Given a number, how do you convert that to letters?
You can just translate your existing logic into a t-sql scalar function, like so:
CREATE FUNCTION dbo.fnColumnNameFromIndex(#i int)
RETURNS varchar(3)
AS
BEGIN
DECLARE #dividend int, #letters varchar(3), #modulo int
SET #dividend = #i
SET #letters = ''
WHILE #dividend > 0
BEGIN
SET #modulo = (#dividend - 1) % 26
SET #letters = CHAR(65 + #modulo) + #letters
SET #dividend = CONVERT(int, (#dividend - #modulo) / 26 )
END
RETURN #letters;
END
GO
Call it like this:
SELECT dbo.fnColumnNameFromIndex(2000)
...admittedly not set logic, but I don't see this as a set logic problem.
Related
I need to create a consecutive sequence of varchar(5) (always 5 chars only) code starting from PREVIOUS code.
For example
'00000', '00001', '00002'...'00009', '0000A', '0000B'...'0000Z', '00010','00011'...'ZZZZZ'.
So if I have #PREVIOUS_CODE = '00000', #NEXT_CODE will be '00001'.
If I have #PREVIOUS_CODE = '00009', #NEXT_CODE will be '0000A'
If I have #PREVIOUS_CODE = '0000Z', #NEXT_CODE will be '00010'
So I need something like that
USE [DATABASE]
GO
CREATE PROCEDURE [dbo].[spGetNextCode]
#PREVIOUS_CODE VARCHAR(5)
AS
DECLARE #NEXT_CODE VARCHAR(5)
DO STUFF
...
SELECT #NEXT_CODE AS NEXT_CODE
GO
Any Help?
Just keep an integer counter in the same table and convert it. I'm using the following SQL Server function in one of my applications:
CREATE FUNCTION [dbo].[GetAlphanumericCode]
(
#number BIGINT,
#leadingzeroes INT = 0
)
RETURNS varchar(255)
AS
BEGIN
DECLARE #charPool varchar(36)
SET #charPool = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ'
DECLARE #result varchar(255)
IF #number < 0
RETURN ''
IF #number = 0
SET #result = '0'
ELSE BEGIN
SET #result = ''
WHILE (#number > 0)
BEGIN
SET #result = substring(#charPool, #number % 36 + 1, 1) + #result
SET #number = #number / 36
END
END
IF #leadingzeroes > 0 AND len(#result) < #leadingzeroes
SET #result = right(replicate('0', #leadingzeroes) + #result, #leadingzeroes)
RETURN #result
END
It should be a trivial task to rewrite it as a stored procedure
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
I need to pull a specific substring from a string of the form:
foo=abc;bar=def;baz=ghi
For example, how would I get the value of "bar" from that string?
You can use charindex and substring. For example, to search for the value of "baz":
declare #str varchar(128)
set #str = 'foo=abc;bar=def;baz=ghi'
-- Make sure #str starts and ends with a ;
set #str = ';' + #str + ';'
select substring(#str,
charindex(';baz=',#str) + len(';baz='),
charindex('=',#str,charindex(';baz=',#str)) - charindex(';baz=',#str) - 1)
Or for the value of "foo" at the start of the string:
select substring(#str,
charindex(';foo=',#str) + len(';foo='),
charindex('=',#str,charindex(';foo=',#str)) - charindex(';foo=',#str) - 1)
Here's a UDF to accomplish this (more readable version inspired by BlackTigerX's answer):
create function dbo.FindValueInString(
#search varchar(256),
#name varchar(30))
returns varchar(30)
as
begin
declare #name_start int
declare #name_length int
declare #value_start int
declare #value_end int
set #search = ';' + #search
set #name_start = charindex(';' + #name + '=',#search)
if #name_start = 0
return NULL
set #name_length = len(';' + #name + '=')
set #value_start = #name_start + #name_length
set #value_end = charindex(';', #search, #value_start)
return substring(#search, #value_start, #value_end - #value_start)
end
As you can see, this isn't easy in Sql Server :) Better do this in the client language, or normalize your database so the substrings go in their own columns.
