I have these year number ranges
1993-1997
1923-1935
1998-2015
I'm trying to produce this shortened version of these year ranges.
1993-7
1923-35
1998-2015
So far my query looks like this. But its not working on the 2nd and 3rd samples, 1923-1935, and 1998-2015.
declare #bookyear varchar(50)
declare #year1 char(4)
declare #year2 char(4)
set #bookyear = '1993-1997'
set #year1 = substring(#bookyear, 1, charindex('-', #bookyear)-1)
set #year2 = substring(#bookyear, charindex('-', #bookyear) + 1, len(#bookyear))
select cast(#year1 as varchar(50)) + '-'+ substring(#year2, 4, 1)
Note: Year is always in 4 digits.
I am assuming you want a single digit year if its within the same decade, and double digit year when its a different decade. In which case use a case statement to compare the decade component and then display the appropriate number of digits e.g.
declare #bookyear varchar(50), #year1 char(4), #year2 char(4);
set #bookyear = '1993-1997';
set #year1 = substring(#bookyear, 1, charindex('-', #bookyear)-1);
set #year2 = substring(#bookyear, charindex('-', #bookyear) + 1, len(#bookyear));
select cast(#year1 as varchar(50)) + '-'
+ case when substring(#Year1,1,3) = substring(#Year2,1,3) then substring(#year2, 4, 1)
when substring(#Year1,1,2) = substring(#Year2,1,2) then substring(#year2, 3, 2)
else substring(#year2, 1, 4) end;
Assuming your input strings will always have same length i.e. 9 characters
drop table if exists t
create table t (d varchar(9))
insert into t values
('1993-1997')
,('1923-1935')
,('1998-2015')
,('2095-2115')
SQLFIDDLE
select d
, case
when LEFT(d, 3) = LEFT(RIGHT(d , 4), 3) then LEFT(d, 5) + RIGHT(d, 1)
when LEFT(d, 2) = LEFT(RIGHT(d , 4), 2) then LEFT(d, 5) + RIGHT(d, 2)
when LEFT(d, 1) = LEFT(RIGHT(d , 4), 1) then LEFT(d, 5) + RIGHT(d, 3)
ELSE d
end
FROM t
There are actually 2 ways to address this.
Looking for the same from the right.
Or looking for differences from the left.
Example snippet:
declare #bookyear varchar(9);
set #bookyear = '1993-1997';
--
-- looking for different digits from left to right
--
set #bookyear = left(#bookyear,5) +
case
when left(#bookyear,1) != substring(#bookyear,6,1) then right(#bookyear,4)
when left(#bookyear,2) != substring(#bookyear,6,2) then right(#bookyear,3)
when left(#bookyear,3) != substring(#bookyear,6,3) then right(#bookyear,2)
else right(#bookyear,1)
end;
select #bookyear as bookyear1;
-- reset variable
set #bookyear = '1993-1997';
--
-- looking for same digits from right to left
--
set #bookyear = left(#bookyear,5) +
case
when left(#bookyear,3) = substring(#bookyear,6,3) then substring(#bookyear,9,1)
when left(#bookyear,2) = substring(#bookyear,6,2) then substring(#bookyear,8,2)
when left(#bookyear,1) = substring(#bookyear,6,1) then substring(#bookyear,7,3)
else substring(#bookyear,6,4)
end;
select #bookyear as bookyear2;
A test snippet using a table variable:
declare #Test table (bookyear varchar(9));
insert into #Test (bookyear) values
('1993-1997'),
('1923-1935'),
('1998-2015'),
('2095-2115');
select
bookyear,
left(bookyear,5) +
case
when left(bookyear,1) != substring(bookyear,6,1) then right(bookyear,4)
when left(bookyear,2) != substring(bookyear,6,2) then right(bookyear,3)
when left(bookyear,3) != substring(bookyear,6,3) then right(bookyear,2)
else right(bookyear,1)
end as bookyear1,
left(bookyear,5) +
case
when left(bookyear,3) = substring(bookyear,6,3) then substring(bookyear,9,1)
when left(bookyear,2) = substring(bookyear,6,2) then substring(bookyear,8,2)
when left(bookyear,1) = substring(bookyear,6,1) then substring(bookyear,7,3)
else substring(bookyear,6,4)
end as bookyear2
from #Test;
Returns:
bookyear bookyear1 bookyear2
1993-1997 1993-7 1993-7
1923-1935 1923-35 1923-35
1998-2015 1998-2015 1998-2015
2095-2115 2095-115 2095-115
A test on rextester here
Related
In a procedure in SQL Server 2008 I need to replace a substring between 2 identifiers. I do not know the complete strings to be replaced, or if the second identifier exists.
