I'm trying to replace a combination of numbers using the REPLACE function in SQL Server but a specific ones inside a specific row. For example:
row value = '17,171,217,317,28' and will do it like this:
UPDATE TABLENAME SET COLUMN_NAME = REPLACE(column value,'17,','')
but that will replace all the 17 found it in the column row.
How do I specify to only replace the starting 17 found in the row?
Try using this:
UPDATE TABLENAME SET COLUMN_NAME = REPLACE(','+column value,',17,','')
Note that I agree not to store your data like that, violates 1NF (First Normal Form)
If I understand the requirement correctly:
DECLARE #ReplaceValue varchar(32) = '17,';
UPDATE dbo.TABLENAME
SET COLUMN_NAME = STUFF(COLUMN_NAME, 1, LEN(#ReplaceValue), '')
WHERE COLUMN_NAME LIKE #ReplaceValue + '%';
Example db<>fiddle
If you can fix your database design to not use comma separated lists, that is the best solution. If that is not an option, the following approach may work.
I assume that you want to remove whole values at either the start, middle, or end of the list, such as "17,171,217,317,28", "13,17,19", or "11,13,17". If the same value exists more than once, all should be removed. Example: "17,23,17,217,17" would become "23,217".
This can be done by adding extra delimiters at the start/end, replacing the delimited value (delimiter + value + delimiter) with a single delimiter, and then stripping the extra delimiters just added. If there is a chance that duplicate consecutive values may exist, existing delimiters will need to be temporarily doubled up and the replace would change to simply remove any matching (delimiter + value + delimiter) sequance.
Logic to handle all of the edge cases can get complex. The following appears to be fairly robust over a variety of test data.
DECLARE #ValueToRemove VARCHAR(100) = '17'
DECLARE #Delim VARCHAR(10) = ','
UPDATE #Data
SET UpdatedValueList =
TRIM(#Delim FROM
REPLACE(REPLACE(REPLACE(
#Delim + ValueList + #Delim -- Add leading and trailing delimiters
, #Delim, #Delim + #Delim) -- Double up delimiters
, #Delim + #ValueToRemove + #Delim, '') -- Remove value
, #Delim + #Delim, #Delim) -- Undo doubled delimiters
) -- Trim leading and trailing delimiters
Results:
ValueList
UpdatedValueList
17,171,217,317,28
171,217,317,28
13,17,19
13,19
11,13,17
11,13
17,23,17,217,17
23,217
17,18
18
16,17
16
16,17,17,18
16,18
17
16
16
The TRIM(char FROM string) syntax is available in SQL Server 2019 and later. For older versions, trimming leading and trailing delimiters gets tricky. One approach involves a STUFF/REVERSE/STUFF/REVERSE sequence, with a special case needed to handle an empty result.
UPDATE #Data
SET UpdatedValueList =
ISNULL(REVERSE(STUFF(REVERSE(STUFF(
REPLACE(REPLACE(REPLACE(
#Delim + ValueList + #Delim -- Add leading and trailing delimiters
, #Delim, #Delim + #Delim) -- Double up delimiters
, #Delim + #ValueToRemove + #Delim, '') -- Remove value
, #Delim + #Delim, #Delim) -- Undo doubled delimiters
, 1, LEN(#Delim), '')) -- Trim leading delimiter and reverse
, 1, LEN(#Delim), '')) -- Trim trailing delimiter and undo reverse
, '') -- Special case for empty result (when prior STUFF yielded null)
SELECT * FROM #Data
See this db<>fiddle for a demo.
If preserving order does not matter, you can use STRING_SPLIT() and STRING_AGG().
UPDATE #Data
SET UpdatedValueList = ISNULL((
SELECT STRING_AGG(S.Value, ',')
FROM STRING_SPLIT(ValueList, ',') S
WHERE S.Value <> #ValueToRemove
), '')
See this db<>fiddle fr a demo.
Caution. Although The examples tend to preserve order in small lists, this is not guaranteed in the function documentation. These functions are only available in SQL Server 2017 and later.
I am facing to an issue that I cannot update or replace some characters in my database.
Here is how this text look like in my column when I retrieve it:
As you can see, there is an unknown characters between 'master' and 'degree' which I cannot even paste it here.
