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I want to know a flexible way to extract the string between two '-'. The issue is that '-' may or may not exist in the source string. I have the following code which works fine when '-' exists twice in the source string, marking the start and end of the extracted string. But it throws an error "Invalid length parameter passed to the LEFT or SUBSTRING function" when there is only one '-' or none at all or if the string is blank. Can someone please help? Thanks
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
Desired Output: ExtractThis
If there is one dash only e.g. 'BLAH90-ExtractThisWOW' then the output should be everything after the first dash i.e. ExtractThisWOW. If there are no dashes then the string will have a blank space instead e.g. 'BLAH90 ExtractThisWOW' and should return everything after the blank space i.e. ExtractThisWOW.
You can try something like this.
When there is no dash, it starts at the space if there is one or take the whole string if not.
Then I look if there is only one dash or 2
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
declare #dash_pos integer = CHARINDEX('-',#string)
SELECT CASE
WHEN #dash_pos = 0 THEN
RIGHT(#string,LEN(#string)-CHARINDEX(' ',#string))
ELSE (
CASE
WHEN #dash_pos = LEN(#string)-CHARINDEX('-',REVERSE(#string))+1
THEN RIGHT(#string,LEN(#string)-#dash_pos)
ELSE SUBSTRING(#string,#dash_pos+1, CHARINDEX('-',#string,#dash_pos+1) -
#dash_pos -1)
END
)
END as My_String
Try this. If there are two dashes, it'll take what is inside. If there is only one or none, it'll keep the original string.
declare #string varchar(100) = 'BLAH-90ExtractThisWOW'
declare #dash_index1 int = case when #string like '%-%' then CHARINDEX('-', #string) else -1 end
declare #dash_index2 int = case when #string like '%-%'then len(#string) - CHARINDEX('-', reverse(#string)) + 1 else -1 end
SELECT case
when #dash_index1 <> #dash_index2 then SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1)
else #string end
as My_String
Take your existing code:
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
insert one line, like so:
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SET #string = #string + '--'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
and you're done. (If NULL, you will get NULL returned. Also, this will return all data based on the FIRST dash found in the string, regardless of however many dashes are in the string.)
In SQL how can I remove (from displaying on my report no deleting from database) the first 3 characters (CN=) and everything after the comma that is followed by "OU" so that I am left with the name and last name in the same column? for example:
CN=Tom Chess,OU=records,DC=1234564786_data for testing, 1234567
CN=Jack Bauer,OU=records,DC=1234564786_data for testing, 1234567
CN=John Snow,OU=records,DC=1234564786_data for testing, 1234567
CN=Anna Rodriguez,OU=records,DC=1234564786_data for testing, 1234567
Desired display:
Tom Chess
Jack Bauer
John Snow
Anna Rodriguez
I tried playing with TRIM but I don't know how to do it without declaring the position and with names and last names having different lengths I really don't know how to handle that.
Thank you in advance
Update: I wonder about an approach of using Locate to match the position of the comma and then feed that to a sub-string. Not sure if a approach like would work and not sure how to put the syntax together. What do you think? will it be a feasible approach?
You can try this one SUBSTRING(ColumnName, 4, CHARINDEX(',', ColumnName) - 4)
In Postgres, you could use split_part() assuming no name contains a ,
select substr(split_part(the_column, ',', 1), 4)
from ...
Db2 11.x for LUW:
with tab (str) as (values
' CN = Tom Chess , OU = records,DC=1234564786_data for testing, 1234567'
, 'CN=Jack Bauer,OU=records,DC=1234564786_data for testing, 1234567'
, 'CN=John Snow,OU=records,DC=1234564786_data for testing, 1234567'
, 'CN=Anna Rodriguez,OU=records,DC=1234564786_data for testing, 1234567'
)
select REGEXP_REPLACE(str, '^\s*CN\s*=\s*(.*)\s*,\s*OU\s*=.*', '\1')
from tab;
Note, that such a regex pattern allows an arbitrary number of spaces as in the 1-st record of example above.
In Oracle 11g, it might work.
