Related
How to draw triangles in SQL Server as shown below?
I want to implement it by using WHILE loops, but I am unable to print 20 '*' in a single line in SQL Server.
How can I achieve this?
Use REPLICATE inside a WHILE. I think, you can achieve your desired output if you do it correctly?
DECLARE #i INT = 20
WHILE(#i>0)
BEGIN
PRINT REPLICATE('* ', #i);
SET #i = #i - 1;
END
You can use REPLICATE to repeat a character a certain number of times. To generate a sequence of numbers from 1 to 20 you don't need a WHILE anyway - SQL doesn't really need the WHILE statement to work with data.
Number sequences are always useful which is why almost every SQL developer creates a Numbers table.
If you don't already have one, a quick and dirty way to generate 20 numbers is to select the top 20 rows from a systems table, and use ROW_NUMBER to calculate row numbers eg:
select top 20 replicate('*',21-row_number() over (order by id) )
from sys.sysobjects
With a Numbers table, the query is simpler:
select replicate('*',Number )
from dbo.Numbers
where Numbers.Number <= 20
order by Number desc
Numbers tables are extremely useful, eg for sets of elements like 200 days starting from 2017/1/1 :
select dateadd(d,Number,cast('20170101' as date))
from dbo.Numbers
where Numbers.n<= 20
order by Number desc
Try This,
DECLARE #StrLen INT = 20
WHILE #StrLen >= 1
BEGIN
PRINT REPLICATE('*',#StrLen)
SET #StrLen = #StrLen - 1
END
Without replicate() solution
BEGIN
DECLARE #i int = 20
DECLARE #j int
DECLARE #line varchar(max)
WHILE #i > 0
BEGIN
SET #line = ''
SET #j = 0
WHILE #j < #i
BEGIN
SET #line += '* '
SET #j = #j + 1
END
PRINT #line
SET #i = #i - 1
END
END
Using recursive CTE (Press Ctrl+T for results in text)
;WITH A AS (
SELECT REPLICATE('* ', 20) X
UNION ALL
SELECT LEFT(X, LEN(X) - 2)
FROM A
WHERE X > '* '
)
SELECT * FROM A
Another way:
;WITH A AS (
SELECT 20 X
UNION ALL
SELECT X - 1 FROM A
WHERE X > 1
)
SELECT REPLICATE('* ', X) FROM A
SELECT REPLACE('<strong>100</strong><b>.00 GB', '%^(^-?\d*\.{0,1}\d+$)%', '');
I want to replace any markup between two parts of the number with above regex, but it does not seem to work. I'm not sure if it is regex syntax that's wrong because I tried simpler one such as '%[^0-9]%' just to test but it didn't work either. Does anyone know how can I achieve this?
You can use PATINDEX
to find the first index of the pattern (string's) occurrence. Then use STUFF to stuff another string into the pattern(string) matched.
Loop through each row. Replace each illegal characters with what you want. In your case replace non numeric with blank. The inner loop is if you have more than one illegal character in a current cell that of the loop.
DECLARE #counter int
SET #counter = 0
WHILE(#counter < (SELECT MAX(ID_COLUMN) FROM Table))
BEGIN
WHILE 1 = 1
BEGIN
DECLARE #RetVal varchar(50)
SET #RetVal = (SELECT Column = STUFF(Column, PATINDEX('%[^0-9.]%', Column),1, '')
FROM Table
WHERE ID_COLUMN = #counter)
IF(#RetVal IS NOT NULL)
UPDATE Table SET
Column = #RetVal
WHERE ID_COLUMN = #counter
ELSE
break
END
SET #counter = #counter + 1
END
Caution: This is slow though! Having a varchar column may impact. So using LTRIM RTRIM may help a bit. Regardless, it is slow.
Credit goes to this StackOverFlow answer.
EDIT
Credit also goes to #srutzky
Edit (by #Tmdean)
Instead of doing one row at a time, this answer can be adapted to a more set-based solution. It still iterates the max of the number of non-numeric characters in a single row, so it's not ideal, but I think it should be acceptable in most situations.
WHILE 1 = 1 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, '')
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 BREAK;
END;
You can also improve efficiency quite a lot if you maintain a bit column in the table that indicates whether the field has been scrubbed yet. (NULL represents "Unknown" in my example and should be the column default.)
