We are using WHILE loop in SQL Server 2008 instead of Cursor. Now I want to find which method is best to write WHILE loop in my procedure.
Method 1 (Using BREAK Keyword):
DECLARE #V_Counter INT = 1;
WHILE (1 = 1)
BEGIN
PRINT #V_Counter;
SET #V_Counter = #V_Counter + 1;
IF #V_Counter = 4
BEGIN
BREAK;
END
END
Method 2 (Using BOOL VARIABLE):
DECLARE #V_Counter INT = 1, #V_CloseLoop TINYINT = 1;
WHILE (#V_CloseLoop = 1)
BEGIN
PRINT #V_Counter;
SET #V_Counter = #V_Counter + 1;
IF #V_Counter = 4
BEGIN
SET #V_CloseLoop = 0;
END
END
My questions are:
Which method I have to use or both are same?
Is there any other method which I can use?
Thanks in advance...
Which method I have to use or both are same?
No, both are not same.
setting the variable to zero will still execute the lines after that, but nothing else in the loop will be executed after the break.
DECLARE #V_Counter INT = 1;
WHILE (1 = 1)
BEGIN
PRINT #V_Counter;
SET #V_Counter = #V_Counter + 1;
IF #V_Counter = 4
BEGIN
BREAK;
END
PRINT 'STACKOVERFLOW' // this is not executed.
END
DECLARE #V_Counter INT = 1, #V_CloseLoop TINYINT = 1;
WHILE (#V_CloseLoop = 1)
BEGIN
PRINT #V_Counter;
SET #V_Counter = #V_Counter + 1;
IF #V_Counter = 4
BEGIN
SET #V_CloseLoop = 0;
END
PRINT 'STACKOVERFLOW' // this is executed.
END
Using break will always be a clean approach, than setting the variable to zero, as the code will be easy to maintain when we use break than the other way around.
A breakis considered as "bad programming style". So better to use a variable. Except if you have a good reason.
But try to avoid while loops at all. In SQL Server, you can very often replace loops with good SQL statements which do everything in one rush.
a) What are you doing here at all? I see a counter being counted up to 4 ... ;)
b) You could rewrite it like this:
DECLARE #V_Counter INT = 1;
WHILE (#V_Counter < 4)
BEGIN
PRINT #V_Counter;
SET #V_Counter = #V_Counter + 1;
END
Related
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.
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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 want to post a solution to a interesting problem I was facing in T-SQL.
The problem:
Compare two string fields based on a percent match.
In addition, the two strings may have the words in them translocated.
For example: "Joni Bravo" and "Bravo Joni". These two strings should return a match of 100%, which means that position is not relevant. Few more things worth noting are that this code is made to compare strings that have space as delimiter in them. If the first string doesnt have space the match is set to 100% without actual check. This was not developed, because the strings this function is ment to compare always contain two or more words. Also, it is written on MS SQL Server 2017 if that mathers.
So here is the solution, hope this helps anyone :)
gl
/****** Object: UserDefinedFunction [dbo].[STRCOMP] Script Date: 29/03/2018 15:31:45 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[STRCOMP] (
-- Add the parameters for the function here
#name_1 varchar(255),#name_2 varchar(255)
)
RETURNS float
AS
BEGIN
-- Declare the return variable and any needed variable here
declare #p int = 0;
declare #c int = 0;
declare #br int = 0;
declare #p_temp int = 0;
declare #emergency_stop int = 0;
declare #fixer int = 0;
declare #table1_temp table (
row_id int identity(1,1),
str1 varchar (255));
declare #table2_temp table (
row_Id int identity(1,1),
str2 varchar (255));
declare #n int = 1;
declare #count int = 1;
declare #result int = 0;
declare #total_result float = 0;
declare #result_temp int = 0;
declare #variable float = 0.0;
--clean the two strings from unwanted symbols and numbers
set #name_1 = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(#name_1,'!',''),' ',' '),'1',''),'2',''),'3',''),'4',''),'5',''),'0',''),'6',''),'7',''),'8',''),'9','');
set #name_2 = REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(#name_2,'!',''),' ',' '),'1',''),'2',''),'3',''),'4',''),'5',''),'0',''),'6',''),'7',''),'8',''),'9','');
--check if the first string has more than one words inside. If the string does
--not have more than one words, return 100%
set #c = charindex(' ',substring(#name_1,#p,len(#name_1)));
IF(#c = 0)
BEGIN
RETURN 100.00
END;
--main logic of the operation. This is based on sound indexing and comparing the
--outcome. This loops through the string whole words and determines their soundex
--code and then compares it one against the other to produce a definitive number --showing the raw match between the two strings #name_1 and #name_2.
