How can I replace the $SV### value in this string example if I don't know what the numbers following will be? Or even just get the 1234544 into another string variable.
I tried this but it doesn't replace anything:
declare #string varchar(100) = 'F4 Obstructed reach beyond. [$SV1234544" provided.]'
SELECT REPLACE(#string,PATINDEX('%$SV[0-9+]%',#string),'test')
Thank you!
The below code will work with the expectation that the number will always be prefixed with "$SV". It goes through a series of iterations to locate the numeric value after "$SV" and then replaces the "$SV" and the numeric number with whatever text you want it replaced with. This code uses the substring, patindex, concat, and replace functions.
DECLARE #string varchar(100) = 'F4 Obstructed reach beyond. [$SV1234544" provided.]'
DECLARE #SubString Varchar(100) = SUBSTRING(#string, PATINDEX('%$SV[0-9]%', #string), LEN(#string))
SELECT REPLACE(#string,CONCAT('$SV',LEFT(SUBSTRING(#SubString, PATINDEX('%[0-9.-]%', #SubString), 8000),PATINDEX('%[^0-9.-]%', SUBSTRING(#SubString, PATINDEX('%[0-9.-]%', #SubString), 8000) + 'X') -1)),'test')
I would like to remove all characters before and after a string in the select statement.
In the example below I would like to remove everything before and including /Supply> and after and including >/
Note the remaining part will be a fixed number of characters.
Any help would be much appreciated
Eg.
abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd
Would become:
hhfhjgglldppprrr
If your input always has exactly two instances of ">" you could use PARSENAME.
declare #SomeValue varchar(100) = 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
select PARSENAME(replace(#SomeValue, '>', '.'), 2)
This will not work correctly if your data also contains any periods (.). We can deal with that if needed with a couple of replace statements. Still very simple and easy to maintain with the same caveat of exactly 2 >.
declare #SomeOtherValue varchar(100) = 'abs/Supply>hhfhjgg.lldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
select replace(PARSENAME(replace(replace(#SomeOtherValue, '.', '~!##'), '>', '.'), 2), '~!##', '.')
You can use PATINDEX() to identify the position of the patterns you are looking for (/Supply> and >/) then remove them based on the length of the string:
SELECT LEFT(RIGHT(col,LEN(col) - PATINDEX('%/Supply>%',col) -7), PATINDEX('%>/%', RIGHT(col,LEN(col) - PATINDEX('%Supply>%',col) -7))-1)
Simply replace col in the above with your column name.
Example below with test string abs/Supply>keep>/remove
First remove everything before and including /Supply>:
SELECT RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7)
This will give keep>/remove
Then remove everything after and including >/:
SELECT LEFT('keep>/remove',PATINDEX('%>/%','keep>/remove') - 1)
This will give keep, the part of the string you want.
Here is the combined version, same as above, just includes the test string instead of col so you can run it easily:
SELECT LEFT(RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7), PATINDEX('%>/%', RIGHT('abs/Supply>keep>/remove',LEN('abs/Supply>keep>/remove') - PATINDEX('%/Supply>%','abs/Supply>keep>/remove') -7))-1)
This will give keep. You can also replace the string above with the one in your question, I just used a different test string because it is shorter and makes the code more readable.
try this:
DECLARE #inputStr VARCHAR(max)= 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
DECLARE #startString VARCHAR(100)='/Supply>'
DECLARE #EndString VARCHAR(100)='>/'
DECLARE #LenStartString INT = LEN(#startString)
DECLARe #TempInputString VARCHAR(max)='';
DECLARE #StartIndex INT
DECLARE #EndIndex INT
SELECT #StartIndex = CHARINDEX(#startString,#inputStr)+#LenStartString
SELECT #TempInputString = STUFF(#inputStr, 1, #StartIndex, '')
SELECT SUBSTRING(#TempInputString,0,CHARINDEX(#EndString,#TempInputString))
In Single Line
DECLARE #inputStr VARCHAR(max)= 'abs/Supply>hhfhjgglldppprrr>/llllllldsfsjhfhhhfdhudfhfhdhdfhfsd'
DECLARE #startString VARCHAR(100)='/Supply>'
DECLARE #EndString VARCHAR(100)='>/'
SELECT SUBSTRING(STUFF(#inputStr, 1, CHARINDEX(#startString,#inputStr)+LEN(#startString), ''),0,CHARINDEX(#EndString,STUFF(#inputStr, 1,CHARINDEX(#startString,#inputStr)+LEN(#startString), '')))
My string looks something like this:
\\\abcde\fghijl\akjfljadf\\
\\xyz\123
I want to select everything between the 1st set and next set of slashes
Desired result:
abcde
xyz
EDITED: To clarify, the special character is always slashes - but the leading characters are not constant, sometimes there are 3 slashes and other times there are only 2 slashes, followed by texts, and then followed by 1 or more slashes, some more texts, 1 or more slash, so on and so forth. I'm not using any adapter at all, just looking for a way to select this substring in my SQL query
Please advise.
