Replace Last character in SQL Server 2008 - sql

I am working with SQL server 2008, and facing problem about character replacement.
If I use
SELECT REPLACE(MYWORD,0,1) FROM MYTABLE
It is replacing all 0 into 1, I just want to replace Last character Like MYWORD = "ERMN0" so it will be MYWORD = "ERMN1"

using STUFF, which, IMO, ends up being most readable:
DECLARE #MyWORD VARCHAR(20) = 'ABCDEF123'
SELECT STUFF(#MyWORD, LEN(#MyWORD), 1, '2')
output:
ABCDEF122

You may use combination of LEFT, RIGHT, and CASE.
You need to use CASE to check the most RIGHT character whether it's a 0 or not and replace it with 1. And at last, combine it with the LEFT part (after being separated from the last character) of the MYWORD string.
However, depending on your requirement, it may have a drawback.
When there is a word ending with 10, it would also be replaced.
SELECT LEFT(MYWORD,LEN(MYWORD)-1) + CASE RIGHT(MYWORD,1) WHEN '0' THEN '1' ELSE RIGHT(MYWORD,1) END

Try this.
SELECT LEFT('ERMN0', Len('ERMN0')-1)
+ Replace(RIGHT('ERMN0', 1), 0, 1)
OUTPUT : ERMN1
In your case
SELECT LEFT(MYWORD, Len(MYWORD)-1)
+ Replace(RIGHT(MYWORD, 1), 0, 1) as [REPLACED] FROM MYTABLE

Try this
SELECT SUBSTRING(MYWORD, 1, LEN(MYWORD) - 1) +
REPLACE(SUBSTRING(MYWORD, LEN(MYWORD), LEN(MYWORD)), 0, 1) FROM MYTABLE

This will work
SELECT LEFT ('ERMN0' , Len('ERMN0') -1 ) + REPLACE(Right('ERMN0', 1), '0','1')
Or in your case
SELECT LEFT (MYWORD , Len(MYWORD) -1 ) + REPLACE(Right(MYWORD, 1), '0','1') AS MYWORD FROM MYTABLE

this is also use full to replace letters from end
It is used from replacing characters from end 1,2 or N
Declare #Name nvarchar(20) = 'Bollywood'
select #Name = REPLACE(#Name, SUBSTRING(#Name, len(#Name) - 1, 2), 'as')
SELECT #Name
output is "Bollywoas"
Here best part is you can repalce as many character from last you needed.

Related

SQL Server function with count doesn't work as intended

I am working on a project for class & course management. I want to create a function that automatically creates a course code based on the course name and then adds a number based on whether there already exist a course name with the same first 4 characters.
Here is what my function looks like:
CREATE FUNCTION [dbo].[fxGenerateCourseCode]
(#course_name VARCHAR(50))
RETURNS VARCHAR(8)
AS
BEGIN
DECLARE #course_code VARCHAR(8), #count_course INT
SET #count_course = (SELECT COUNT(1) FROM dbo.COURSE_TB
WHERE SUBSTRING(course_name, 1, 4) = SUBSTRING(#course_name, 1, 4)) + 1
SELECT #course_code = UPPER(SUBSTRING(#course_name, 1, 4)) + ' - ' + '00' + CAST(#count_course AS VARCHAR)
RETURN #course_code
END
The problem is, whenever I execute this function, it is not returning the right count values and it doesn't include the 00 either.
For example, executing this:
SELECT dbo.fxGenerateCourseCode('french')
returns FREN-0 rather than FREN-001 (which is what I expect).
However, when I execute the code manually, it gives me the result that I expect:
DECLARE #course_code VARCHAR(8), #count_course INT, #course_name VARCHAR(50)
SET #course_name = 'french'
SET #count_course = (SELECT COUNT(1) + 1 FROM dbo.PARAMETRES_COURS
WHERE SUBSTRING(nom_cours, 1, 4) = SUBSTRING(#course_name, 1, 4))
SELECT course_code = UPPER(SUBSTRING(#course_name, 1, 4)) + ' - ' + '00' + CAST(#count_course AS VARCHAR)
Result: FREN-001.
I have tried changing the way I write the function with not much result. Can someone help me know where the problem here is coming from? Am I doing something wrong in the function?
Thank you
I suspect that you are putting the value into a column with a length of 6. That said, the problem would appear to be the spaces around the hyphen. Also, always include a length for varchar, because the defaut length varies by context and it might not do what you want:
You can simplify your code and make it more accurate:
declare #course_code varchar(8), #count_course int;
select #count_course = count(1) + 1
from dbo.COURSE_TB
where left(course_name, 4) = left(#course_name, 4);
select #course_code = concat(left(upper(#course_name, 4)), '-', '00', #count_course);
Note that this still doesn't do what you want, which is padding the number. This version only works for 9 courses. You seem to want:
select #course_code = concat(left(upper(#course_name, 4)), '-',
right(concat('00', #count_course), 3)
);
Note the use of the concat() function, so you don't have to worry about lengths of strings when doing a conversion.