If the terminator identifier is not present, I would need the end of the string to be considered one. Also, the identifiers could be the same.
DECLARE #startIdenfier VARCHAR(10) = 'the'
DECLARE #endIdenfier VARCHAR(10) = 'the'
DECLARE #newString VARCHAR(20) = 'new string that'
Sample input / output:
'This is the old string that the process needs to be applied on.' ->
'This is the new string that the process needs to be applied on.'
'This is the old string' ->
'This is the new string that'
SET #endIdenfier = 'I'
'This is the old string that I don't like' ->
'This is the new string that I don't like''
It is a generic replace and I havn't been able to find the proper way to do it, since the REPLACE function does not accept indexes.
EDIT: This community is awesome. I am sorry I cannot select multiple accepted solutions, but I thank you all for your help. I'll try all solutions already posted (besides the already accepted one which I tested) and up-vote individually.
--Find the start index plus the length of the string found (plus one for the space)
SET #startIdx = CHARINDEX(#startIdentifier, #initialString, 1) + LEN(#startIdentifier) + 1
--Find the next occurrence of the end identifier (minus one for the space)
SET #endIdx = CHARINDEX(#endIdentifier, #initialString, #startIdx) - 1;
--end not found?
IF #endIdx = -1 SET #endIdx = LEN(#initialString) + 1;
--Use the STUFF function to remove the old chars from endindex-startindex, and insert the new string at the startindex
SET #results = STUFF(#initialString, #startIdx, #endIdx - #startIdx, #newString)
In full:
DECLARE #startIdenfier Varchar(10)
SET #startIdenfier = 'the'
DECLARE #endIdenfier Varchar(10)
SET #endIdenfier = 'the'
DECLARE #newString Varchar(100)
SET #newString = 'new string that'
DECLARE #initialString VARCHAR(256) = 'this is the old string that the process needs to be applied on';
DECLARE #startIdx INT;
SET #startIdx = CHARINDEX(#startIdenfier, #initialString, 1) + LEN(#startIdenfier) + 1;
DECLARE #endIdx INT;
SET #endIdx= CHARINDEX(#endIdenfier, #initialString, #startIdx) - 1;
IF #endIdx = -1 SET #endIdx = LEN(#initialString) + 1;
DECLARE #results VARCHAR(256);
SET #results = STUFF(#initialString, #startIdx, #endIdx - #startIdx, #newString);
SELECT #results
You could do something like this:
/*Declare necessary variables*/
DECLARE #startIndex INT
DECLARE #endIndex INT
DECLARE #startReplace INT
DECLARE #lengthReplace INT
DECLARE #replaceString VARCHAR(500)
/*Get the index of the start/end idenfier*/
SELECT #startIndex = CHARINDEX ( #startIdenfier , #originalString)
SELECT #endIndex = CHARINDEX ( #startIdenfier , #originalString, #startIndex+1)
/*In case the end idenfier doesn't exist*/
IF #endIndex = 0
SET #endIndex = LEN(#originalString) + 1
SET #startReplace = #startIndex + len(#startIdenfier)
SET #lengthReplace = #endIndex - #startReplace
SELECT STUFF(#originalString, #startReplace, #lengthReplace, #newString)
I think you will need to incorporate the PATINDEX function and create the pattern by using your start and end identifiers. That will satisfy the condition where both start and end identifiers are present...
DECLARE #OldString nvarchar(max)
DECLARE #NewString nvarchar(max)
DECLARE #StartLocation bigint
DECLARE #Pattern nvarchar(200) = '%' + #StartIdentifier + ' % ' + #EndIdentifer + '%'
SELECT #StartLocation = PATINDEX(#Pattern, 'old complete string')
If the pattern is located, then you can get the string to be replaced by substringing the 'old complete string', starting at position (#StartLocation + Length of #StartIdentifier + 1). To determine the length for SUBSTRING, you need to locate the position of #EndIdentifier, using CHARINDEX of the old complete string beginning at ( #StartLocation + Length of #StartIdentifier + 1). Subtract ( #StartLocation + Length of StartIdentifier + 1) from result of CHARINDEX.