REGEXP_SUBSTR(REGEXP_SUBSTR(COLUMN_NAME, '[^CN=]+',1,1),'[^,OU]+',1,1)
I think there has to be a loop to handle this. Here's SQL Server function that will parse this out. (I know the question didn't specify SQL Server, but it's an example of how it can be done.)
select dbo.ScrubFieldValue(value) from table will return what you're looking for
CREATE FUNCTION ScrubFieldValue
(
#Input varchar(8000)
)
RETURNS varchar(8000)
AS
BEGIN
DECLARE #retval varchar(8000)
DECLARE #charidx int
DECLARE #remaining varchar(8000)
DECLARE #current varchar(8000)
DECLARE #currentLength int
select #retval = ''
select #remaining = #Input
select #charidx = CHARINDEX('CN=', #remaining,2)
while(LEN(#remaining) > 0)
BEGIN
--strip current row from remaining
if (#charidx > 0)
BEGIN
select #current = SUBSTRING(#remaining, 1, #charidx - 1)
END
else
BEGIN
select #current = #remaining
END
select #currentLength = LEN(#current)
-- get current name
select #current = SUBSTRING(#current, 4, CHARINDEX(',OU', #current)-4)
select #retval = #retval + #current + ' '
-- strip off current from remaining
select #remaining =substring(#remaining,#currentLength + 1,
LEN(#remaining) - #currentLength)
select #charidx = CHARINDEX('CN=', #remaining,2)
END
RETURN #retval
END
On my version of DB2 for Z/OS CHARINDEX throws a syntax error. Here are two ways to work around that.
SUBSTRING(ColumnName, 4, INSTR(ColumnName,',',1) - 4)
SUBSTRING(ColumnName, 4, LOCATE_IN_STRING(ColumnName,',') - 4)
I should add that the version is V12R1
If input str is wellformed (i.e. looks like your sample data without any additional tokens such as space), you could use something like:
substr(str,locate('CN=', str)+length('CN='), locate(',', str)-length('CN=')-1)
If your Db2 version support REGEXP, that's a better choice.
String = "Vishal Hariharan" .. Replace first "H" with "A" SQL
I have one scenario where I want replace only first "H" character among others "H" positions and remaining keep as it is.
There are many simplistic answers for this on Stack Overflow that don't take into account the case where the character being searched for doesn't exist in the string. In these cases, the SQL will return NULL if the replacement character does not exist in the string. Try this instead:
declare #name varchar(max), #ToReplace varchar(max), #ReplaceWith varchar(max)
set #name = 'Vishal Hariharan'
set #ToReplace = 'H'
set #ReplaceWith = 'A'
select #name
select stuff(#name, charindex(#ToReplace,#name),1, CASE WHEN charindex(#ToReplace,#name) <> 0 THEN #ReplaceWith ELSE #ToReplace END)
The CASE statement simply checks if the character exists in the string, and if it doesn't, it replaces the matching character with itself.
DECLARE #findChar varchar(max)='H'
DECLARE #RepalceCharacter varchar(max)='A'
DECLARE #OriginText varchar(max)='Vishal Hariharan'
IF (CharIndex(#findChar, #OriginText)<>0)
SELECT Stuff(#OriginText, CharIndex(#findChar, #OriginText), Len(#findChar), #RepalceCharacter)
ELSE
SELECT #OriginText
I need a SQL query to get the value between two known strings (the returned value should start and end with these two strings).
An example.
"All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought."
In this case the known strings are "the dog" and "immediately". So my query should return "the dog had been very bad and required harsh punishment immediately"
I've come up with this so far but to no avail:
SELECT SUBSTRING(#Text, CHARINDEX('the dog', #Text), CHARINDEX('immediately', #Text))
#Text being the variable containing the main string.
Can someone please help me with where I'm going wrong?
The problem is that the second part of your substring argument is including the first index.
You need to subtract the first index from your second index to make this work.