I have a generalized solution that works for this problem:
CREATE FUNCTION [dbo].[fn_StringBetween]
(
#BaseString varchar(max),
#StringDelim1 varchar(max),
#StringDelim2 varchar(max)
)
RETURNS varchar(max)
AS
BEGIN
DECLARE #at1 int
DECLARE #at2 int
DECLARE #rtrn varchar(max)
SET #at1 = CHARINDEX(#StringDelim1, #BaseString)
IF #at1 > 0
BEGIN
SET #rtrn = SUBSTRING(#BaseString, #at1
+ LEN(#StringDelim1), LEN(#BaseString) - #at1)
SET #at2 = CHARINDEX(#StringDelim2, #rtrn)
IF #at2 > 0
SET #rtrn = LEFT(#rtrn, #at2 - 1)
END
RETURN #rtrn
END
so if you run (just wrap your original string to be searched with ';' at beginning and end):
PRINT dbo.fn_StringBetween(';foo=abc;bar=def;baz=ghi;', ';bar=', ';')
you will get 'def' returned.
look into the PATINDEX function. It has wildcard matching which should help you..
you can use this function
alter function FindValue(#txt varchar(200), #find varchar(200))
returns varchar(200)
as
begin
declare
#firstPos int,
#lastPos int
select #firstPos = charindex(#find, #txt), #lastPos = charindex(';', #txt, #firstPos+5)
select #lastPos = len(#txt)+1 where #lastPos = 0
return substring(#txt, #firstPos+len(#find)+1, #lastPos-#firstPos-len(#find)-1)
end
select dbo.FindValue('foo=abc;bar=def;baz=ghi', 'bar')
update: was not using the length of #find
this is assuming that the string will have the same string format just substitute the column name for the 'foo=abc;bar=def;baz=ghi'
select substring('foo=abc;bar=def;baz=ghi',patindex('%bar=%','foo=abc;bar=def;baz=ghi')+4, len('foo=abc;bar=def;baz=ghi')-patindex('%;baz=%','foo=abc;bar=def;baz=ghi')-4)
This can achieve in simple way
DECLARE #str VARCHAR(30)
DECLARE #start INT
SET #str='foo=abc;bar=def;baz=ghi'
SET #start=CHARINDEX('bar',#str)
PRINT SUBSTRING(#str,#start,3)
I have a string 'some.file.name',I want to grab 'some.file'.
To do that,I need to find the last occurrence of '.' in a string.
My solution is :
declare #someStr varchar(20)
declare #reversedStr varchar(20)
declare #index int
set #someStr = '001.002.003'
set #reversedStr = reverse(#someStr)
set #index = len(#someStr) - charindex('.',#reversedStr)
select left(#someStr,#index)
Well,isn't it too complicated?I was just intented to using 'some.file' in a where-clause.
Anyone has a good idea?
What do you need to do with it?? Do you need to grab the characters after the last occurence of a given delimiter?
If so: reverse the string and search using the normal CHARINDEX:
declare #test varchar(100)
set #test = 'some.file.name'
declare #reversed varchar(100)
set #reversed = REVERSE(#test)
select
REVERSE(SUBSTRING(#reversed, CHARINDEX('.', #reversed)+1, 100))
You'll get back "some.file" - the characters up to the last "." in the original file name.
There's no "LASTCHARINDEX" or anything like that in SQL Server directly. What you might consider doing in SQL Server 2005 and up is great a .NET extension library and deploy it as an assembly into SQL Server - T-SQL is not very strong with string manipulation, whereas .NET really is.