Joining on numeric part of string

It's been a while...I'd like to get your advice on the most efficient way to join on only the number part of a field that may be prefixed and/or suffixed with up to 2 letters. Here's a simplified snippet of what I'm trying to do:
SELECT a, b, c
FROM table 1 t1
LEFT JOIN table 2 t2 ON t1.PolicyCode = t2.sPolicyID,
Where t2.sPolicyID could begin and/or end with up to 2 letters. Some examples: TG73100, S7286674, 2344506R, etc. We only want to join to just its numeric part in between the letters, i.e. 73100, 7286674 or 2344506 from the examples.
Could someone please advise on a simple way of doing this?
Here is one way:
LEFT JOIN table 2 t2 ON t1.PolicyCode =
LEFT(SUBSTRING(t2.sPolicyID, PATINDEX('%[0-9]%', t2.sPolicyID), 50),
PATINDEX('%[^0-9]%',
SUBSTRING(t2.sPolicyID, PATINDEX('%[0-9]%', t2.sPolicyID), 50)
+ 'a') -1)
To break this down, there are 4 main parts.
1: Find the position of the first number with PATINDEX:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[0-9]%', #spolicyID)
--Returns 3
2: Use SUBSTRING() to cut off everything before the first letter:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50)
--Returns 123123xx
If we hardcoded the 3 that we know is returned from the first part, it would look like this:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT SUBSTRING(#spolicyID, 3), 50)
--50 is the number of characters to extract, set to something
--higher than the max string length to be safe
Of course, we don't want to hardcode it since it can change, but that makes seeing the different functions a bit easier.
3: Find the position of the next letter using PATINDEX again:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[^0-9]%', SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50) + 'a')
--Returns 7 since it is looking at 123123xx
--The first x is in the 7th position
Note that we added an a onto the string. This is because if we had a string with no letters at the end, it would throw an error as the length 0 would be returned to SUBSTRING. You could add any letter or letters to the end and it would work, we are just making sure there is at least one. Try removing the + 'a' and using a string like xx123123 to see the error.
If we hardcoded the 123123xx from step 2 it would look like this (again just for easy example):
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
SELECT PATINDEX('%[^0-9]%', '123123xx' + 'a')
4: Use LEFT() to return everything before the trailing letters, leaving us with only the numbers in between:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
LEFT(SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50),PATINDEX('%[^0-9]%', SUBSTRING(#spolicyID, PATINDEX('%[0-9]%', #spolicyID), 50) + 'a') -1)
--Need to add `-1` because step 3 PATINDEX returns 7
--as the position of first trailing letter, and
--we want the 6 characters before that
And again hardcoded from step 2 and 3 for easy viewing:
DECLARE #spolicyID VARCHAR(20) = 'xx123123xx'
LEFT('123123xx', 7-1)

How to remove the first character if it is a specific character in SQL

I currently have a table Telephone it has entries like the following:
9073456789101
+773456789101
0773456789101
What I want to do is remove only the 9 from the start of all the entries that have a 9 there but leave the others as they are.
any help would be greatly appreciated.
While all other answer are probably also working, I'd suggest to try and use STUFF function to easily replace a part of the string.
UPDATE Telephone
SET number = STUFF(number,1,1,'')
WHERE number LIKE '9%'
SQLFiddle DEMO
Here is the code and a SQLFiddle
SELECT CASE
WHEN substring(telephone_number, 1, 1) <> '9'
THEN telephone_number
ELSE substring(telephone_number, 2, LEN(telephone_number))
END
FROM Telephone
Update Telephone set number = RIGHT(number,LEN(number)-1) WHERE number LIKE '9%';
I recently solved a similar problem with a combination of RIGHT(), LEN() & PATINDEX(). PATINDEX will return the integer 1 when it finds a 9 as the first character and 0 otherwise. This method allows all records to be returned at once without a CASE WHEN statement.
SELECT
RIGHT(number, LEN(number) - PATINDEX('9%', number))
FROM Telephone
UPDATE dbo.Telephone
SET column_name = SUBSTRING(column_name, 2, 255)
WHERE column_name LIKE '9%';
Stuff is a great function for this. However, using it with an update statement with a where clause is great, but what if I was doing an insert, and I needed all of the rows inserted in one pass. The below will remove the first character if it is a period, does not use the slower case statement, and converts nulls to an empty string.
DECLARE #Attachment varchar(6) = '.GIF',
#Attachment2 varchar(6)
SELECT
#Attachment2 = ISNULL(ISNULL(NULLIF(LEFT(#Attachment, 1), '.'), '') + STUFF(#Attachment, 1, 1, ''), '')
SELECT
#Attachment2
DECLARE #STR nvarchar(200) = 'TEST'
SET #STR = STUFF(#STR,1,1,'')
PRINT #STR
Result will be "EST"
You can use replace in select statement instead of where or update
SELECT REPLACE(REPLACE('_'+number,'_9',''),'_','') FROM #tbl