DECLARE #done bit = 0;
WHILE #done = 0 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table
WHERE COALESCE(Scrubbed_Column, 0) = 0)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, ''),
Scrubbed_Column = 0
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 SET #done = 1;
-- if Scrubbed_Column is still NULL, then the PATINDEX
-- must have given 0
UPDATE table
SET Scrubbed_Column = CASE
WHEN Scrubbed_Column IS NULL THEN 1
ELSE NULLIF(Scrubbed_Column, 0)
END;
END;
If you don't want to change your schema, this is easy to adapt to store intermediate results in a table valued variable which gets applied to the actual table at the end.
Instead of stripping out the found character by its sole position, using Replace(Column, BadFoundCharacter, '') could be substantially faster. Additionally, instead of just replacing the one bad character found next in each column, this replaces all those found.
WHILE 1 = 1 BEGIN
UPDATE dbo.YourTable
SET Column = Replace(Column, Substring(Column, PatIndex('%[^0-9.-]%', Column), 1), '')
WHERE Column LIKE '%[^0-9.-]%'
If ##RowCount = 0 BREAK;
END;
I am convinced this will work better than the accepted answer, if only because it does fewer operations. There are other ways that might also be faster, but I don't have time to explore those right now.
In a general sense, SQL Server does not support regular expressions and you cannot use them in the native T-SQL code.
You could write a CLR function to do that. See here, for example.
For those looking for a performant and easy solution and are willing to enable CLR:
CREATE database TestSQLFunctions
go
use TestSQLFunctions
go
ALTER database TestSQLFunctions set trustworthy on
EXEC sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE
go
CREATE ASSEMBLY [SQLFunctions]
AUTHORIZATION [dbo]
FROM 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
WITH PERMISSION_SET = SAFE
go
CREATE FUNCTION RegexReplace(
#input nvarchar(max),
#pattern nvarchar(max),
#replacement nvarchar(max)
) RETURNS nvarchar (max)
AS EXTERNAL NAME SQLFunctions.[SQLFunctions.Regex].Replace;
go
-- outputs This is a test
SELECT dbo.RegexReplace('This is a test 12345','[0-9]','')
Content of the DLL:
I stumbled across this post looking for something else but thought I'd mention a solution I use which is far more efficient - and really should be the default implementation of any function when used with a set based query - which is to use a cross applied table function. Seems the topic is still active so hopefully this is useful to someone.
Example runtime on a few of the answers so far based on running recursive set based queries or scalar function, based on 1m rows test set removing the chars from a random newid, ranges from 34s to 2m05s for the WHILE loop examples and from 1m3s to {forever} for the function examples.
Using a table function with cross apply achieves the same goal in 10s. You may need to adjust it to suit your needs such as the max length it handles.
Function:
CREATE FUNCTION [dbo].[RemoveChars](#InputUnit VARCHAR(40))
RETURNS TABLE
AS
RETURN
(
WITH Numbers_prep(Number) AS
(
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
)
,Numbers(Number) AS
(
SELECT TOP (ISNULL(LEN(#InputUnit),0))
row_number() OVER (ORDER BY (SELECT NULL))
FROM Numbers_prep a
CROSS JOIN Numbers_prep b
)
SELECT
OutputUnit
FROM
(
SELECT
substring(#InputUnit,Number,1)
FROM Numbers
WHERE substring(#InputUnit,Number,1) like '%[0-9]%'
ORDER BY Number
FOR XML PATH('')
) Sub(OutputUnit)
)
Usage:
UPDATE t
SET column = o.OutputUnit
FROM ##t t
CROSS APPLY [dbo].[RemoveChars](t.column) o
Here is a function I wrote to accomplish this based off of the previous answers.
CREATE FUNCTION dbo.RepetitiveReplace
(
#P_String VARCHAR(MAX),
#P_Pattern VARCHAR(MAX),
#P_ReplaceString VARCHAR(MAX),
#P_ReplaceLength INT = 1
)
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #Index INT;
-- Get starting point of pattern
SET #Index = PATINDEX(#P_Pattern, #P_String);
while #Index > 0
begin
--replace matching charactger at index
SET #P_String = STUFF(#P_String, PATINDEX(#P_Pattern, #P_String), #P_ReplaceLength, #P_ReplaceString);
SET #Index = PATINDEX(#P_Pattern, #P_String);
end
RETURN #P_String;
END;
[Gist][1]
[1]: https://gist.github.com/jkdba/ca13fe8f2a9855c4bdbfd0a5d3dfcda2
Edit:
Originally I had a recursive function here which does not play well with sql server as it has a 32 nesting level limit which would result in an error like the below any time you attempt to make 32+ replacements with the function. Instead of trying to make a server level change to allow more nesting (which could be dangerous like allow never ending loops) switching to a while loop makes a lot more sense.