SELECT SUBSTRING(#Text, CHARINDEX('the dog', #Text)
, CHARINDEX('immediately',#text) - CHARINDEX('the dog', #Text) + Len('immediately'))
I think what Evan meant was this:
SELECT SUBSTRING(#Text, CHARINDEX(#First, #Text) + LEN(#First),
CHARINDEX(#Second, #Text) - CHARINDEX(#First, #Text) - LEN(#First))
An example is this: You have a string and the character $
String :
aaaaa$bbbbb$ccccc
Code:
SELECT SUBSTRING('aaaaa$bbbbb$ccccc',CHARINDEX('$','aaaaa$bbbbb$ccccc')+1, CHARINDEX('$','aaaaa$bbbbb$ccccc',CHARINDEX('$','aaaaa$bbbbb$ccccc')+1) -CHARINDEX('$','aaaaa$bbbbb$ccccc')-1) as My_String
Output:
bbbbb
You need to adjust for the LENGTH in the SUBSTRING. You were pointing it to the END of the 'ending string'.
Try something like this:
declare #TEXT varchar(200)
declare #ST varchar(200)
declare #EN varchar(200)
set #ST = 'the dog'
set #EN = 'immediately'
set #TEXT = 'All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought.'
SELECT SUBSTRING(#Text, CHARINDEX(#ST, #Text), (CHARINDEX(#EN, #Text)+LEN(#EN))-CHARINDEX(#ST, #Text))
Of course, you may need to adjust it a bit.
I had a similar need to parse out a set of parameters stored within an IIS logs' csUriQuery field, which looked like this: id=3598308&user=AD\user¶meter=1&listing=No needed in this format.
I ended up creating a User-defined function to accomplish a string between, with the following assumptions:
If the starting occurrence is not found, a NULL is returned, and
If the ending occurrence is not found, the rest of the string is returned
Here's the code:
CREATE FUNCTION dbo.str_between(#col varchar(max), #start varchar(50), #end varchar(50))
RETURNS varchar(max)
WITH EXECUTE AS CALLER
AS
BEGIN
RETURN substring(#col, charindex(#start, #col) + len(#start),
isnull(nullif(charindex(#end, stuff(#col, 1, charindex(#start, #col)-1, '')),0),
len(stuff(#col, 1, charindex(#start, #col)-1, ''))+1) - len(#start)-1);
END;
GO
For the above question, the usage is as follows:
DECLARE #a VARCHAR(MAX) = 'All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought.'
SELECT dbo.str_between(#a, 'the dog', 'immediately')
-- Yields' had been very bad and required harsh punishment '
Try this and replace '[' & ']' with your string
SELECT SUBSTRING(#TEXT,CHARINDEX('[',#TEXT)+1,(CHARINDEX(']',#TEXT)-CHARINDEX('[',#TEXT))-1)
I have a feeling you might need SQL Server's PATINDEX() function. Check this out:
Usage on Patindex() function
So maybe:
SELECT SUBSTRING(#TEXT, PATINDEX('%the dog%', #TEXT), PATINDEX('%immediately%',#TEXT))
SELECT
SUBSTRING( '123#yahoo.com', charindex('#','123#yahoo.com',1) + 1, charindex('.','123#yahoo.com',1) - charindex('#','123#yahoo.com',1) - 1 )
DECLARE #Text VARCHAR(MAX), #First VARCHAR(MAX), #Second VARCHAR(MAX)
SET #Text = 'All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought.'
SET #First = 'the dog'
SET #Second = 'immediately'
SELECT SUBSTRING(#Text, CHARINDEX(#First, #Text),
CHARINDEX(#Second, #Text) - CHARINDEX(#First, #Text) + LEN(#Second))
You're getting the starting position of 'punishment immediately', but passing that in as the length parameter for your substring.
You would need to substract the starting position of 'the dog' from the charindex of 'punishment immediately', and then add the length of the 'punishment immediately' string to your third parameter. This would then give you the correct text.
Here's some rough, hacky code to illustrate the process:
DECLARE #text VARCHAR(MAX)
SET #text = 'All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought.'
DECLARE #start INT
SELECT #start = CHARINDEX('the dog',#text)
DECLARE #endLen INT
SELECT #endLen = LEN('immediately')
DECLARE #end INT
SELECT #end = CHARINDEX('immediately',#text)
SET #end = #end - #start + #endLen
SELECT #end
SELECT SUBSTRING(#text,#start,#end)
Result: the dog had been very bad and required harsh punishment immediately
Among the many options is to create a simple function.
Can keep your code cleaner.
Gives the ability to handle errors if the start or end marker/string is not present.