SELECT #OldString = SUBSTRING('complete old string', #StartLocation + LEN(#StartIdentifier) + 1, CHARINDEX(' ' + #EndIdentifier, 'old complete string', #StartLocation + LEN(#StartIdentifier) + 1) - (#StartLocation + LEN(#StartIdentifier) + 1)))
You can at this point do a straight forward REPLACE to get the new string.
SELECT #NewCompleteString = REPLACE('old complete string', #OldString, #NewString)
"If there is no 'terminator' identifier, I would need the end of the
string to be considered one."
If the initial pattern was not located, we fall back to search for the #StartIdentifier only. For this you can reset the pattern to contain the #StartIdentifier only...
SELECT #Pattern = '%' + #StartIdentifier + ' %'
SELECT #StartLocation = PATINDEX(#Pattern, 'old complete string')
If the pattern is located then you can get the old string to replace by SUBSTRING starting at ( #StartLocation + Length of #StartIdentifier + 1 ), with a length of 'old complete string' length - ( #StartLocation + Length of #StartIdentifier + 1 )...
SELECT #OldString = SUBSTRING('old complete string', #StartLocation + LEN(#StartIdentifier) + 1, LEN('old complete string') - (#StartLocation + LEN(#StartIdentifier) + 1))
You can then REPLACE...
SELECT #NewCompleteString = REPLACE('old complete string', #OldString, #NewString)
Something like this...
DECLARE #initialString VARCHAR(32) = '1234567890123456789'
DECLARE #startIdentifier VARCHAR(32) = '34'
DECLARE #endIdentifier VARCHAR(32) = '34'
DECLARE #newString VARCHAR(32) = 'ABC'
DECLARE #headChars INT = CHARINDEX(#startIdentifier, #initialString, 1)
IF #headChars > 0
SET #headChars = #headChars + LEN(#startIdentifier) - 1
DECLARE #bodyChars INT = CHARINDEX(#endIdentifier, #initialString, #headChars + 1)
IF #bodyChars > 0
SET #bodyChars = LEN(#initialString) - #bodyChars + 1
SELECT
LEFT(#initialString, #headChars)
+ #newString
+ RIGHT(#initialString, #bodyChars)
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=b8a179d0e63840dfa60905d9951e4b22
For example...
'1234567890123456789'
'34' => Start # 1 => Found # 3 => keep chars 1->4
'34' => Start # 5 => Found # 13 => keep chars 13->end
Just use STUFF and CHARINDEX. Figure out:
the position at which replacement begins (position of the + length of the)
the position at which replacement ends (position of the starting from see above)
And subtract the positions to calculate the number of characters to replace.
DECLARE #string VARCHAR(100) = 'This is the old string that the process needs to be applied on.'
DECLARE #replace VARCHAR(100) = 'NEW STRING THAT '
DECLARE #delim1 VARCHAR(100) = 'the '
DECLARE #delim2 VARCHAR(100) = 'the '
DECLARE #pos1 INT = CHARINDEX(#delim1, #string) + DATALENGTH(#delim1)
DECLARE #pos2 INT = ISNULL(NULLIF(CHARINDEX(#delim2, #string, #pos1), 0), DATALENGTH(#string) + 1)
SELECT STUFF(#string, #pos1, #pos2 - #pos1, #replace)
-- "This is the NEW STRING THAT the process needs to be applied on."
SET #delim2 = 'xxx'
SET #pos1 = CHARINDEX(#delim1, #string) + DATALENGTH(#delim1)
SET #pos2 = ISNULL(NULLIF(CHARINDEX(#delim2, #string, #pos1), 0), DATALENGTH(#string) + 1)
SELECT STUFF(#string, #pos1, #pos2 - #pos1, #replace)
-- "This is the NEW STRING THAT "
Note: spaces should be a part of search delimiters instead of the logic. the should not match them and threfore.
