I found this code at this website and is working well in SQL SERVER
CREATE FUNCTION [dbo].[CalculaDigitoRut]
(
#rut int
)
RETURNS char(1)
AS
BEGIN
DECLARE #digito int
DECLARE #contador int
DECLARE #multiplo int
DECLARE #acumulador int
DECLARE #ret char
set #contador = 2
set #acumulador = 0
while (#rut <> 0)
begin
set #multiplo = (#Rut % 10) * #contador
set #acumulador = #acumulador + #multiplo
set #rut = #rut / 10
set #contador = #contador + 1
if (#contador = 8)
set #contador = 2
end
set #digito = 11 - (#acumulador % 11)
if (#digito = 10)
return ('K')
if (#digito = 11)
return ('0')
return (#digito)
END
Expected result
RUT -- validation_digit
10214564 K
3781561 6
3433444 7
3066256 3
I tried to adapt it to the MariaDB (10.3.30-MariaDB) syntax and after a few tries there weren't syntax erros but the function isn't returning the expected result.
Heres the new code for MariaDB
CREATE FUNCTION Calculate_national_id_verifier (rut INT)
RETURNS CHAR(1)
BEGIN
DECLARE digito INT;
DECLARE contador INT;
DECLARE multiplo INT;
DECLARE acumulador INT;
DECLARE ret CHAR;
SET contador = 2;
SET acumulador = 0;
WHILE rut <> 0 DO
SET multiplo = (rut % 10) * contador;
SET acumulador = acumulador + multiplo;
SET rut = rut / 10;
SET contador = contador + 1; -- 3
IF (contador = 8) THEN
SET contador = 2;
END IF;
END WHILE;
SET digito = 11 - (acumulador % 11);
IF (digito = 10) THEN
RETURN ('K');
ELSEIF (digito = 11) THEN
RETURN ('0');
ELSE
RETURN (digito);
END IF;
END;
This problem seems very dumb, I've been comparing it for hours and I can't find where is the problem and why there is a difference between both functions.
I'm using DBeaver 21.3.5 as a client to run the queries.
Here are the actual results
RUT -- validation_digit
10214564 6
3781561 K
3433444 7
3066256 5
This is my firts time asking something so sorry if I made a mistake. I couldn't find the answer.
LOCAL in_txt
in_txt='mohammad'
txt_len=LEN(in_txt)
rev_txt=''
FOR ii=1 TO txt_len
w_chr=SUBSTR(in_txt,ii,1)
w_asc=IIF(ASC(w_chr)=32,32,ASC(w_chr)+1)
* ?' #'+w_chr+'='+CHR(w_asc)+'*'+STR(ASC(w_chr))+'>'+STR(ASC(w_chr)+1)+'*'+CHR(ASC(w_chr))+'>'+CHR(ASC(w_chr)+1)
rev_txt=rev_txt+ CHR(w_asc)
ENDFOR
return rev_txt
I understand solve my problem * character is using comment in foxpro.
Here is the equivalent SQL
DECLARE #in_txt varchar(100)
DECLARE #txt_Len int
DECLARE #rev_txt varchar(100)=''
DECLARE #i int=0
DECLARE #w_chr char(1)
DECLARE #w_asc int
SET #in_txt='mohammad'
SET #txt_len=LEN(#in_txt)
WHILE #i < #txt_len
BEGIN
SET #i = #i + 1
SET #w_chr=SUBSTRING(#in_txt,#i,1)
SET #w_asc = ASCII(#w_chr)
IF #w_asc <> 32 SET #w_asc=#w_asc+1
PRINT '#'+#w_chr+'='+CHAR(#w_asc)+'*'+STR(ASCII(#w_chr))+'>'+STR(ASCII(#w_chr)+1)+'*'+CHAR(ASCII(#w_chr))+'>'+CHAR(ASCII(#w_chr)+1)
SET #rev_txt = #rev_txt + CHAR(#w_asc)
END
PRINT #rev_Txt
It appears to be a very simple encryption of the text. The Print statement in the loop (which is the equivalent of the ? in Foxpro) is the debugger statement
It is a badly written piece of VFP code. * marks the line as a comment and basically it is replacing all the characters in the string except the char 32 (SPACE) with the next in ASCII chart. That is not something you could easily convert in MS SQL though, because in VFP a string could have any character in it including char(0). Assuming you don't have char 0:
Declare #in_txt varbinary(MAX);
DECLARE #rev_txt varbinary(MAX);
set #in_txt = CAST('mohammad' AS VARBINARY(MAX));
declare #i int;
declare #w_char int;
set #i = 1;
SET #rev_txt = CAST('' AS VARBINARY(MAX));
WHILE #i <= Len(#in_txt)
Begin
set #w_char = ASCII(SUBSTRING(#in_txt,#i,1));
SET #rev_txt = #rev_txt +
CAST(#w_char + case when #w_char = 32 then 0 else 1 end AS BINARY(1));
SET #i = #i +1;
END
SELECT #rev_txt;
enter image description here
I wanted to draw a dimaond like this in Sql. Can you help me please
I have find this code in queryexamples.com there are to many examples like this
Declare #i int = 0,#max int = 11,#c int = 1
While (#i<#max)
Begin
Print space(abs((#max-#c)/2))+ Replicate('X',#c)
Set #i = #i + 1
if(#i>(#max/2))
Set #c -= 2
else
Set #c += 2
End
I'm using #subno as Input. And I had to find the odd and even numbers. 16 is not a fix and it can be any other number. My question is how to find the odd and even number of my input?