Maximum stored procedure, function, trigger, or view nesting level exceeded (limit 32).
Wrapping the solution inside a SQL function could be useful if you want to reuse it.
I'm even doing it at the cell level, that's why I'm putting this as a different answer:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(300))
RETURNS VARCHAR(300)
BEGIN
DECLARE #str VARCHAR(300) = #string;
DECLARE #Pattern VARCHAR (20) = '%[^a-zA-Z0-9]%';
DECLARE #Len INT;
SELECT #Len = LEN(#String);
WHILE #Len > 0
BEGIN
SET #Len = #Len - 1;
IF (PATINDEX(#Pattern,#str) > 0)
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
A more speedy approach for large strings would look something like this:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(MAX))
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #str VARCHAR(MAX) = #string;
DECLARE #Pattern VARCHAR (MAX) = '%[^a-zA-Z0-9]%';
WHILE PATINDEX(#Pattern,#str) > 0
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
RETURN #str
END
I've created this function to clean up a string that contained non numeric characters in a time field. The time contained question marks when they did not added the minutes, something like this 20:??. Function loops through each character and replaces the ? with a 0 :
CREATE FUNCTION [dbo].[CleanTime]
(
-- Add the parameters for the function here
#intime nvarchar(10)
)
RETURNS nvarchar(5)
AS
BEGIN
-- Declare the return variable here
DECLARE #ResultVar nvarchar(5)
DECLARE #char char(1)
-- Add the T-SQL statements to compute the return value here
DECLARE #i int = 1
WHILE #i <= LEN(#intime)
BEGIN
SELECT #char = CASE WHEN substring(#intime,#i,1) like '%[0-9:]%' THEN substring(#intime,#i,1) ELSE '0' END
SELECT #ResultVar = concat(#ResultVar,#char)
set #i = #i + 1
END;
-- Return the result of the function
RETURN #ResultVar
END
I think this solution is faster and simple. I use always CTE/recursive because WHILE is so slow on SQL Server.
I use it in projects I work with and large databases.
/*
Function: dbo.kSql_ReplaceRegExp
Create Date: 20.02.2021
Author: Karcan Ozbal
Description: The given string value will be replaced according to the given regexp/pattern.
Parameter(s): #Value : Value/Text to REPLACE.
#RegExp : The regexp/pattern to be used for REPLACE operation.
Usage: select dbo.kSql_ReplaceRegExp('2T3EST5','%[0-9]%')
Output: 'TEST'
*/
ALTER FUNCTION [dbo].[kSql_ReplaceRegExp](
#Value nvarchar(max),
#RegExp nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #Result nvarchar(max)
;WITH CTE AS (
SELECT NUM = 1, VALUE = #Value, IDX = PATINDEX(#RegExp, #Value)
UNION ALL
SELECT NUM + 1, VALUE = REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''), IDX = PATINDEX(#RegExp, REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''))
FROM CTE
WHERE IDX > 0
)
SELECT TOP(1) #Result = VALUE
FROM CTE
ORDER BY NUM DESC
OPTION (maxrecursion 0)
RETURN #Result
END
If you are doing this just for a parameter coming into a Stored Procedure, you can use the following:
declare #badIndex int
set #badIndex = PatIndex('%[^0-9]%', #Param)
while #badIndex > 0
set #Param = Replace(#Param, Substring(#Param, #badIndex, 1), '')
set #badIndex = PatIndex('%[^0-9]%', #Param)
I thought this was clearer:
ALTER FUNCTION [dbo].[func_ReplaceChars](
#Value nvarchar(max),
#Chars nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #cLen int = len(#Chars);
DECLARE #curChar int = 0;
WHILE #curChar<#cLen
BEGIN
set #Value = replace(#Value,substring(#Chars,#curChar,1),'');
set #curChar = #curChar + 1;
END;
RETURN #Value
END
I'm using this code similar to several codes above:
DROP FUNCTION [dbo].[fnCleanString]
GO
CREATE FUNCTION [dbo].[fnCleanString] (#input VARCHAR(max), #Pattern
VARCHAR (20))
RETURNS VARCHAR(max)
BEGIN
DECLARE #str VARCHAR(max) = #input;
DECLARE #Len INT;
DECLARE #INDEX INT;
SELECT #Len = LEN(#input);
WHILE #Len > 0
BEGIN
SET #INDEX = PATINDEX(#Pattern,#str);
IF (#INDEX > 0)
BEGIN
SET #str=REPLACE(#str,SUBSTRING(#str,#INDEX, 1), '');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
You can use it like this:
SELECT CleanName = dbo.[fnCleanString](Name, '%[0-9]%') from YourTable
I think a simpler and faster approach is iterate by each character of the alphabet:
DECLARE #i int
SET #i = 0
WHILE(#i < 256)
BEGIN
IF char(#i) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.')