Thanks in advance.
You could do a cross join to find the second position of the backslash. And then, use substring function to get the string between 2nd and 3rd backslash of the text like this:
SELECT substring(string, 3, (P2.Pos - 2)) AS new_string
FROM strings
CROSS APPLY (
SELECT (charindex('\', replace(string, '\\', '\')))
) AS P1(Pos)
CROSS APPLY (
SELECT (charindex('\', replace(string, '\\', '\'), P1.Pos + 1))
) AS P2(Pos)
SQL Fiddle Demo
UPDATE
In case, when you have unknown number of backslashes in your string, you could just do something like this:
DECLARE #string VARCHAR(255) = '\\\abcde\fghijl\akjfljadf\\'
SELECT left(ltrim(replace(#string, '\', ' ')),
charindex(' ',ltrim(replace(#string, '\', ' ')))-1) AS new_string
SQL Fiddle Demo2
Use substring, like this (only works for the specified pattern of two slashes, characters, then another slash):
declare #str varchar(100) = '\\abcde\cc\xxx'
select substring(#str, 3, charindex('\', #str, 3) - 3)
Replace #str with the column you actually want to search, of course.
The charindex returns the location of the first slash, starting from the 3rd character (i.e. skipping the first two slashes). Then the substring returns the part of your string starting from the 3rd character (again, skipping the first two slashes), and continuing until just before the next slash, as determined by charindex.
Edit: To make this work with different numbers of slashes at the beginning, use patindex with regex to find the first alphanumeric character, instead of hardcoding that it should be the third character. Example:
declare #str varchar(100) = '\\\1Abcde\cc\xxx'
select substring(#str, patindex('%[a-zA-Z0-9]%', #str), charindex('\', #str, patindex('%[a-zA-Z0-9]%', #str)) - patindex('%[a-zA-Z0-9]%', #str))
APH's solution works better if your string always follows the pattern as described. However this will get the text despite the pattern.
declare #str varchar(100) = '\\abcde\fghijl\akjfljadf\\'
declare #srch char(1) = '\'
select
SUBSTRING(#str,
(CHARINDEX(#srch,#str,(CHARINDEX(#srch,#str,1)+1))+1),
CHARINDEX(#srch,#str,(CHARINDEX(#srch,#str,(CHARINDEX(#srch,#str,1)+1))+1))
- (CHARINDEX(#srch,#str,(CHARINDEX(#srch,#str,1)+1))+1)
)
Sorry for the formatting.
Edited to correct user paste error. :)
I have the following string: 'BOB*', how do I trim the * so it shows up as 'BOB'
I tried the RTRIM('BOB*','*') but does not work as says needs only 1 parameter.
Another pretty good way to implement Oracle's TRIM char FROM string in MS SQL Server is the following:
First, you need to identify a char that will never be used in your string, for example ~
You replace all spaces with that character
You replace the character * you want to trim with a space
You LTrim + RTrim the obtained string
You replace back all spaces with the trimmed character *
You replace back all never-used characters with a space
For example:
REPLACE(REPLACE(LTrim(RTrim(REPLACE(REPLACE(string,' ','~'),'*',' '))),' ','*'),'~',' ')
CREATE FUNCTION dbo.TrimCharacter
(
#Value NVARCHAR(4000),
#CharacterToTrim NVARCHAR(1)
)
RETURNS NVARCHAR(4000)
AS
BEGIN
SET #Value = LTRIM(RTRIM(#Value))
SET #Value = REVERSE(SUBSTRING(#Value, PATINDEX('%[^'+#CharacterToTrim+']%', #Value), LEN(#Value)))
SET #Value = REVERSE(SUBSTRING(#Value, PATINDEX('%[^'+#CharacterToTrim+']%', #Value), LEN(#Value)))
RETURN #Value
END
GO
--- Example
----- SELECT dbo.TrimCharacter('***BOB*********', '*')
----- returns 'BOB'
If you want to remove all asterisks then it's obvious:
SELECT REPLACE('Hello*', '*', '')
However, If you have more than one asterisk at the end and multiple throughout, but are only interested in trimming the trailing ones, then I'd use this:
DECLARE #String VarChar(50) = '**H*i****'
SELECT LEFT(#String, LEN(REPLACE(#String, '*', ' '))) --Returns: **H*i
I updated this answer to include show how to remove leading characters:
SELECT RIGHT(#String, LEN(REPLACE(REVERSE(#String), '*', ' '))) --Returns: H*i****
LEN() has a "feature" (that looks a lot like a bug) where it does not count trailing spaces.