WHILE (#br != 2 or #emergency_stop = 20)
BEGIN
insert into #table1_temp(str1)
select substring (#name_1,#p,#c);
set #p = len(substring (#name_1,#p,#c))+2;
set #p = #p + #p_temp - #fixer;
set #p_temp = #p;
set #c = CASE WHEN charindex(' ',substring(#name_1,#p,len(#name_1))) = 0 THEN len(#name_1) ELSE charindex(' ',substring(#name_1,#p,len(#name_1))) END;
set #fixer = 1;
set #br = CASE WHEN charindex(' ',substring(#name_1,#p,len(#name_1))) = 0 THEN #br + 1 ELSE 0 END;
set #emergency_stop = #emergency_stop +1;
END;
set #p = 0;
set #br = 0;
set #emergency_stop = 0;
set #fixer = 0;
set #p_temp = 0;
set #c = charindex(' ',substring(#name_2,#p,len(#name_2)));
WHILE (#br != 2 or #emergency_stop = 20)
BEGIN
insert into #table2_temp(str2)
select substring (#name_2,#p,#c);
set #p = len(substring (#name_2,#p,#c))+2;
set #p = #p + #p_temp - #fixer;
set #p_temp = #p;
set #c = CASE WHEN charindex(' ',substring(#name_2,#p,len(#name_2))) = 0 THEN len(#name_2) ELSE charindex(' ',substring(#name_2,#p,len(#name_2))) END;
set #fixer = 1;
set #br = CASE WHEN charindex(' ',substring(#name_2,#p,len(#name_2))) = 0 THEN #br + 1 ELSE 0 END;
set #emergency_stop = #emergency_stop +1;
END;
WHILE((select str1 from #table1_temp where row_id = #n) is not null)
BEGIN
set #count = 1;
set #result = 0;
WHILE((select str2 from #table2_temp where row_id = #count) is not null)
BEGIN
set #result_temp = DIFFERENCE((select str1 from #table1_temp where row_id = #n),(select str2 from #table2_temp where row_id = #count));
IF(#result_temp > #result)
BEGIN
set #result = #result_temp;
END;
set #count = #count + 1;
END;
set #total_result = #total_result + #result;
set #n = #n + 1;
END;
--gather the results and transform them in a percent match.
set #variable = (select #total_result / (select max(row_count) from (
select max(row_id) as row_count from #table1_temp
union
select max(row_id) as row_count from #table2_temp) a));
RETURN #variable/4 * 100;
END
GO
PS: I decided to write it in a user-defined function just for the needs of my project.
This was my first attempt at a SQL function. I wrote it in VB and it works like a charm. When I translated it to SQL Server, it returns not what I expect. What the function is intended to do is to return a percentage match of two strings.
How I expected it to function is this:
accept two strings, compare the two, and based on my rating values, return a percentage of the match value...matchscore/max possiblescore
The length of the larger string is multiplied by 3. This is the max possiblescore.
Go character by character of the first string and find that character in the second string.
If the character is found in the same position in the second string, add three to the matchscore and move on to the next letter.
If the character is found in the second word, but not in the same position, add one to match score and move on to the next character.
If the character is not found in the second string, add nothing and move on to the next character.
Divide the matchscore by the max possiblescore. This returns a decimal value. I read RETURN only returns an integer, so I multiplied the division result by 100.
An example of what I expected is I compare "CAT" to "CART". My expected return is 7/12...0.58. Instead, I get 0. If I compare "CAT" to "CAT", I expect 9/9...1.00. Instead, I get 2.
(Note from 9/17/2014: I appreciate your input. I used what you suggested, and did one more major change, that doesn't affect what I asked about, other than getting the correct final answer is that I got rid of the second While Loop. Instead, I search for #strLetter in #strWord2. If it is found, then, I look to see if it is in the same position in #strWord2 as #strWord1. If it is, then I add 3, if not, I add 1. This sped up the function and made the count accurate.