Lastly, subno includes ( . ) dot at any position. e.g 123456.789123 I need to find the odd and even number of "123456" and the odd and even number of "789123" the dot ( . ) is the separator.
Once you find the odd for the left side, sum them up together. Once you find the even number for left side sum it up as well and then add the total odd to the total even values. That goes the same for the right side eg "789123".
Please help me. this is my 2nd week of trying to find the solution. Once you find all the total values for each side, multiply them together. example "123456" - total value of odd and even * the "789123" total value of odd and even.
It is for the the check digit validation. Validating the subscriber number. after validating through the calculation it should match the calculated reference number to the valid check digit number. It's the business rule. Kind of algorithm
create procedure ProcedureName
(#subno VARCHAR(16), --Input the 16 subscriber number
#result INT OUT,
)
as
begin
IF(LEN(#subno) <> 16)
SET #result = 1 -- INVALID RESULT
ELSE
IF(#subno % 2 = 0)
SET #result = #subno - even numbers
ELSE
SET #result = #subno --odd numbers
end
Please see below my sample work
-- this is the sample
create procedure ProcedureName
(
#subno VARCHAR(20), --Subscriber no
#result INT OUT, --result is invalid for 1, valid for 0
#payamt int
)
as
DECLARE #WA VARCHAR(2)
DECLARE #Weights varchar(9)
DECLARE #I INT
DECLARE #WD INT
DECLARE #WP INT
DECLARE #A INT
DECLARE #B INT
DECLARE #R INT
DECLARE #WR INT
SET #WR = 0
SET #R = 0
SET #A = 0
SET #B = 0
SET #WP = 0
SET #I = 0
BEGIN
IF (LEN(#subNo) = 7) AND (SUBSTRING(#subno,1,1) = '2') OR (SUBSTRING(#subno,1,1) = '9')
BEGIN
SET #result = 0 --VALID
END
ELSE IF(LEN(#subno) = 8) AND (SUBSTRING(#subno,1,1) = '2') OR
(SUBSTRING(#subno,1,1) = '9')
BEGIN
SET #result = 0 --VALID
END
ELSE IF(LEN(#subno) = 9)
BEGIN
SET #WA = SUBSTRING(#subno,1,2)
IF(#WA = '65')
set #result = 1 -- INVALID
else
BEGIN
SET #Weights = '12121212'
SET #WA = SUBSTRING(#subno,9,1)
SET #WD = 0
SET #I = 1
WHILE #I<9
BEGIN
SET #WP = cast(SUBSTRING(#Weights, #I,1)as int) * cast(SUBSTRING(#subno, #I, 1) as int)
IF(#WP > 9)
BEGIN
SET #A = SUBSTRING(CAST(#WP AS VARCHAR),1,1)
SET #B = SUBSTRING(CAST(#WP AS VARCHAR),2,1)
SET #WP = CAST(#A AS INT) + CAST(#B AS INT)
END
SET #WD = #WP + #WD
SET #I = #I + 1
END
SET #R = #WD % 10
IF(#R <> 0)
SET #WR = 10 - #R
ELSE
SET #WR = #R
IF(#WR <> CAST(#WA AS INT))
BEGIN
SET #result = 1 -- INVALID
END
ELSE
BEGIN
SET #result = 0 -- VALID
END
END
END
ELSE IF (LEN(#subno) = 10)
BEGIN
SET #I =1
SET #WD = 0
SET #Weights = '121212121'
SET #WA = SUBSTRING(#subno,10,1)
WHILE(#I < 10)
BEGIN
SET #WP = CAST(SUBSTRING(#Weights, #I, 1)AS INT) * CAST(SUBSTRING(#subno, #I, 1) AS INT)
IF(#WP > 9)
BEGIN
SET #A = SUBSTRING(CAST(#WP AS VARCHAR),1,1)
SET #B = SUBSTRING(CAST(#WP AS VARCHAR),2,1)
SET #WP = CAST(#A AS INT) + CAST(#B AS INT)
END
SET #WD = #WP + #WD
SET #I = #I + 1
END
SET #R = #WD % 10
IF(#R <> 0)
SET #WR = 10 - #R
ELSE
SET #WR = #R
IF (#WR<> #WA)
BEGIN
SET #result = 1 -- INVALID
END
ELSE
BEGIN
SET #result = 0 -- VALID
END
END
ELSE
SET #result = 1 -- INVALID
END
Split the values which u get . Then iterate iver each side and add them. Please see the sample below.
