_GENERATE_SNAME(p_name varchar2) return varchar2 is
l_idx number;
l_name_ln := length(trim(p_name));
begin
select max(substr(trim(UPPER(a.repsname)), 1, -length(trim(UPPER(a.repsname)) + l_name_ln))
into l_idx
from MMSTREPHDR a
where substr(trim(UPPER(a.repsname)), 1, l_name_ln) = trim(UPPER(vl_sname));
if l_idx is null then
-- mean name is unique
return vl_sname;
else
return vl_sname ||(l_idx + 1);
end if;
exception
when others then
return vl_sname;
end fun_generate_sname;