Here is the code:
CREATE FUNCTION [dbo].[CompareWords]
(#strWord1 VARCHAR(2000), #strWord2 VARCHAR(2000))
RETURNS DECIMAL
AS
BEGIN
SET #strWord1 = UPPER(#strWord1)
SET #strWord2 = UPPER(#strWord2)
DECLARE #intLength INT
IF LEN(#strWord1) >= LEN(#strWord2)
BEGIN
SET #intLength = LEN(#strWord1)
END
ELSE
BEGIN
SET #intLength = LEN(#strWord2)
END
DECLARE #iWordLoop1 INT
DECLARE #iWordLoop2 INT
DECLARE #intWordLoop2 INT
DECLARE #intWordScore INT
DECLARE #intLetterScore INT
SET #intWordScore = 0
SET #intWordLoop2 = Len(#strWord2)
DECLARE #strLetter VARCHAR(1000)
DECLARE #count1 INT
SET #count1 = 0
SET #iWordLoop1 = Len(#strWord1)
WHILE (#count1 < #iWordLoop1)
BEGIN
SET #strLetter = SUBSTRING(#strWord1, #count1+1, 1)
SET #intLetterScore = 0
DECLARE #count2 INT
SET #count2 = 0
SET #iWordLoop2 = Len(#strWord2)
WHILE (#count2 < #iWordLoop2)
BEGIN
If #strLetter = SUBSTRING(#strWord2, #count2+1, 1)
BEGIN
If #iWordLoop1 = #iWordLoop2
BEGIN
SET #intLetterScore = 3
SET #iWordLoop2 = Len(#strWord2)
END
ELSE
BEGIN
SET #intLetterScore = 1
END
END
SET #intWordScore = #intWordScore + #intLetterScore
SET #count2 = (#count2 + 1)
END
SET #count1 = (#count1 + 1)
END
DECLARE #sinScore DEC
SET #sinScore = (#intWordScore / (3 * #intLength)) * 100
RETURN #sinSCore
END;
The most significant changes I made were to
reset the intLetterScore to 0 after it's been used in the intWordScore calculation. Without it being reset, the same value was being used each time the inner loop and the character was not matched.
move the multiplication by 100 into the
brackets in the calculation of sinScore.
As referred to in a previous post, because you are doing integer multiplication, the decimal portion is truncated from the calculation. By growing the wordScore by a factor of 100, it is much more likely to be larger than the length and yield a result which is non-zero.
Multiplying outside the brackets has the multiplies the integer result of the division score by length. If this answer is already zero, then the multiplication result is also zero.
Other changes I made are commented in the code: the variable intWordLoop2 has no effect on the calculation and can be removed; strLetter can be declared as a Char(1) instead of VarChar(1000).
CREATE FUNCTION [dbo].[CompareWords]
(#strWord1 VARCHAR(2000), #strWord2 VARCHAR(2000))
RETURNS DECIMAL
AS
BEGIN
SET #strWord1 = UPPER(#strWord1)
SET #strWord2 = UPPER(#strWord2)
--Set #intLength (maxLength as len of word1 or word2)
DECLARE #intLength INT --maxLength
IF LEN(#strWord1) >= LEN(#strWord2)
BEGIN
SET #intLength = LEN(#strWord1)
END
ELSE
BEGIN
SET #intLength = LEN(#strWord2)
END
DECLARE #iWordLoop1 INT, #iWordLoop2 INT--, #intWordLoop2 INT --This variable doesn't impact the calculation
DECLARE #intWordScore INT
DECLARE #intLetterScore INT
SET #intWordScore = 0
--SET #intWordLoop2 = Len(#strWord2)--this value is not used anywhere else, so removing makes no difference.
--DECLARE #strLetter VARCHAR(1000)
DECLARE #strLetter CHAR(1)--there is no need for 1000 characters since we're only ever assigning a single character to this
DECLARE #count1 INT
SET #count1 = 0
SET #iWordLoop1 = Len(#strWord1)
WHILE (#count1 < #iWordLoop1)
BEGIN
SET #strLetter = SUBSTRING(#strWord1, #count1+1, 1)
SET #intLetterScore = 0
DECLARE #count2 INT
SET #count2 = 0
SET #iWordLoop2 = Len(#strWord2)
WHILE (#count2 < #iWordLoop2)
BEGIN
If #strLetter = SUBSTRING(#strWord2, #count2+1, 1)
BEGIN
If #iWordLoop1 = #iWordLoop2
BEGIN
SET #intLetterScore = 3
SET #iWordLoop2 = Len(#strWord2)
END
ELSE
BEGIN
SET #intLetterScore = 1
END
END
SET #intWordScore = #intWordScore + #intLetterScore
SET #intLetterScore = 0
SET #count2 = (#count2 + 1)
END
SET #count1 = (#count1 + 1)
END
DECLARE #sinScore DEC
SET #sinScore = (#intWordScore*100 / (3 * #intLength))
RETURN #sinSCore
END;
select dbo.comparewords ('Cat','cart')
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))