LEFT('BOB*', LEN('BOB*')-1)
should do it.
If you wanted behavior similar to how RTRIM handles spaces i.e. that "B*O*B**" would turn into "B*O*B" without losing the embedded ones then something like -
REVERSE(SUBSTRING(REVERSE('B*O*B**'), PATINDEX('%[^*]%',REVERSE('B*O*B**')), LEN('B*O*B**') - PATINDEX('%[^*]%', REVERSE('B*O*B**')) + 1))
Should do it.
If you only want to remove a single '*' character from the value when the value ends with a '*', a simple CASE expression will do that for you:
SELECT CASE WHEN RIGHT(foo,1) = '*' THEN LEFT(foo,LEN(foo)-1) ELSE foo END AS foo
FROM (SELECT 'BOB*' AS foo)
To remove all trailing '*' characters, then you'd need a more complex expression, making use of the REVERSE, PATINDEX, LEN and LEFT functions.
NOTE: Be careful with the REPLACE function, as that will replace all occurrences of the specified character within the string, not just the trailing ones.
How about.. (in this case to trim off trailing comma or period)
For a variable:
-- Trim commas and full stops from end of City
WHILE RIGHT(#CITY, 1) IN (',', '.'))
SET #CITY = LEFT(#CITY, LEN(#CITY)-1)
For table values:
-- Trim commas and full stops from end of City
WHILE EXISTS (SELECT 1 FROM [sap_out_address] WHERE RIGHT([CITY], 1) IN (',', '.'))
UPDATE [sap_out_address]
SET [CITY] = LEFT([CITY], LEN([CITY])-1)
WHERE RIGHT([CITY], 1) IN (',', '.')
An other approach ONLY if you want to remove leading and trailing characters is the use of TRIM function.
By default removes white spaces but have te avility of remove other characters if you specify its.
SELECT TRIM('=' FROM '=SPECIALS=') AS Result;
Result
--------
SPECIALS
Unfortunately LTRIM and RTRIM does not work in the same way and only removes white spaces instead of specified characters like TRIM does if you specify its.
Reference and more examples:
https://database.guide/how-to-remove-leading-and-trailing-characters-in-sql-server/
RRIM() LTRIM() only remove spaces try http://msdn.microsoft.com/en-us/library/ms186862.aspx
Basically just replace the * with empty space
REPLACE('TextWithCharacterToReplace','CharacterToReplace','CharacterToReplaceWith')
So you want
REPLACE ('BOB*','*','')
I really like Teejay's answer, and almost stopped there. It's clever, but I got the "almost too clever" feeling, as, somehow, your string at some point will actually have a ~ (or whatever) in it on purpose. So that's not defensive enough for me to put into production.
I like Chris' too, but the PATINDEX call seems like overkill.
Though it's probably a micro-optimization, here's one without PATINDEX:
CREATE FUNCTION dbo.TRIMMIT(#stringToTrim NVARCHAR(MAX), #charToTrim NCHAR(1))
RETURNS NVARCHAR(MAX)
AS
BEGIN
DECLARE #retVal NVARCHAR(MAX)
SET #retVal = #stringToTrim
WHILE 1 = charindex(#charToTrim, reverse(#retVal))
SET #retVal = SUBSTRING(#retVal,0,LEN(#retVal))
WHILE 1 = charindex(#charToTrim, #retVal)
SET #retVal = SUBSTRING(#retVal,2,LEN(#retVal))
RETURN #retVal
END
--select dbo.TRIMMIT('\\trim\asdfds\\\', '\')
--trim\asdfds
Returning a MAX nvarchar bugs me a little, but that's the most flexible way to do this..
I've used a similar approach to some of the above answers of using pattern matching and reversing the string to find the first non-trimmable character, then cutting that off. The difference is this version does less work than those above, so should be a little more efficient.
This creates RTRIM functionality for any specified character.
It includes an additional step set #charToFind = case... to escape the chosen character.
There is currently an issue if #charToReplace is a right crotchet (]) as there appears to be no way to escape this.