This function also allows for trimming leading or trailing whitespace as an option.
SELECT dbo.GetStringBetweenMarkers('123456789', '234', '78', 0, 1)
Yields:
56
--Code to create the function
USE [xxxx_YourDB_xxxx]
GO
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[GetStringBetweenMarkers] (#FullString varchar(max), #StartMarker varchar(500), #EndMarker varchar(500), #TrimLRWhiteSpace bit, #ReportErrorInResult bit)
RETURNS varchar(max)
AS
BEGIN
--Purpose is to simply return the string between 2 string markers. ew 2022-11-06
--Will perform a LTRIM and RTRIM if #TrimLRWhiteSpace = 1
--Will report errors of either marker not being found in the RETURNed string if #ReportErrorInResult = 1.
-- When #ReportErrorInResult = 0, if the start marker isn't found, will return everything from the start of the #FullString to the left of the end marker.
-- When #ReportErrorInResult = 0, if the end marker isn't found, SQL will return an error of "Invalid length parameter passed to the LEFT or SUBSTRING function."
DECLARE #ReturnString VARCHAR(max) = ''
DECLARE #StartOfStartMarker INT = CHARINDEX(#StartMarker, #FullString)
DECLARE #StartOfTarget INT = CHARINDEX(#StartMarker, #FullString) + LEN(#StartMarker)
DECLARE #EndOfTarget INT = CHARINDEX(#EndMarker, #FullString, #StartOfTarget)
--If a marker wasn't found, put that into the
IF #ReportErrorInResult = 1
BEGIN
IF #EndOfTarget = 0 SET #ReturnString = '[ERROR: EndMarker not found.]'
IF #StartOfStartMarker = 0 SET #ReturnString = '[ERROR: StartMarker not found.]'
IF #StartOfStartMarker = 0 AND #EndOfTarget = 0 SET #ReturnString = '[ERROR: Both StartMarker and EndMarker not found.]'
END
--If not reporting errors, and start marker not found (i.e. CHARINDEX = 0) we would start our string at the LEN(#StartMarker).
-- This would give an odd result. Best to just provide from 0, i.e. the start of the #FullString.
IF #ReportErrorInResult = 0 AND #StartOfStartMarker = 0 SET #StartOfTarget = 0
--Main action
IF #ReturnString = '' SET #ReturnString = SUBSTRING(#FullString, #StartOfTarget, #EndOfTarget - #StartOfTarget)
IF #TrimLRWhiteSpace = 1 SET #ReturnString = LTRIM(RTRIM(#ReturnString))
RETURN #ReturnString
--Examples
-- SELECT '>' + dbo.GetStringBetweenMarkers('123456789','234','78',0,1) + '<' AS 'Result-Returns what is in between markers w/ white space'
-- SELECT '>' + dbo.GetStringBetweenMarkers('1234 56 789','234','78',0,1) + '<' AS 'Result-Without trimming white space'
-- SELECT '>' + dbo.GetStringBetweenMarkers('1234 56 789','234','78',1,1) + '<' AS 'Result-Will trim white space with a #TrimLRWhiteSpace = 1'
-- SELECT '>' + dbo.GetStringBetweenMarkers('abcdefgh','ABC','FG',0,1) + '<' AS 'Result-Not Case Sensitive'
-- SELECT '>' + dbo.GetStringBetweenMarkers('abc_de_fgh','_','_',0,1) + '<' AS 'Result-Using the same marker for start and end'
--Errors are returned if start or end marker are not found
-- SELECT '>' + dbo.GetStringBetweenMarkers('1234 56789','zz','78',0,1) + '<' AS 'Result-Start not found'
-- SELECT '>' + dbo.GetStringBetweenMarkers('1234 56789','234','zz',0,1) + '<' AS 'Result-End not found'
-- SELECT '>' + dbo.GetStringBetweenMarkers('1234 56789','zz','zz',0,1) + '<' AS 'Result-Niether found'
--If #ReportErrorInResult = 0
-- SELECT '>' + dbo.GetStringBetweenMarkers('123456789','zz','78',0,0) + '<' AS 'Result-Start not found-Returns from the start of the #FullString'
-- SELECT '>' + dbo.GetStringBetweenMarkers('123456789','34','zz',0,0) + '<' AS 'Result-End found-should get "Invalid length parameter passed to the LEFT or SUBSTRING function."'