UPDATE Table SET Column = replace(Column, char(#i), '')
SET #i = #i + 1
END
SELECT REPLACE('<strong>100</strong><b>.00 GB', '%^(^-?\d*\.{0,1}\d+$)%', '');
I want to replace any markup between two parts of the number with above regex, but it does not seem to work. I'm not sure if it is regex syntax that's wrong because I tried simpler one such as '%[^0-9]%' just to test but it didn't work either. Does anyone know how can I achieve this?
You can use PATINDEX
to find the first index of the pattern (string's) occurrence. Then use STUFF to stuff another string into the pattern(string) matched.
Loop through each row. Replace each illegal characters with what you want. In your case replace non numeric with blank. The inner loop is if you have more than one illegal character in a current cell that of the loop.
DECLARE #counter int
SET #counter = 0
WHILE(#counter < (SELECT MAX(ID_COLUMN) FROM Table))
BEGIN
WHILE 1 = 1
BEGIN
DECLARE #RetVal varchar(50)
SET #RetVal = (SELECT Column = STUFF(Column, PATINDEX('%[^0-9.]%', Column),1, '')
FROM Table
WHERE ID_COLUMN = #counter)
IF(#RetVal IS NOT NULL)
UPDATE Table SET
Column = #RetVal
WHERE ID_COLUMN = #counter
ELSE
break
END
SET #counter = #counter + 1
END
Caution: This is slow though! Having a varchar column may impact. So using LTRIM RTRIM may help a bit. Regardless, it is slow.
Credit goes to this StackOverFlow answer.
EDIT
Credit also goes to #srutzky
Edit (by #Tmdean)
Instead of doing one row at a time, this answer can be adapted to a more set-based solution. It still iterates the max of the number of non-numeric characters in a single row, so it's not ideal, but I think it should be acceptable in most situations.
WHILE 1 = 1 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, '')
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 BREAK;
END;
You can also improve efficiency quite a lot if you maintain a bit column in the table that indicates whether the field has been scrubbed yet. (NULL represents "Unknown" in my example and should be the column default.)
DECLARE #done bit = 0;
WHILE #done = 0 BEGIN
WITH q AS
(SELECT ID_Column, PATINDEX('%[^0-9.]%', Column) AS n
FROM Table
WHERE COALESCE(Scrubbed_Column, 0) = 0)
UPDATE Table
SET Column = STUFF(Column, q.n, 1, ''),
Scrubbed_Column = 0
FROM q
WHERE Table.ID_Column = q.ID_Column AND q.n != 0;
IF ##ROWCOUNT = 0 SET #done = 1;
-- if Scrubbed_Column is still NULL, then the PATINDEX
-- must have given 0
UPDATE table
SET Scrubbed_Column = CASE
WHEN Scrubbed_Column IS NULL THEN 1
ELSE NULLIF(Scrubbed_Column, 0)
END;
END;
If you don't want to change your schema, this is easy to adapt to store intermediate results in a table valued variable which gets applied to the actual table at the end.
Instead of stripping out the found character by its sole position, using Replace(Column, BadFoundCharacter, '') could be substantially faster. Additionally, instead of just replacing the one bad character found next in each column, this replaces all those found.
WHILE 1 = 1 BEGIN
UPDATE dbo.YourTable
SET Column = Replace(Column, Substring(Column, PatIndex('%[^0-9.-]%', Column), 1), '')
WHERE Column LIKE '%[^0-9.-]%'
If ##RowCount = 0 BREAK;
END;
I am convinced this will work better than the accepted answer, if only because it does fewer operations. There are other ways that might also be faster, but I don't have time to explore those right now.