I have this SQL Data:
AT_SendMail_v1
AT_Certificate_v10
AT_Certificate_v100
And I want this output:
AT_SendMail_v2
AT_Certificate_v11
AT_Certificate_v101
I have this code but it doesn't work for the numbers that they more than 19:
SELECT CASE
WHEN ISNUMERIC(RIGHT([internalname],1)) = 1
THEN LEFT([internalname],LEN([internalname])-1) + cast((CONVERT(INT, RIGHT([internalname],1)) + 1) as varchar(100))
ELSE [internalname] + '_v1'
END
,[id]
FROM [task]
thanks :)
Assuming that the number is always the end of the string:
using patindex():
select
name
, new_name = case when patindex('%[_]v[1234567890]%',name) > 0
then left(name,(patindex('%[_]v[1234567890]%',name)+1))
+ convert(varchar(10),convert(int
,right(name,len(name)-(patindex('%[_]v[1234567890]%',name)+1)))
+1)
else name+'_v1'
end
from task
rextester demo: http://rextester.com/AWPXB86451
returns:
+----------------------+----------------------+
| name | new_name |
+----------------------+----------------------+
| AT_SendMail_v1 | AT_SendMail_v2 |
| AT_Certificate_v10 | AT_Certificate_v11 |
| AT_Certificate_v_v50 | AT_Certificate_v_v51 |
+----------------------+----------------------+
You could use SUBSTRING to get number after _v, after it check If It is numeric using ISNUMERIC, after It increase that number by 1, convert It to string and concat string.
DECLARE #internalname NVARCHAR(40)
SET #internalname = 'AT_SendMail_v19'
SELECT
CASE WHEN ISNUMERIC(SUBSTRING(#internalname, CHARINDEX('_v', #internalname) + 2, LEN(#internalname))) = 1
THEN SUBSTRING(#internalname,+2, CHARINDEX('_v',#internalname)) + CAST(SUBSTRING(#internalname, CHARINDEX('_v', #internalname) + 2, LEN(#internalname)) + 1 AS NVARCHAR(60))
ELSE #internalname + '_v1'
END
Try this :
declare #test varchar(50) = 'AT_SendMail_v1';
select Output = case when ISNUMERIC( substring ( #test , charindex('_v',#test)+2 , len(#test)) )=1 then
left (#test , charindex('_v',#test)+1) + cast( cast(substring ( #test , charindex('_v',#test)+2 , len(#test)) as int)+1 as varchar )
else
#test+'_v1' end;
Demo.
Here is one way to increment the numbers, you can then build it into your string...
DECLARE #str NVARCHAR(20) = 'AT_SendMail_v1';
DECLARE #intCharIndex INT = CHARINDEX('_v',#str);
IF(#intCharIndex = 0)
BEGIN
SELECT 'No string found';
END
IF(#intCharIndex > 0)
BEGIN
DECLARE #int INT = LEN(#str) - #intCharIndex - 1;
DECLARE #intVersion INT = RIGHT(#str,#int) + 1;
SELECT LEFT(#str,#intCharIndex) + 'v' + RTRIM(CAST(#intVersion AS NVARCHAR(4)));
END
that was the best code for me, thanks :)
SELECT
CASE
WHEN CHARINDEX('_v', [internalname], 1) <> 0 THEN LEFT([internalname], CHARINDEX('_v', [internalname], 1) + 1) + CAST(CAST(RIGHT([internalname], LEN([internalname]) - CHARINDEX('_v', [internalname], 1) - 1) AS INT) + 1 AS VARCHAR(100))
ELSE [internalname] + '_v1'
END
,[id]
FROM [task]
How about this that caters for ANY number after the _v
Select Left(internalName, Len(internalName) - CharIndex('v_', Reverse(internalName)) + 1) + Cast((Cast((Right(internalName, CharIndex('v_', Reverse(internalName)) -1)) As Integer) + 1) As NVarchar(32))
From task
I have a stored procedure like this
declare #Tmp varchar(60)
declare #Hasil varchar(60)
declare #PFID varchar(50)
declare #CntPFID int
declare #PrevPFID varchar(60)
DECLARE #CheckLetter CHAR(5)
set #PFID = 'PF-FB/ALL/009/MKX/VI/16'
set #CheckLetter = (UPPER(RIGHT((CAST(SUBSTRING(#PFID, 10, 4) AS VARCHAR(5))), 3)))
select #CntPFID = count(pfid)
from pf
where pfid like '%'+ right(#PFID, 9)
and pfid like ''+ left(#PFID, 13) +'%'
if #CntPFID = 1
begin
set #Tmp = (UPPER(RIGHT((CAST(SUBSTRING(#PFID, 10, 4) AS VARCHAR(5))), 3))) + '-' + '01'
set #Hasil = replace(#PFID, replace(#CheckLetter, ' ', ''), #Tmp)
end
else
begin
set #Tmp = (UPPER(RIGHT((CAST(SUBSTRING(#PFID, 10, 4) AS VARCHAR(5))), 3))) + '-0' + cast(#CntPFID as varchar(3))
set #Hasil = replace(#PFID, replace(#CheckLetter, ' ', ''), #Tmp)
end
select #Hasil
I was stuck when the record data already reach 'PF-FB/ALL/009-09/MKX/VI/16' the next number becoming PF-FB/ALL/009-010/MKX/VI/16 (there is three digits = 010) while I want it will be PF-FB/ALL/009-10/MKX/VI/16 (2 digits running).