A very simple way is:
SELECT
RIGHT(#str, CHARINDEX('.', REVERSE(#str)) - 1)
This will also work:
DECLARE
#test VARCHAR(100)
SET #test = 'some.file.name'
SELECT
LEFT(#test, LEN(#test) - CHARINDEX('.', REVERSE(#test)))
Take one ')'
declare #test varchar(100)
set #test = 'some.file.name'
select left(#test,charindex('.',#test)+charindex('.',#test)-1)
CREATE FUNCTION [dbo].[Instr] (
-------------------------------------------------------------------------------------------------
-- Name: [dbo].[Instr]
-- Purpose: Find The Nth Value Within A String
-------------------------------------------------------------------------------------------------
-- Revisions:
-- 25-FEB-2011 - HESSR - Initial Revision
-------------------------------------------------------------------------------------------------
-- Parameters:
-- 1) #in_FindString - NVARCHAR(MAX) - INPUT - Input Find String
-- 2) #in_String - NVARCHAR(MAX) - INPUT - Input String
-- 3) #in_StartPos - SMALLINT - INPUT - Position In The String To Start Looking From
-- (If Start Position Is Negative, Search Begins At The End Of The String)
-- (Negative 1 Starts At End Position 1, Negative 3 Starts At End Position Minus 2)
-- 4) #in_Nth - SMALLINT - INPUT - Nth Occurrence To Find The Location For
-------------------------------------------------------------------------------------------------
-- Returns: SMALLINT - Position Of String Segment (Not Found = 0)
-------------------------------------------------------------------------------------------------
#in_FindString NVARCHAR(MAX),
#in_String NVARCHAR(MAX),
#in_StartPos SMALLINT = NULL,
#in_Nth SMALLINT = NULL
)
RETURNS SMALLINT
AS
BEGIN
DECLARE #loc_FindString NVARCHAR(MAX);
DECLARE #loc_String NVARCHAR(MAX);
DECLARE #loc_Position SMALLINT;
DECLARE #loc_StartPos SMALLINT;
DECLARE #loc_Nth SMALLINT;
DECLARE #loc_Idx SMALLINT;
DECLARE #loc_FindLength SMALLINT;
DECLARE #loc_Length SMALLINT;
SET #loc_FindString = #in_FindString;
SET #loc_String = #in_String;
SET #loc_Nth = ISNULL(ABS(#in_Nth), 1);
SET #loc_FindLength = LEN(#loc_FindString+N'.') - 1;
SET #loc_Length = LEN(#loc_String+N'.') - 1;
SET #loc_StartPos = ISNULL(#in_StartPos, 1);
SET #loc_Idx = 0;
IF (#loc_StartPos = ABS(#loc_StartPos))
BEGIN
WHILE (#loc_Idx < #loc_Nth)
BEGIN
SET #loc_Position = CHARINDEX(#loc_FindString,#loc_String,#loc_StartPos);
IF (#loc_Position > 0)
SET #loc_StartPos = #loc_Position + #loc_FindLength
ELSE
SET #loc_Idx = #loc_Nth;
SET #loc_Idx = #loc_Idx + 1;
END;
END
ELSE
BEGIN
SET #loc_StartPos = ABS(#loc_StartPos);
SET #loc_FindString = REVERSE(#in_FindString);
SET #loc_String = REVERSE(#in_String);
WHILE (#loc_Idx < #loc_Nth)
BEGIN
SET #loc_Position = CHARINDEX(#loc_FindString,#loc_String,#loc_StartPos);
IF (#loc_Position > 0)
SET #loc_StartPos = #loc_Position + #loc_FindLength
ELSE
SET #loc_Idx = #loc_Nth;
SET #loc_Idx = #loc_Idx + 1;
END;
IF (#loc_Position > 0)
SET #loc_Position = #loc_Length - #loc_Position + (1 - #loc_FindLength) + 1;
END;
RETURN (#loc_Position);
END;
GO
Here is a shorter version
DECLARE #someStr varchar(20)
set #someStr = '001.002.003'
SELECT REVERSE(Substring(REVERSE(#someStr),CHARINDEX('.', REVERSE(#someStr))+1,20))