In a general sense, SQL Server does not support regular expressions and you cannot use them in the native T-SQL code.
You could write a CLR function to do that. See here, for example.
For those looking for a performant and easy solution and are willing to enable CLR:
CREATE database TestSQLFunctions
go
use TestSQLFunctions
go
ALTER database TestSQLFunctions set trustworthy on
EXEC sp_configure 'clr enabled', 1
RECONFIGURE WITH OVERRIDE
go
CREATE ASSEMBLY [SQLFunctions]
AUTHORIZATION [dbo]
FROM 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WITH PERMISSION_SET = SAFE
go
CREATE FUNCTION RegexReplace(
#input nvarchar(max),
#pattern nvarchar(max),
#replacement nvarchar(max)
) RETURNS nvarchar (max)
AS EXTERNAL NAME SQLFunctions.[SQLFunctions.Regex].Replace;
go
-- outputs This is a test
SELECT dbo.RegexReplace('This is a test 12345','[0-9]','')
Content of the DLL:
I stumbled across this post looking for something else but thought I'd mention a solution I use which is far more efficient - and really should be the default implementation of any function when used with a set based query - which is to use a cross applied table function. Seems the topic is still active so hopefully this is useful to someone.
Example runtime on a few of the answers so far based on running recursive set based queries or scalar function, based on 1m rows test set removing the chars from a random newid, ranges from 34s to 2m05s for the WHILE loop examples and from 1m3s to {forever} for the function examples.
Using a table function with cross apply achieves the same goal in 10s. You may need to adjust it to suit your needs such as the max length it handles.
Function:
CREATE FUNCTION [dbo].[RemoveChars](#InputUnit VARCHAR(40))
RETURNS TABLE
AS
RETURN
(
WITH Numbers_prep(Number) AS
(
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
)
,Numbers(Number) AS
(
SELECT TOP (ISNULL(LEN(#InputUnit),0))
row_number() OVER (ORDER BY (SELECT NULL))
FROM Numbers_prep a
CROSS JOIN Numbers_prep b
)
SELECT
OutputUnit
FROM
(
SELECT
substring(#InputUnit,Number,1)
FROM Numbers
WHERE substring(#InputUnit,Number,1) like '%[0-9]%'
ORDER BY Number
FOR XML PATH('')
) Sub(OutputUnit)
)
Usage:
UPDATE t
SET column = o.OutputUnit
FROM ##t t
CROSS APPLY [dbo].[RemoveChars](t.column) o
Here is a function I wrote to accomplish this based off of the previous answers.
CREATE FUNCTION dbo.RepetitiveReplace
(
#P_String VARCHAR(MAX),
#P_Pattern VARCHAR(MAX),
#P_ReplaceString VARCHAR(MAX),
#P_ReplaceLength INT = 1
)
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #Index INT;
-- Get starting point of pattern
SET #Index = PATINDEX(#P_Pattern, #P_String);
while #Index > 0
begin
--replace matching charactger at index
SET #P_String = STUFF(#P_String, PATINDEX(#P_Pattern, #P_String), #P_ReplaceLength, #P_ReplaceString);
SET #Index = PATINDEX(#P_Pattern, #P_String);
end
RETURN #P_String;
END;
[Gist][1]
[1]: https://gist.github.com/jkdba/ca13fe8f2a9855c4bdbfd0a5d3dfcda2
Edit:
Originally I had a recursive function here which does not play well with sql server as it has a 32 nesting level limit which would result in an error like the below any time you attempt to make 32+ replacements with the function. Instead of trying to make a server level change to allow more nesting (which could be dangerous like allow never ending loops) switching to a while loop makes a lot more sense.
Maximum stored procedure, function, trigger, or view nesting level exceeded (limit 32).
Wrapping the solution inside a SQL function could be useful if you want to reuse it.