Is there any dynamically way without using condition else if > 9 ... ?
The expression you need to change: '-0' + cast(#CntPFID as varchar(3)), try this way:
declare #CntPFID int;
set #CntPFID = 9;
select '-' + right('00'+ cast(#CntPFID as varchar(3)),2)
set #CntPFID = 11;
select '-' + right('00'+ cast(#CntPFID as varchar(3)),2)
I need to create this table function. The function needs to return single words from passed parameters like: hello, hhuu, value
The table function should return:
hello,
hhuu,
value
But I am always getting some errors, please could you help me?
you can write as:
DECLARE #input_char VARCHAR(255)
SET #input_char = 'hello, hhuu, value'
;WITH cte AS (
SELECT
CAST('<r>' + REPLACE(#input_char, ' ', '</r><r>') + '</r>' AS XML) AS input_char
)
SELECT
rtrim( LTRIM (xTable.xColumn.value('.', 'VARCHAR(MAX)')) ) AS input_char
FROM cte
CROSS APPLY input_char.nodes('//r') AS xTable(xColumn)
Please give a look at this article:
http://www.codeproject.com/Tips/625872/Convert-a-CSV-delimited-string-to-table-column-in
and you could use ' ' (space) as delimiter.
SELECT * FROM dbo.CSVtoTable('hello, hhuu, value', ' ')
I have used the following function many times. It is a bit lengthy forgive me, but it has become a great tool for me.
CREATE Function [dbo].[ParseText2Table]
(
#p_SourceText varchar(MAX)
,#p_Delimeter varchar(100) = ',' --default to comma delimited.
)
RETURNS #retTable TABLE
(
POSITION INT
,Int_Value bigint
,Num_value REAL--Numeric(18,3)
,txt_value varchar(MAX)
)
AS
BEGIN
DECLARE #tmpTable TABLE
(
Position2 INT IDENTITY(1,1) PRIMARY KEY
,Int_Value bigint
,Num_value REAL--Numeric(18,3)
,txt_value varchar(MAX)
)
DECLARE #w_Continue INT
,#w_StartPos INT
,#w_Length INT
,#w_Delimeter_pos INT
,#w_tmp_int bigint
,#w_tmp_num REAL--numeric(18,3)
,#w_tmp_txt varchar(MAX)
,#w_Delimeter_Len INT
IF len(#p_SourceText) = 0
BEGIN
SET #w_Continue = 0 -- force early exit
END
ELSE
BEGIN
-- if delimiter is ' ' change
IF #p_Delimeter = ' '
BEGIN
SET #p_SourceText = replace(#p_SourceText,' ','ÿ')
SET #p_Delimeter = 'ÿ'
END
-- parse the original #p_SourceText array into a temp table
SET #w_Continue = 1
SET #w_StartPos = 1
SET #p_SourceText = RTRIM( LTRIM( #p_SourceText))
SET #w_Length = DATALENGTH( RTRIM( LTRIM( #p_SourceText)))
SET #w_Delimeter_Len = len(#p_Delimeter)
END
WHILE #w_Continue = 1
BEGIN
SET #w_Delimeter_pos = CHARINDEX( #p_Delimeter
,(SUBSTRING( #p_SourceText, #w_StartPos
,((#w_Length - #w_StartPos) + #w_Delimeter_Len)))
)
IF #w_Delimeter_pos > 0 -- delimeter(s) found, get the value
BEGIN
SET #w_tmp_txt = LTRIM(RTRIM( SUBSTRING( #p_SourceText, #w_StartPos
,(#w_Delimeter_pos - 1)) ))
IF dbo.isReallyNumeric(#w_tmp_txt) = 1 --and not #w_tmp_txt in('.', '-', '+', '^')
BEGIN