declare #v varchar (16) , #num1 varchar(20) , #num2 varchar(20)
set #v = '1234567.78906656'
select #num1 = substring(#v,0,charindex('.',#v))
select #num2 = substring(#v,charindex('.',#v)+1,len(#v))
--select #num1 = convert(int, substring(#v,0,charindex('.',#v)))
--select #num2 = substring(#v,charindex('.',#v)+1,len(#v))
declare #index int = 1 ,#len INT , #char CHAR
declare #TotalOddL int = 0
declare #TotalEvenL int = 0
DECLARE #FullTotL INT = 0
declare #TotalOddR int = 0
declare #TotalEvenR int = 0
DECLARE #FullTotR INT = 0
DECLARE #TEMP INT
set #len= LEN(#num1)
WHILE #index <= #len
BEGIN
set #char = SUBSTRING(#num1, #index, 1)
SET #TEMP = cast(#char as int)
IF(#TEMP % 2 = 0)
SET #TotalEvenL = #TotalEvenL + #char
else
SET #TotalOddL = #TotalOddL + #char
SET #FullTotL = #TotalEvenL + #TotalOddL
SET #index= #index+ 1
END
Select 'LeftSide total' , #FullTotL
Select 'Left Side odd' , #TotalOddL
Select 'Left Side Even' , #TotalEvenL
SET #index = 1
set #len= LEN(#num2)
WHILE #index <= #len
BEGIN
set #char = SUBSTRING(#num2, #index, 1)
SET #TEMP = cast(#char as int)
IF(#TEMP % 2 = 0)
SET #TotalEvenR= #TotalEvenR + #char
else
SET #TotalOddR = #TotalOddR + #char
SET #FullTotR = #TotalEvenR + #TotalOddR
SET #index= #index+ 1
END
select 'TotalRSide' , #FullTotR
select 'RsideOdd' , #TotalOddR
select 'RSideEven' , #TotalEvenR
select 'Multiplied value' , #FullTotR * #FullTotL
create or replace procedure prc_even_odd(i_number in number, o_result out varchar2)
as
begin
if (mod(i_number,2) = 0) then
o_result := 'EVEN';
else
o_result := 'ODD';
end prc_even;
Need help to create a function that returns TRUE or FALSE. TRUE - if type 1 or 3 words (like '__hello_', '_hello', '_hello my frend' - spaces should be cut), if the condition is not fulfilled FALSE
CREATE FUNCTION dbo.nazvFac(#f nvarchar(30))
RETURNS bit
AS
BEGIN
DECLARE #l int = 1, #s nvarchar(30), #i int = 0, #b bit
WHILE LTRIM(RTRIM(LEN(#f))) >= #l --error here, but I do not know how to fix it
BEGIN
SET #s = SUBSTRING(#f, #l, 1)
IF #s BETWEEN 'А' AND 'я'
SET #l += 1
ELSE IF #s = ' '
BEGIN
SET #l -= 1
SET #s = SUBSTRING(#f, #l, 1)
SET #s = RTRIM(#s)
SET #l += 2
SET #i += 1
END
ELSE
BREAK
END
IF #i = 0 OR #i = 2
SET #b = 'TRUE'
ELSE
SET #b = 'FALSE'
RETURN #b
END
GO
WHILE LTRIM(RTRIM(LEN(#f))) >= #l --error here, but I do not know how to fix it
LEN() returns an int, which you are then passing to a string function: RTRIM().
You want to return TRUE only if there are one or three words? This should do it:
CREATE FUNCTION dbo.nazvFac(#f NVARCHAR(30))
RETURNS BIT
AS
BEGIN
DECLARE #l INT, #b BIT
SET #l = LEN(#f) - LEN(REPLACE(#f, ' ', '')) + 1
IF #l == 1 OR #l == 3
SET #b = 'TRUE'
ELSE
SET #b = 'FALSE'
RETURN #b
END
Also, JC. is right about the len() error.
You should trim the string and then check the length.
CREATE FUNCTION dbo.nazvFac(#f NVARCHAR(30))
RETURNS BIT
AS
BEGIN
DECLARE #l INT = 1, #s NVARCHAR(30), #i INT = 0, #b BIT
WHILE LEN(LTRIM(RTRIM(#f))) >= #l --I think youn need to trim the string and then check length
BEGIN
SET #s = SUBSTRING(#f, #l, 1)
IF #s BETWEEN 'А' AND 'я'
SET #l += 1
ELSE IF #s = ' '
BEGIN
SET #l -= 1
SET #s = SUBSTRING(#f, #l, 1)
SET #s = RTRIM(#s)
SET #l += 2
SET #i += 1
END
ELSE
BREAK
END
IF #i = 0 OR #i = 2
SET #b = 'TRUE'
ELSE
SET #b = 'FALSE'
RETURN #b
END
GO