I'm even doing it at the cell level, that's why I'm putting this as a different answer:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(300))
RETURNS VARCHAR(300)
BEGIN
DECLARE #str VARCHAR(300) = #string;
DECLARE #Pattern VARCHAR (20) = '%[^a-zA-Z0-9]%';
DECLARE #Len INT;
SELECT #Len = LEN(#String);
WHILE #Len > 0
BEGIN
SET #Len = #Len - 1;
IF (PATINDEX(#Pattern,#str) > 0)
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
A more speedy approach for large strings would look something like this:
CREATE FUNCTION [dbo].[fnReplaceInvalidChars] (#string VARCHAR(MAX))
RETURNS VARCHAR(MAX)
BEGIN
DECLARE #str VARCHAR(MAX) = #string;
DECLARE #Pattern VARCHAR (MAX) = '%[^a-zA-Z0-9]%';
WHILE PATINDEX(#Pattern,#str) > 0
BEGIN
SELECT #str = STUFF(#str, PATINDEX(#Pattern,#str),1,'');
END
RETURN #str
END
I've created this function to clean up a string that contained non numeric characters in a time field. The time contained question marks when they did not added the minutes, something like this 20:??. Function loops through each character and replaces the ? with a 0 :
CREATE FUNCTION [dbo].[CleanTime]
(
-- Add the parameters for the function here
#intime nvarchar(10)
)
RETURNS nvarchar(5)
AS
BEGIN
-- Declare the return variable here
DECLARE #ResultVar nvarchar(5)
DECLARE #char char(1)
-- Add the T-SQL statements to compute the return value here
DECLARE #i int = 1
WHILE #i <= LEN(#intime)
BEGIN
SELECT #char = CASE WHEN substring(#intime,#i,1) like '%[0-9:]%' THEN substring(#intime,#i,1) ELSE '0' END
SELECT #ResultVar = concat(#ResultVar,#char)
set #i = #i + 1
END;
-- Return the result of the function
RETURN #ResultVar
END
I think this solution is faster and simple. I use always CTE/recursive because WHILE is so slow on SQL Server.
I use it in projects I work with and large databases.
/*
Function: dbo.kSql_ReplaceRegExp
Create Date: 20.02.2021
Author: Karcan Ozbal
Description: The given string value will be replaced according to the given regexp/pattern.
Parameter(s): #Value : Value/Text to REPLACE.
#RegExp : The regexp/pattern to be used for REPLACE operation.
Usage: select dbo.kSql_ReplaceRegExp('2T3EST5','%[0-9]%')
Output: 'TEST'
*/
ALTER FUNCTION [dbo].[kSql_ReplaceRegExp](
#Value nvarchar(max),
#RegExp nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #Result nvarchar(max)
;WITH CTE AS (
SELECT NUM = 1, VALUE = #Value, IDX = PATINDEX(#RegExp, #Value)
UNION ALL
SELECT NUM + 1, VALUE = REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''), IDX = PATINDEX(#RegExp, REPLACE(VALUE, SUBSTRING(VALUE,IDX,1),''))
FROM CTE
WHERE IDX > 0
)
SELECT TOP(1) #Result = VALUE
FROM CTE
ORDER BY NUM DESC
OPTION (maxrecursion 0)
RETURN #Result
END
If you are doing this just for a parameter coming into a Stored Procedure, you can use the following:
declare #badIndex int
set #badIndex = PatIndex('%[^0-9]%', #Param)
while #badIndex > 0
set #Param = Replace(#Param, Substring(#Param, #badIndex, 1), '')
set #badIndex = PatIndex('%[^0-9]%', #Param)
I thought this was clearer:
ALTER FUNCTION [dbo].[func_ReplaceChars](
#Value nvarchar(max),
#Chars nvarchar(50)
)
RETURNS nvarchar(max)
AS
BEGIN
DECLARE #cLen int = len(#Chars);
DECLARE #curChar int = 0;
WHILE #curChar<#cLen
BEGIN
set #Value = replace(#Value,substring(#Chars,#curChar,1),'');
set #curChar = #curChar + 1;
END;
RETURN #Value
END
I'm using this code similar to several codes above:
DROP FUNCTION [dbo].[fnCleanString]
GO
CREATE FUNCTION [dbo].[fnCleanString] (#input VARCHAR(max), #Pattern
VARCHAR (20))
RETURNS VARCHAR(max)
BEGIN
DECLARE #str VARCHAR(max) = #input;
DECLARE #Len INT;
DECLARE #INDEX INT;
SELECT #Len = LEN(#input);
WHILE #Len > 0
BEGIN
SET #INDEX = PATINDEX(#Pattern,#str);
IF (#INDEX > 0)
BEGIN
SET #str=REPLACE(#str,SUBSTRING(#str,#INDEX, 1), '');
END
ELSE
BEGIN
BREAK;
END
END
RETURN #str
END
You can use it like this:
SELECT CleanName = dbo.[fnCleanString](Name, '%[0-9]%') from YourTable
I think a simpler and faster approach is iterate by each character of the alphabet:
DECLARE #i int
SET #i = 0
WHILE(#i < 256)
BEGIN
IF char(#i) NOT IN ('0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '.')