I also tried to update and replace it with below code (I cannot paste that two vertical lines here since they are not supported in any browser and I am not sure what they are, Please see the picture above to see what is in my SQL statement).
begin transaction
update gm_desc set projdesc=replace(projdesc,'%â s%','') where projdesc like '%âs%' and proposalno = '15-149-01'
You can see the real SQL Statement here:
I tried to update, or replace it but I cannot do it. The update statement successfully works but I still see that weird special charters. I would be appreciate to help me.
Here's a scalar-valued function which removes all non-alphanumeric characters (preserves spaces) from a string.
Hopefully it helps!
dbfiddle
create function dbo.get_alphanumeric_str
(
#string varchar(max)
)
returns varchar(max)
as
begin
declare #ret varchar(max);
with nums as (
select 1 as n
union all select n+1 from nums
where n < 256
)
select #ret = replace(stuff(
(
select '' + substring(#string, nums.n, 1)
from nums
where patindex('%[^0-9A-Za-z ]%', substring(#string, nums.n,1)) = 0
for xml path('')
), 1, 0, ''
), ' ', ' ')
option (MAXRECURSION 256)
return #ret;
end
Usage
select dbo.get_alphanumeric_str('Helloᶄ âWorld 1234⅊⅐')
Returns: Hello World 1234
How it works
The nums CTE is just to get a list of numbers (you can set the maximum of 256 to a higher value if your strings are longer; n.b. option (MAXRECURSION n) is for this CTE but has to be placed at the query)
The stuff essentially iterates through the string, using the list of numbers above and extracts a substring of length 1; each of these chars are checked if they match the [^0-9A-Za-z ] regex group (0-9 all digits, A-Za-z all letters both lower and upper case, and a single space character)
If they match, patindex() should return 0; i.e. index zero.
Use replace(string, ' ', ' ') for the space character as the xml path returns a special encoding, see this question.
Use a binary collation for accented characters; see this answer
I am a bit new to this site but I have looked an many possible answers to my question but none of them has answered my need. I have a feeling it's a good challenge. Here it goes.
In one of our tables we list what is used to run a report this can mean that we can have a short EXEC [svr1].[dbo].[stored_procedure] or "...From svr1.dbo.stored_procedure...".
My goal is to get the stored procedure name out of this string (column). I have tried to get the string between '[' and ']' but that breaks when there are no brackets. I have been at this for a few days and just can't seem to find a solution.
Any assistance you can provide is greatly appreciated.
Thank you in advance for entertaining this question.
almostanexpert
Considering the ending character of your sample sentences is space, or your sentences end without trailing ( whether space or any other character other than given samples ), and assuming you have no other dots before samples, the following would be a clean way which uses substring(), len(), charindex() and replace() together :
with t(str) as
(
select '[svr1].[dbo].[stored_procedure]' union all
select 'before svr1.dbo.stored_procedure someting more' union all
select 'abc before svr1.dbo.stored_procedure'
), t2(str) as
(
select replace(replace(str,'[',''),']','') from t
), t3(str) as
(
select substring(str,charindex('.',str)+1,len(str)) from t2
)
select
substring(
str,
charindex('.',str)+1,
case
when charindex(' ',str) > 0 then
charindex(' ',str)
else
len(str)
end - charindex('.',str)
) as "Result String"
from t3;
Result String
----------------
stored_procedure
stored_procedure
stored_procedure
Demo
With the variability of inputs you seem to have we will need to plan for a few scenarios. The below code assumes that there will be exactly two '.' characters before the stored_procedure, and that [stored_procedure] will either end the string or be followed by a space if the string continues.
SELECT TRIM('[' FROM TRIM(']' FROM --Trim brackets from final result if they exist
SUBSTR(column || ' ', --substr(string, start_pos, length), Space added in case proc name is end of str
INSTR(column || ' ', '.', 1, 2)+1, --start_pos: find second '.' and start 1 char after
INSTR(column || ' ', ' ', INSTR(column || ' ', '.', 1, 2), 1)-(INSTR(column || ' ', '.', 1, 2)+1))
-- Len: start after 2nd '.' and go until first space (subtract 2nd '.' index to get "Length")
))FROM TABLE;
Working from the middle out we'll start with using the SUBSTR function and concatenating a space to the end of the original string. This allows us to use a space to find the end of the stored_procedure even if it is the last piece of the string.