--set #w_tmp_int = cast( cast(#w_tmp_txt as real) as bigint)--numeric) as bigint)
SET #w_tmp_int = CASE WHEN (CAST(#w_tmp_txt AS REAL) BETWEEN -9223372036854775808 AND 9223372036854775808) THEN CAST( CAST(#w_tmp_txt AS REAL) AS bigint) ELSE NULL END
SET #w_tmp_num = CAST( #w_tmp_txt AS REAL)--numeric(18,3))
END
ELSE
BEGIN
SET #w_tmp_int = NULL
SET #w_tmp_num = NULL
END
SET #w_StartPos = #w_Delimeter_pos + #w_StartPos + (#w_Delimeter_Len- 1)
END
ELSE -- No more delimeters, get last value
BEGIN
SET #w_tmp_txt = LTRIM(RTRIM( SUBSTRING( #p_SourceText, #w_StartPos
,((#w_Length - #w_StartPos) + #w_Delimeter_Len)) ))
IF dbo.isReallyNumeric(#w_tmp_txt) = 1 --and not #w_tmp_txt in('.', '-', '+', '^')
BEGIN
--set #w_tmp_int = cast( cast(#w_tmp_txt as real) as bigint)--as numeric) as bigint)
SET #w_tmp_int = CASE WHEN (CAST(#w_tmp_txt AS REAL) BETWEEN -9223372036854775808 AND 9223372036854775808) THEN CAST( CAST(#w_tmp_txt AS REAL) AS bigint) ELSE NULL end
SET #w_tmp_num = CAST( #w_tmp_txt AS REAL)--numeric(18,3))
END
ELSE
BEGIN
SET #w_tmp_int = NULL
SET #w_tmp_num = NULL
END
SELECT #w_Continue = 0
END
INSERT INTO #tmpTable VALUES( #w_tmp_int, #w_tmp_num, #w_tmp_txt )
END
INSERT INTO #retTable SELECT Position2, Int_Value ,Num_value ,txt_value FROM #tmpTable
RETURN
END
Here are the supporting functions for above as well:
CREATE FUNCTION dbo.isReallyInteger
(
#num VARCHAR(64)
)
RETURNS BIT
BEGIN
IF LEFT(#num, 1) = '-'
SET #num = SUBSTRING(#num, 2, LEN(#num))
RETURN CASE
WHEN PATINDEX('%[^0-9-]%', #num) = 0
AND CHARINDEX('-', #num) <= 1
AND #num NOT IN ('.', '-', '+', '^')
AND LEN(#num)>0
AND #num NOT LIKE '%-%'
THEN
1
ELSE
0
END
END
CREATE FUNCTION dbo.isReallyNumeric
(
#num VARCHAR(64)
)
RETURNS BIT
BEGIN
IF LEFT(#num, 1) = '-'
SET #num = SUBSTRING(#num, 2, LEN(#num))
DECLARE #pos TINYINT
SET #pos = 1 + LEN(#num) - CHARINDEX('.', REVERSE(#num))
RETURN CASE
WHEN PATINDEX('%[^0-9.-]%', #num) = 0
AND #num NOT IN ('.', '-', '+', '^')
AND LEN(#num)>0
AND #num NOT LIKE '%-%'
AND
(
((#pos = LEN(#num)+1)
OR #pos = CHARINDEX('.', #num))
)
THEN
1
ELSE
0
END
END
Usage Examples:
--Single Character Delimiter
--select * from dbo.ParseText2Table('100|120|130.56|Yes|Cobalt|Blue','|')
--select txt_value from dbo.ParseText2Table('100 120 130.56 Yes Cobalt Blue',' ') where Position = 3
--select * from dbo.ParseText2Table('100,120,130.56,Yes,Cobalt,Blue,,',',')
/*
POSITION Int_Value Num_value txt_value
----------- ----------- -------------------- --------------
1 100 100.000 100
2 120 120.000 120
3 131 130.560 130.56
4 NULL NULL Yes
5 NULL NULL Cobalt Blue
*/
I just had to scour the internet for this code yet again so I figured I would put it here so I can find it a little faster the next time and hopefully you found it a little faster too :)
Check this.
