Related
I have the following code...
TRIM(LEADING '/' FROM ci.Long_description_1) as 'Description',
I am getting the error message
Incorrect syntax near '/'
Whats the best way of writing this?
Thanks
If you are using SQL Server 2017+, you may use TRIM() with a small trick (by default TRIM() removes the specified characters from the start and the end of the string):
SELECT LEFT(
TRIM('/' FROM Long_description_1 + '?'),
LEN(TRIM('/' FROM Long_description_1 + '?')) - 1
) AS Description
FROM (VALUES
(NULL),
('abcd'),
('1234/'),
('////abcd'),
('/folder/subfolder/x.yz')
) ci (Long_description_1)
Result:
Description
----------------------
abcd
1234/
abcd
folder/subfolder/x.yz
I think you want to remove '/' if it is the first character of Long_description_1 column value. TRIM function in SQL Server will not work the way you are expecting.
For this you can write your query like following.
SELECT CASE
WHEN CHARINDEX('/', ci.Long_description_1) = 1
THEN RIGHT(ci.Long_description_1, LEN(ci.Long_description_1) - 1)
ELSE ci.Long_description_1
END AS [Description]
FROM YouTable
In SQL Query, I need the values as below using select query of my column.
Result has to be the text after the first space ' ' and before the first '('
Source Column
create Table Test_Table (Column1 Varchar(50))
Insert into Test_Table Values
('0636 KAVITHI (LOC)'),
('0638 SRI KRISHNA (NAT)'),
('0639 SELVAM'),
('0643 GOOD SERVICE (LOC)'),
('0644 FINA CARE EVENT (LOC)')
I need get the string found between first ' ' and the '('
Expected Result
KAVITHI
SRI KRISHNA
SELVAM
GOOD SERVICE
FINA CARE EVENT
Another approach without using an OUTER APPLY.
SELECT CASE WHEN Column1 LIKE '%(%'
THEN SUBSTRING(RIGHT(Column1,LEN(Column1)-CHARINDEX(' ',Column1)),0,
CHARINDEX('(',RIGHT(Column1,LEN(Column1)-CHARINDEX(' ',Column1)),0))
ELSE RIGHT(Column1,LEN(Column1)-CHARINDEX(' ',Column1))
END AS Trimmed
FROM Test_Table
OUTPUT
Trimmed
KAVITHI
SRI KRISHNA
SELVAM
GOOD SERVICE
FINA CARE EVENT
SQL Fiddle: http://sqlfiddle.com/#!3/69dd1/20/0
CHARINDEX() can be used to find the position of specific characters.
OUTER APPLY can be used to find the position of the space and brace characters, and store them in a place that you can re-use them.
SUBSTRING() can be used to find the text between the space and the brace.
EDIT: Added CASE to cope with values that contain no (.
SELECT
SUBSTRING(
test_table.column1, -- the field we're searching
stats.idx_space + 1, -- starting from the character after the first space
CASE
WHEN stats.idx_brace > stats.idx_space
THEN stats.idx_brace
ELSE stats.idx_eos
END
-
stats.idx_space -- for as many characters as there are between the space and the brace
)
FROM
test_table
OUTER APPLY
(
SELECT
CHARINDEX(' ', test_table.column1) AS idx_space, -- position of the first space
CHARINDEX('(', test_table.column1) AS idx_brace, -- position of the first brace
LEN(test_table.column1) AS idx_eos -- position of the end-of-string
)
AS stats
EDIT: A single "line", as requested.
Do note that forcing this as a single line does make this harder to read, maintain and adapt. One of APPLY's strongest use-cases is to maintain DRY (Don't Repeat Yourself) principles.
This query repeats several parts several times:
- find the first space repeated 2 times
- find the first brace repeated 3 times
SELECT
SUBSTRING(
test_table.column1,
CHARINDEX(' ', test_table.column1) + 1,
CASE
WHEN CHARINDEX('(', test_table.column1) > CHARINDEX(' ', test_table.column1)
THEN CHARINDEX('(', test_table.column1)
ELSE LEN(test_table.column1)
END
-
CHARINDEX('(', test_table.column1)
)
FROM
test_table
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
I have a table in a SQL Server database with an NTEXT column. This column may contain data that is enclosed with double quotes. When I query for this column, I want to remove these leading and trailing quotes.
For example:
"this is a test message"
should become
this is a test message
I know of the LTRIM and RTRIM functions but these workl only for spaces. Any suggestions on which functions I can use to achieve this.
I have just tested this code in MS SQL 2008 and validated it.
Remove left-most quote:
UPDATE MyTable
SET FieldName = SUBSTRING(FieldName, 2, LEN(FieldName))
WHERE LEFT(FieldName, 1) = '"'
Remove right-most quote: (Revised to avoid error from implicit type conversion to int)
UPDATE MyTable
SET FieldName = SUBSTRING(FieldName, 1, LEN(FieldName)-1)
WHERE RIGHT(FieldName, 1) = '"'
I thought this is a simpler script if you want to remove all quotes
UPDATE Table_Name
SET col_name = REPLACE(col_name, '"', '')
You can simply use the "Replace" function in SQL Server.
like this ::
select REPLACE('this is a test message','"','')
note: second parameter here is "double quotes" inside two single quotes and third parameter is simply a combination of two single quotes. The idea here is to replace the double quotes with a blank.
Very simple and easy to execute !
My solution is to use the difference in the the column values length compared the same column length but with the double quotes replaced with spaces and trimmed in order to calculate the start and length values as parameters in a SUBSTRING function.