Uppercase first two characters in a column in a db table

I've got a column in a database table (SQL Server 2005) that contains data like this:
TQ7394
SZ910284
T r1534
su8472
I would like to update this column so that the first two characters are uppercase. I would also like to remove any spaces between the first two characters. So T q1234 would become TQ1234.
The solution should be able to cope with multiple spaces between the first two characters.
Is this possible in T-SQL? How about in ANSI-92? I'm always interested in seeing how this is done in other db's too, so feel free to post answers for PostgreSQL, MySQL, et al.
Here is a solution:
EDIT: Updated to support replacement of multiple spaces between the first and the second non-space characters
/* TEST TABLE */
DECLARE #T AS TABLE(code Varchar(20))
INSERT INTO #T SELECT 'ab1234x1' UNION SELECT ' ab1234x2'
UNION SELECT ' ab1234x3' UNION SELECT 'a b1234x4'
UNION SELECT 'a b1234x5' UNION SELECT 'a b1234x6'
UNION SELECT 'ab 1234x7' UNION SELECT 'ab 1234x8'
SELECT * FROM #T
/* INPUT
code
--------------------
ab1234x3
ab1234x2
a b1234x6
a b1234x5
a b1234x4
ab 1234x8
ab 1234x7
ab1234x1
*/
/* START PROCESSING SECTION */
DECLARE #s Varchar(20)
DECLARE #firstChar INT
DECLARE #secondChar INT
UPDATE #T SET
#firstChar = PATINDEX('%[^ ]%',code)
,#secondChar = #firstChar + PATINDEX('%[^ ]%', STUFF(code,1, #firstChar,'' ) )
,#s = STUFF(
code,
1,
#secondChar,
REPLACE(LEFT(code,
#secondChar
),' ','')
)
,#s = STUFF(
#s,
1,
2,
UPPER(LEFT(#s,2))
)
,code = #s
/* END PROCESSING SECTION */
SELECT * FROM #T
/* OUTPUT
code
--------------------
AB1234x3
AB1234x2
AB1234x6
AB1234x5
AB1234x4
AB 1234x8
AB 1234x7
AB1234x1
*/
UPDATE YourTable
SET YourColumn = UPPER(
SUBSTRING(
REPLACE(YourColumn, ' ', ''), 1, 2
)
)
+
SUBSTRING(YourColumn, 3, LEN(YourColumn))
UPPER isn't going to hurt any numbers, so if the examples you gave are completely representative, there's not really any harm in doing:
UPDATE tbl
SET col = REPLACE(UPPER(col), ' ', '')
The sample data only has spaces and lowercase letters at the start. If this holds true for the real data then simply:
UPPER(REPLACE(YourColumn, ' ', ''))
For a more specific answer I'd politely ask you to expand on your spec, otherwise I'd have to code around all the other possibilities (e.g. values of less than three characters) without knowing if I was overengineering my solution to handle data that wouldn't actually arise in reality :)
As ever, once you've fixed the data, put in a database constraint to ensure the bad data does not reoccur e.g.
ALTER TABLE YourTable ADD
CONSTRAINT YourColumn__char_pos_1_uppercase_letter
CHECK (ASCII(SUBSTRING(YourColumn, 1, 1)) BETWEEN ASCII('A') AND ASCII('Z'));
ALTER TABLE YourTable ADD
CONSTRAINT YourColumn__char_pos_2_uppercase_letter
CHECK (ASCII(SUBSTRING(YourColumn, 2, 1)) BETWEEN ASCII('A') AND ASCII('Z'));
#huo73: yours doesn't work for me on SQL Server 2008: I get 'TRr1534' instead of 'TR1534'.
update Table set Column = case when len(rtrim(substring (Column , 1 , 2))) < 2
then UPPER(substring (Column , 1 , 1) + substring (Column , 3 , 1)) + substring(Column , 4, len(Column)
else UPPER(substring (Column , 1 , 2)) + substring(Column , 3, len(Column) end
This works on the fact that if there is a space then the trim of that part of string would yield length less than 2 so we split the string in three and use upper on the 1st and 3rd char. In all other cases we can split the string in 2 parts and use upper to make the first two chars to upper case.
If you are doing an UPDATE, I would do it in 2 steps; first get rid of the space (RTRIM on a SUBSTRING), and second do the UPPER on the first 2 chars:
// uses a fixed column length - 20-odd in this case
UPDATE FOO
SET bar = RTRIM(SUBSTRING(bar, 1, 2)) + SUBSTRING(bar, 3, 20)
UPDATE FOO
SET bar = UPPER(SUBSTRING(bar, 1, 2)) + SUBSTRING(bar, 3, 20)
If you need it in a SELECT (i.e. inline), then I'd be tempted to write a scalar UDF