The code below can check digit in all GTIN's (EAN8, EAN13, EAN14, UPC/A, UPC/E)
CREATE FUNCTION [dbo].[check_digit]
(
#GTIN VARCHAR(14)
)
RETURNS TINYINT
AS
BEGIN
DECLARE #Index TINYINT,
#Multiplier TINYINT,
#Sum TINYINT,
#checksum TINYINT,
#result TINYINT,
#GTIN_strip VARCHAR(13)
SELECT #GTIN_strip = SUBSTRING(#GTIN,1,LEN(#GTIN)-1);
SELECT #Index = LEN(#GTIN_strip),
#Multiplier = 3,
#Sum = 0
WHILE #Index > 0
SELECT #Sum = #Sum + #Multiplier * CAST(SUBSTRING(#GTIN_strip, #Index, 1) AS TINYINT),
#Multiplier = 4 - #Multiplier,
#Index = #Index - 1
SELECT #checksum = CASE #Sum % 10
WHEN 0 THEN '0'
ELSE CAST(10 - #Sum % 10 AS CHAR(1))
END
IF (SUBSTRING(#GTIN,LEN(#GTIN),1) = #checksum)
RETURN 1; /*true*/
RETURN 0; /*false*/
END
CREATE FUNCTION [dbo].[fn_EAN13CheckDigit] (#Barcode nvarchar(12))
RETURNS nvarchar(13)
AS
BEGIN
DECLARE #SUM int , #COUNTER int, #RETURN varchar(13), #Val1 int, #Val2 int
SET #COUNTER = 1 SET #SUM = 0
WHILE #Counter < 13
BEGIN
SET #VAL1 = SUBSTRING(#Barcode,#counter,1) * 1
SET #VAL2 = SUBSTRING(#Barcode,#counter + 1,1) * 3
SET #SUM = #VAL1 + #SUM
SET #SUM = #VAL2 + #SUM
SET #Counter = #Counter + 2;
END
SET #SUM=(10-(#SUM%10))%10
SET #Return = #BARCODE + CONVERT(varchar,((#SUM)))
RETURN #Return
END
And here's a MySQL implementation as well.
Note it requires a 13 digit UPC/EAN13 as input, so you should LPAD your input as required.
drop function if exists barcodify;
delimiter //
at the last digit.
create function barcodify(upc varchar(15))
returns varchar(15)
sql security invoker
begin
declare i, r, odd, even,result int;
set odd=0;
set even =0;
set i = length(upc);
while i > 0 do
set r = substring(upc, i, 1) ;
if(i % 2 >0) then set odd = odd + r;
else set even = even + r;
end if;
set i = i - 1;
end while;
set result = (3*odd)+even;
if result % 10 =0 then return 0;
else return (10 - (result %10));
end if;
end //
delimiter ;
Here's my implementation of a function in Oracle.
CREATE OR REPLACE FUNCTION GenerateCheckDigit(pUPC IN VARCHAR2) RETURN INT
IS
vCheckDigit INT;
BEGIN
WITH barcodeData AS
(
SELECT pUPC barcode FROM dual
)
SELECT
10-REPLACE(MOD(SUM(SUBSTR(barcode, -LEVEL, 1) * DECODE(MOD(LEVEL-1, 2), 1, 1, 3)),10),0,10) INTO vCheckDigit
FROM barcodeData
CONNECT BY LEVEL <= LENGTH(barcode);
RETURN vCheckDigit;
END;
To calculate the last digit.