UPDATE Table SET Column = replace(Column, char(#i), '')
SET #i = #i + 1
END
i'm pretty new to SQL and stored procedures and i'm a bit stuck - so any help would be appreciated
how do i loop through each row and assign it the random value i'm generating?
Here is my Storedproc:
CREATE PROCEDURE StoredProc8
AS
BEGIN
DECLARE #total INT
DECLARE #Count INT = 0
DECLARE #Random INT = 0
SELECT #total = COUNT(CustomerID) FROM Customers
WHILE(#Count<= #total)
BEGIN
SELECT #Random = 2 * RAND()
EXEC ('update Customers set col1= ' + #Random )
SELECT #Count = #Count+1
END
END
If you simple need to assign 0 or 1 randomly - you can use RAND() with random seed:
UPDATE Customers SET COL1 = RAND(CHECKSUM(NEWID()))*2
Demo: http://sqlfiddle.com/#!3/31699/9
suppose I have a number 62. This is composed of 2 digits
How can add 2 digit together divide by 10 and if result = something like 6.2 just take reminder
declare #Number int,#Result int
set #Number =62
if len(#Number) > 1
set #Result=????=--Add 6 and 2 =8
set #result=#result % 10 --Mod operator
print #result
-- the result should be 2 in this case
what Am I doing wrong?
Thanks a lot
set #tens = floor(#Number / 10);
set #ones = #number - #tens;
set #Result = #tens + #ones;
Or use left and right to access substrings.
I have no way to try it:
DECLARE #Number VARCHAR(2)
SET #Number = '62'
declare #firstNum INT, #secondNum INT
SET #firstNum = CAST(SUBSTRING(#Number, 1, 1) AS INT)
SET #secondNum = CAST(SUBSTRING(#Number, 2, 1) AS INT)
DECLARE #Result int
SET #Result = (#firstNum + #secondNum) % 10
I think #Number % 10is what you are looking for. It returns the last digit of any number. 62 -> 2, 97 -> 7, etc...
Update:
I may have misunderstood the question. Maybe you want 10 % ((#Number / 10) + #Number) instead.
(#Number / 10) + #Number is the sum of the digits of a two-digit number.
If the number is always two digits, then I would use LEFT(#Number,1) and RIGHT(#Number,1) to access each digit. If it's not, comment back and I'll give you some help with the loop you'll need.
More pressing an issue is that you're expecting 2, but the result of 8 mod 10 is 8. If you're looking for 2, the sum is 10 mod 8 (10 % #Result).
In any case, hit me back if this doesn't answer exactly what you wanted.
Try this
declare #myNumber float,#result float
set #myNumber = 62
select
#result = (case when len(#myNumber) = 1 then #myNumber
else (#myNumber /10) + cast(#myNumber as int) % 10 end)
select [Output] = STUFF(
cast(#result as varchar(50))
,1
,charindex('.',cast(#result as varchar(50)))
,'')
Output
2
Hope this helps
And if u just want to add 2 numbers try this(though it has been implemented above)
declare #myNumber int,#result int
set #myNumber = 62
select
Result = (case when len(#myNumber) = 1 then #myNumber
else (#myNumber /10) + #myNumber % 10 end)
Result
8
So I guess you want something like this - parse the string representing your number, adding up the individual digits as integer values. This gives you a total result at the end - then you do whatever you need to do with that. This code works for any length of string (up to 50 characters = 50 digits in your original number):
DECLARE #Number INT
SET #Number = 62
DECLARE #NumString VARCHAR(50)
SET #NumString = CAST(#Number AS VARCHAR(50))
DECLARE #Index INT
SET #Index = 1
DECLARE #Sum INT
SET #Sum = 0
WHILE #Index <= LEN(#NumString)
BEGIN
SET #Sum = #Sum + CAST(SUBSTRING(#NumString, #Index, 1) AS INT)
SET #Index = #Index + 1
END
SELECT #Sum AS 'Sum of all digits'
With the initial value of "62" for #Number, I get a result of 8 - now you can continue on using that value.
If you need this function often, I would probably encapsulate it into a user-defined function so you can call it from everywhere in your code.