Removing leading zeroes from a field in a SQL statement

I am working on a SQL query that reads from a SQLServer database to produce an extract file. One of the requirements to remove the leading zeroes from a particular field, which is a simple VARCHAR(10) field. So, for example, if the field contains '00001A', the SELECT statement needs to return the data as '1A'.
Is there a way in SQL to easily remove the leading zeroes in this way? I know there is an RTRIM function, but this seems only to remove spaces.
select substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
select replace(ltrim(replace(ColumnName,'0',' ')),' ','0')
You can use this:
SELECT REPLACE(LTRIM(REPLACE('000010A', '0', ' ')),' ', '0')
I had the same need and used this:
select
case
when left(column,1) = '0'
then right(column, (len(column)-1))
else column
end
select substring(substring('B10000N0Z', patindex('%[0]%','B10000N0Z'), 20),
patindex('%[^0]%',substring('B10000N0Z', patindex('%[0]%','B10000N0Z'),
20)), 20)
returns N0Z, that is, will get rid of leading zeroes and anything that comes before them.
If you want the query to return a 0 instead of a string of zeroes or any other value for that matter you can turn this into a case statement like this:
select CASE
WHEN ColumnName = substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
THEN '0'
ELSE substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
END
In case you want to remove the leading zeros from a string with a unknown size.
You may consider using the STUFF command.
Here is an example of how it would work.
SELECT ISNULL(STUFF(ColumnName
,1
,patindex('%[^0]%',ColumnName)-1
,'')
,REPLACE(ColumnName,'0','')
)
See in fiddler various scenarios it will cover
https://dbfiddle.uk/?rdbms=sqlserver_2012&fiddle=14c2dca84aa28f2a7a1fac59c9412d48
You can try this - it takes special care to only remove leading zeroes if needed:
DECLARE #LeadingZeros VARCHAR(10) ='-000987000'
SET #LeadingZeros =
CASE WHEN PATINDEX('%-0', #LeadingZeros) = 1 THEN
#LeadingZeros
ELSE
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
END
SELECT #LeadingZeros
Or you can simply call
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
Here is the SQL scalar value function that removes leading zeros from string:
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Vikas Patel
-- Create date: 01/31/2019
-- Description: Remove leading zeros from string
-- =============================================
CREATE FUNCTION dbo.funRemoveLeadingZeros
(
-- Add the parameters for the function here
#Input varchar(max)
)
RETURNS varchar(max)
AS
BEGIN
-- Declare the return variable here
DECLARE #Result varchar(max)
-- Add the T-SQL statements to compute the return value here
SET #Result = #Input
WHILE LEFT(#Result, 1) = '0'
BEGIN
SET #Result = SUBSTRING(#Result, 2, LEN(#Result) - 1)
END
-- Return the result of the function
RETURN #Result
END
GO
To remove the leading 0 from month following statement will definitely work.
SELECT replace(left(Convert(nvarchar,GETDATE(),101),2),'0','')+RIGHT(Convert(nvarchar,GETDATE(),101),8)
Just Replace GETDATE() with the date field of your Table.
To remove leading 0, You can multiply number column with 1
Eg: Select (ColumnName * 1)
select CASE
WHEN TRY_CONVERT(bigint,Mtrl_Nbr) = 0
THEN ''
ELSE substring(Mtrl_Nbr, patindex('%[^0]%',Mtrl_Nbr), 18)
END
you can try this
SELECT REPLACE(columnname,'0','') FROM table
I borrowed from ideas above. This is neither fast nor elegant. but it is accurate.
CASE
WHEN left(column, 3) = '000' THEN right(column, (len(column)-3))
WHEN left(column, 2) = '00' THEN right(a.column, (len(column)-2))
WHEN left(column, 1) = '0' THEN right(a.column, (len(column)-1))
ELSE
END
select ltrim('000045', '0') from dual;
LTRIM
-----
45
This should do.