https://segovoni.medium.com/sql-server-function-to-calculate-gs1-barcode-check-digit-d55b478ff645
CREATE FUNCTION dbo.udf_GetGS1EAN13CheckDigit
(
#ACode AS VARCHAR(12)
)
RETURNS SMALLINT
AS BEGIN
/*
Author: Sergio Govoni
Notes: Calculate the check-digit of a GS1 EAN13 code
Version: 1.0
*/
DECLARE
#tmpCode AS VARCHAR(12)
,#tmpMulSup AS VARCHAR(8000)
,#tmp AS VARCHAR(8000)
,#i AS INT
,#j AS INT
,#z AS INT
,#SumDEven AS INT
,#SumDOdd AS INT
,#List AS VARCHAR(8000)
,#tmpList AS VARCHAR(8000)
,#CheckSum AS SMALLINT
SET #SumDEven = 0
SET #SumDOdd = 0
SET #List = ''
SET #tmpList = ''
SET #tmp = ''
SET #tmpCode = #ACode
/* 0. List builder */
SET #j = LEN(#tmpCode) + 1
SET #i = 1
WHILE (#i <= LEN(#tmpCode)) BEGIN SET #List = #List + '|' + LTRIM(RTRIM(STR(#j))) + ';' + SUBSTRING(#tmpCode, #i, 1) SET #j = (#j - 1) SET #i = (#i + 1) END /* 1. Add up the digits in even position */ SET #i = 1 SET #tmpList = #List WHILE (CHARINDEX('|', #tmpList) > 0)
BEGIN
SET #j = CHARINDEX('|', #tmpList)
SET #z = CHARINDEX(';', #tmpList)
IF (CAST(SUBSTRING(#tmpList, (#j + 1), (#z - (#j + 1))) AS INTEGER) % 2) = 0
BEGIN
SET #SumDEven = #SumDEven + CAST(SUBSTRING(#tmpList, (#z + 1), 1) AS INTEGER)
END
SET #tmpList = SUBSTRING(#tmpList, (#z + 2), LEN(#tmpList))
END
/* 2. Multiply the result of the previous step (the first step) to 3 (three) */
SET #SumDEven = (#SumDEven * 3)
/* 3. Add up the digits in the odd positions */
SET #i = 1
SET #tmpList = #List
WHILE (CHARINDEX('|', #tmpList) > 0)
BEGIN
SET #j = CHARINDEX('|', #tmpList)
SET #z = CHARINDEX(';', #tmpList)
IF (CAST(SUBSTRING(#tmpList, (#j + 1), (#z - (#j + 1))) AS INTEGER) % 2) <> 0
BEGIN
SET #SumDOdd = #SumDOdd + CAST(SUBSTRING(#tmpList, (#z + 1), 1) AS INTEGER)
END
SET #tmpList = SUBSTRING(#tmpList, (#z + 2), LEN(#tmpList))
END
/* 4. Add up the results obtained in steps two and three */
SET #CheckSum = (#SumDEven + #SumDOdd)
/* 5. Subtract the upper multiple of 10 from the result obtained in step four */
IF ((#CheckSum % 10) = 0)
BEGIN
/* If the result of the four step is a multiple of Ten (10), like
Twenty, Thirty, Forty and so on,
the check-digit will be equal to zero, otherwise the check-digit will be
the result of the fifth step
*/
SET #CheckSum = 0
END
ELSE BEGIN
SET #tmpMulSup = LTRIM(RTRIM(STR(#CheckSum)))
SET #i = 0
WHILE #i <= (LEN(#tmpMulSup) - 1)
BEGIN
SET #tmp = #tmp + SUBSTRING(#tmpMulSup, #i, 1)
IF (#i = LEN(#tmpMulSup) - 1)
BEGIN
SET #tmp = LTRIM(RTRIM(STR(CAST(#tmp AS INTEGER) + 1)))
SET #tmp = #tmp + '0'
END
SET #i = (#i + 1)
END
SET #CheckSum = CAST(#tmp AS INTEGER) - #CheckSum
END
RETURN #CheckSum
END;
CREATE FUNCTION sfn_ean_chkdigit(#barcode varchar(20))
RETURNS CHAR(1)
AS
BEGIN
DECLARE #chk_digit int, #chk int
DECLARE #num TABLE (num int)
IF LEN(#barcode) NOT IN (7, 12) RETURN NULL
INSERT INTO #num
SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL
SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL
SELECT 9 UNION ALL SELECT 10 UNION ALL SELECT 11 UNION ALL SELECT 12
SELECT #chk_digit = SUM(CONVERT(int, SUBSTRING(#barcode, LEN(#barcode) - num + 1, 1)) * CASE WHEN num % 2 = 1 THEN 3 ELSE 1 END)
FROM #num
WHERE num <= LEN(#barcode)
SELECT #chk_digit = (10 - (#chk_digit % 10)) % 10
RETURN CHAR(ASCII('0') + #chk_digit)
END
Num unico script
select lpad('123456789012',12,'0') || ( (r+1)*10 - soma )%10
from (select sum(v::int * case when i%2=0 then 3 else 1 end) soma,
round(sum(v::int * case when i%2=0 then 3 else 1 end)/10.0) r
from unnest(string_to_array(lpad('123456789012',12,'0'),null)) with ordinality d(v,i)
) a