.
declare #stringToSearch nvarchar(max) = '****this is****a ** demo*****'
, #charToFind nvarchar(5) = '*'
--escape #charToFind so it doesn't break our pattern matching
set #charToFind = case #charToFind
when ']' then '[]]' --*this does not work / can't find any info on escaping right crotchet*
when '^' then '\^'
--when '%' then '%' --doesn't require escaping in this context
--when '[' then '[' --doesn't require escaping in this context
--when '_' then '_' --doesn't require escaping in this context
else #charToFind
end
select #stringToSearch
, left
(
#stringToSearch
,1
+ len(#stringToSearch)
- patindex('%[^' + #charToFind + ']%',reverse(#stringToSearch))
)
SqlServer2017 has a new way to do it: https://learn.microsoft.com/en-us/sql/t-sql/functions/trim-transact-sql?view=sql-server-2017
SELECT TRIM('0' FROM '00001900'); -> 19
SELECT TRIM( '.,! ' FROM '# test .'); -> # test
SELECT TRIM('*' FROM 'BOB*'); --> BOB
Unfortunately, RTRIM does not support trimming a specific character.
SELECT REPLACE('BOB*', '*', '')
SELECT REPLACE('B*OB*', '*', '')
-------------------------------------
Result : BOB
-------------------------------------
this will replace all asterisk* from the text
Trim with many cases
--id = 100 101 102 103 104 105 106 107 108 109 110 111
select right(id,2)+1 from ordertbl -- 1 2 3 4 5 6 7 8 9 10 11 -- last two positions are taken
select LEFT('BOB', LEN('BOB')-1) -- BO
select LEFT('BOB*',1) --B
select LEFT('BOB*',2) --BO
Try this:
Original
select replace('BOB*','*','')
Fixed to be an exact replacement
select replace('BOB*','BOB*','BOB')
Solution for one char parameter:
rtrim('0000100','0') ->
select left('0000100',len(rtrim(replace('0000100','0',' '))))
ltrim('0000100','0') ->
select right('0000100',len(replace(ltrim(replace('0000100','0',' ')),' ','.')))
#teejay solution is great. But the code below can be more understandable:
declare #X nvarchar(max)='BOB *'
set #X=replace(#X,' ','^')
set #X=replace(#X,'*',' ')
set #X= ltrim(rtrim(#X))
set #X=replace(#X,'^',' ')
Here's a function I used in the past. Note that while you can make it more general purpose by having extra parameters like the character(s) you wish to remove and what you will be replacing the space character(s) with, this greatly increases execution time. Here, I used a pipe to replace spaces AFTER pre-trimming the input. Change varchar to nvarchar if required.
CREATE FUNCTION [dbo].[TrimColons]
(
#strToTrim varchar(500)
)
RETURNS varchar(500)
AS
BEGIN
RETURN REPLACE(REPLACE(LTRIM(RTRIM(REPLACE(REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|'),':',' '))),' ',':'),'|',' ')
/*
Here's a breakdown of this fancy, schmancy, trimmer
LTRIM(RTRIM(#strToTrim)) trims leading & trailing spaces first
REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|') replaces inside spaces with pipe char
REPLACE(REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|'),':',' ') replaces demarc character, the colon, with spaces
LTRIM(RTRIM(REPLACE(REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|'),':',' '))) trims the leading & trailing converted-to-space demarc char (colon)
REPLACE(LTRIM(RTRIM(REPLACE(REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|'),':',' '))),' ',':') replaces the inner space characters back to demar char (colon)
REPLACE(REPLACE(LTRIM(RTRIM(REPLACE(REPLACE(LTRIM(RTRIM(#strToTrim)),' ','|'),':',' '))),' ',':'),'|',' ') replaces the pipe characters back to original space characters
*/
END
DECLARE #String VarChar(50) = '**H*i****', #String2 VarChar(50)
--Assign to new variable #String2
;WITH X AS (
SELECT LEFT(#String, LEN(REPLACE(#String, '*', ' '))) [V1]
)
SELECT TOP 1 #String2 = RIGHT(V1, LEN(REPLACE(REVERSE(V1), '*', ' '))) FROM X
SELECT #String [#String], #String2 [#String2]
--See the intermediate values, v0 original, v1 triming end, and v2 trim the v1 leading
;WITH X AS (
SELECT #String V0, LEFT(#String, LEN(REPLACE(#String, '*', ' '))) [V1]
)
SELECT [V0], [V1], RIGHT([V1], LEN(REPLACE(REVERSE([V1]), '*', ' '))) [v2] FROM X