Next to find our starting position, we use INSTR to search for the second instance of the '.' and start 1 position after.
For the length argument, we find the index of the first space after that second '.' and then subtract that '.' index.
From here we have either [stored_procedure] or stored_procedure. Running the TRIM functions for each bracket will remove them if they exist, and if not will just return the name of the procedure.
Sample inputs based on above description:
'EXEC [svr1].[dbo].[stored_procedure]'
'EXEC [svr1].[dbo].[stored_procedure] FROM TEST'
'svr1.dbo.stored_procedure'
Note: This code is written for Oracle SQL but can be translated to mySQL using similar functions.
I have a relational database with several fixed length character fields. I need to permanently replace all the embedded spaces with another character like - so JOHN DOE would become JOHN-DOE and ITSY BISTSY SPIDER would become ITSY-BISTSY-SPIDER. I can search before hand to make sure there are no strings that would conflict. I just need to be able to print the requested files with no embedded spaces. I would do the replacement in the C code but I want to make sure that there is never a future case where there is a JANE DOE and JANE-DOE in the DB.
By the way I have already made sure that there are no strings with more than one consecutive embedded space or leading spaces only trailing spaces to fill the fixed length fields.
Edit: thanks for all the help!
It looks like when I cut & pasted my question from Word to StackOverflow the trailing spaces got lost so the meaning my question was lost a bit.
I need to replace only the embedded spaces not the trailing spaces!
Note: I am using middle dot to stand in for spaces that don't show well.
Using:
SELECT REPLACE(operator_name, ' ', '-') FROM operator_info ;
the string JOHN·DOE············ became JOHN-DOE------------.
I need JOHN-DOE············.
I am thinking I need to use aliasing and the TRIM command but not sure how.
With whatever REPLACE function is built into your particular database.
MySQL:
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_replace
Oracle:
http://psoug.org/reference/translate_replace.html
SQLServer:
http://msdn.microsoft.com/en-us/library/ms186862.aspx
Edits below based on your comment.
I've done this in SQLServer syntax so please modify the example as needed. The first example really breaks down what's going on and the second one bunches it all into a single ugly query :D
#output in this case contains your final value.
DECLARE #input VARCHAR (100) = ' some test ';
DECLARE #trimmed VARCHAR (100);
DECLARE #replaced VARCHAR (100);
DECLARE #output VARCHAR (100);
-- Get just the inner text without the preceding / trailing spaces.
SET #trimmed = LTRIM (RTRIM (#input));
-- Replace the spaces *inside* the trimmed text with a dash.
SET #replaced = REPLACE (#trimmed, ' ', '-');
-- Take the original text and replace the trimmed version (with the inner spaces) with the dash version.
SET #output = REPLACE (#input, #trimmed, #replaced);
-- Show each step of the process!
SELECT #input AS INPUT,
#trimmed AS TRIMMED,
#replaced AS REPLACED,
#output AS OUTPUT;
And as a SELECT statement.
DECLARE #inputTable TABLE (Value VARCHAR (100) NOT NULL);
INSERT INTO #inputTable (Value)
VALUES (' some test '),
(' another test ');
SELECT REPLACE (Value,
LTRIM (RTRIM (Value)),
REPLACE (LTRIM (RTRIM (Value)), ' ', '-'))
FROM #inputTable;
If you are using MSSQL:
SELECT REPLACE(field_name,' ','-');
Edit: After the requirement about skipping the trailing spaces.
You can try this one-liner:
SELECT REPLACE(RTRIM(#name), ' ', '-') + SUBSTRING(#name, LEN(RTRIM(#name)) + 1, LEN(#NAME))
However I would recommend that you put it into a user defined function instead.
assuming SQL Server:
update TABLE set column = replace (column, ' ','-')
SELECT REPLACE(field_name,' ','-');
Edit: After the requirement about skipping the trailing spaces. You can try this one-liner:
SELECT REPLACE(RTRIM(#name), ' ', '-') + SUBSTRING(#name, LEN(RTRIM(#name)) + 1;
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