END
GO
SELECT SUBSTRING('aaaaa$bbbbb$ccccc',instr('aaaaa$bbbbb$ccccc','$',1,1)+1, instr('aaaaa$bbbbb$ccccc','$',1,2)-1) -instr('aaaaa$bbbbb$ccccc','$',1,1)) as My_String
Hope this helps :
Declared a variable , in case of any changes need to be made thats only once .
declare #line varchar(100)
set #line ='Email_i-Julie#mail.com'
select SUBSTRING(#line ,(charindex('-',#line)+1), CHARINDEX('#',#line)-charindex('-',#line)-1)
I needed to get (099) 0000111-> (099) | 0000111 like two different columns.
SELECT
SUBSTRING(Phone, CHARINDEX('(', Phone) + 0, (2 + ((LEN(Phone)) - CHARINDEX(')', REVERSE(Phone))) - CHARINDEX('(', Phone))) AS CodePhone,
LTRIM(SUBSTRING(Phone, CHARINDEX(')', Phone) + 1, LEN(Phone))) AS NumberPhone
FROM
Suppliers
WHERE
Phone LIKE '%(%)%'
DECLARE #text VARCHAR(MAX)
SET #text = 'All I knew was that the dog had been very bad and required harsh punishment immediately regardless of what anyone else thought.'
DECLARE #pretext AS nvarchar(100) = 'the dog'
DECLARE #posttext AS nvarchar(100) = 'immediately'
SELECT
CASE
WHEN CHARINDEX(#posttext, #Text) - (CHARINDEX(#pretext, #Text) + len(#pretext)) < 0
THEN ''
ELSE SUBSTRING(#Text,
CHARINDEX(#pretext, #Text) + LEN(#pretext),
CHARINDEX(#posttext, #Text) - (CHARINDEX(#pretext, #Text) + LEN(#pretext)))
END AS betweentext
I'm a few years behind, but here's what I did to get a string between characters, that are not the same and also in the even you don't find the ending character, to still give you the substring
BEGIN
DECLARE #TEXT AS VARCHAR(20)
SET #TEXT='E101465445454-1'
SELECT SUBSTRING(#TEXT, CHARINDEX('E', #TEXT)+1, CHARINDEX('-',#TEXT)) as 'STR',
CAST(CHARINDEX('E', #TEXT)+1 AS INT) as 'val1', CAST(CHARINDEX('-', #TEXT) AS INT) as 'val2',
(CAST(CHARINDEX('-',#TEXT) AS INT) - CAST(CHARINDEX('E',#TEXT)+1 AS INT)) as 'SUBTR', LEN(#TEXT) as 'LEN'
SELECT CASE WHEN (CHARINDEX('-', #TEXT) > 0) THEN
SUBSTRING(#TEXT, CHARINDEX('E', #TEXT)+1, (CAST(CHARINDEX('-',#TEXT) AS INT) - CAST(CHARINDEX('E',#TEXT)+1 AS INT)))
ELSE
SUBSTRING(#TEXT, CHARINDEX('E', #TEXT)+1,LEN(#TEXT)- CHARINDEX('E', #TEXT))
END
END
Try it and comment for any improvements or if it does the job
select substring(#string,charindex('#first',#string)+1,charindex('#second',#string)-(charindex('#first',#string)+1))
Let us consider we have a string DUMMY_DATA_CODE_FILE and we want to find out the substring between 2nd and 3rd underscore(_). Then we use query something like this.
select SUBSTRING('DUMMY_DATA_CODE_FILE',charindex('_', 'DUMMY_DATA_CODE_FILE', (charindex('_','DUMMY_DATA_CODE_FILE', 1))+1)+1, (charindex('_', 'DUMMY_DATA_CODE_FILE', (charindex('_','DUMMY_DATA_CODE_FILE', (charindex('_','DUMMY_DATA_CODE_FILE', 1))+1))+1)- charindex('_', 'DUMMY_DATA_CODE_FILE', (charindex('_','DUMMY_DATA_CODE_FILE', 1))+1)-1)) as Code
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.