The advantage of doing it this way is that you can remove any leading or trailing character even if it occurs multiple times whilst leaving any characters that are contained within the text.
Here is my answer with some test data:
SELECT
x AS before
,SUBSTRING(x
,LEN(x) - (LEN(LTRIM(REPLACE(x, '"', ' ')) + '|') - 1) + 1 --start_pos
,LEN(LTRIM(REPLACE(x, '"', ' '))) --length
) AS after
FROM
(
SELECT 'test' AS x UNION ALL
SELECT '"' AS x UNION ALL
SELECT '"test' AS x UNION ALL
SELECT 'test"' AS x UNION ALL
SELECT '"test"' AS x UNION ALL
SELECT '""test' AS x UNION ALL
SELECT 'test""' AS x UNION ALL
SELECT '""test""' AS x UNION ALL
SELECT '"te"st"' AS x UNION ALL
SELECT 'te"st' AS x
) a
Which produces the following results:
before after
-----------------
test test
"
"test test
test" test
"test" test
""test test
test"" test
""test"" test
"te"st" te"st
te"st te"st
One thing to note that when getting the length I only need to use LTRIM and not LTRIM and RTRIM combined, this is because the LEN function does not count trailing spaces.
I know this is an older question post, but my daughter came to me with the question, and referenced this page as having possible answers. Given that she's hunting an answer for this, it's a safe assumption others might still be as well.
All are great approaches, and as with everything there's about as many way to skin a cat as there are cats to skin.
If you're looking for a left trim and a right trim of a character or string, and your trailing character/string is uniform in length, here's my suggestion:
SELECT SUBSTRING(ColName,VAR, LEN(ColName)-VAR)
Or in this question...
SELECT SUBSTRING('"this is a test message"',2, LEN('"this is a test message"')-2)
With this, you simply adjust the SUBSTRING starting point (2), and LEN position (-2) to whatever value you need to remove from your string.
It's non-iterative and doesn't require explicit case testing and above all it's inline all of which make for a cleaner execution plan.
The following script removes quotation marks only from around the column value if table is called [Messages] and the column is called [Description].
-- If the content is in the form of "anything" (LIKE '"%"')
-- Then take the whole text without the first and last characters
-- (from the 2nd character and the LEN([Description]) - 2th character)
UPDATE [Messages]
SET [Description] = SUBSTRING([Description], 2, LEN([Description]) - 2)
WHERE [Description] LIKE '"%"'
You can use following query which worked for me-
For updating-
UPDATE table SET colName= REPLACE(LTRIM(RTRIM(REPLACE(colName, '"', ''))), '', '"') WHERE...
For selecting-
SELECT REPLACE(LTRIM(RTRIM(REPLACE(colName, '"', ''))), '', '"') FROM TableName
you could replace the quotes with an empty string...
SELECT AllRemoved = REPLACE(CAST(MyColumn AS varchar(max)), '"', ''),
LeadingAndTrailingRemoved = CASE
WHEN MyTest like '"%"' THEN SUBSTRING(Mytest, 2, LEN(CAST(MyTest AS nvarchar(max)))-2)
ELSE MyTest
END
FROM MyTable
Some UDFs for re-usability.
Left Trimming by character (any number)
CREATE FUNCTION [dbo].[LTRIMCHAR] (#Input NVARCHAR(max), #TrimChar CHAR(1) = ',')
RETURNS NVARCHAR(max)
AS
BEGIN
RETURN REPLACE(REPLACE(LTRIM(REPLACE(REPLACE(#Input,' ','¦'), #TrimChar, ' ')), ' ', #TrimChar),'¦',' ')
END
Right Trimming by character (any number)
CREATE FUNCTION [dbo].[RTRIMCHAR] (#Input NVARCHAR(max), #TrimChar CHAR(1) = ',')
RETURNS NVARCHAR(max)
AS
BEGIN
RETURN REPLACE(REPLACE(RTRIM(REPLACE(REPLACE(#Input,' ','¦'), #TrimChar, ' ')), ' ', #TrimChar),'¦',' ')
END
Note the dummy character '¦' (Alt+0166) cannot be present in the data (you may wish to test your input string, first, if unsure or use a different character).
To remove both quotes you could do this
SUBSTRING(fieldName, 2, lEN(fieldName) - 2)
you can either assign or project the resulting value
You can use TRIM('"' FROM '"this "is" a test"') which returns: this "is" a test
CREATE FUNCTION dbo.TRIM(#String VARCHAR(MAX), #Char varchar(5))
RETURNS VARCHAR(MAX)
BEGIN
RETURN SUBSTRING(#String,PATINDEX('%[^' + #Char + ' ]%',#String)
,(DATALENGTH(#String)+2 - (PATINDEX('%[^' + #Char + ' ]%'
,REVERSE(#String)) + PATINDEX('%[^' + #Char + ' ]%',#String)
)))
END
GO
Select dbo.TRIM('"this is a test message"','"')
Reference : http://raresql.com/2013/05/20/sql-server-trim-how-to-remove-leading-and-trailing-charactersspaces-from-string/
I use this:
UPDATE DataImport
SET PRIO =
CASE WHEN LEN(PRIO) < 2
THEN
(CASE PRIO WHEN '""' THEN '' ELSE PRIO END)
ELSE REPLACE(PRIO, '"' + SUBSTRING(PRIO, 2, LEN(PRIO) - 2) + '"',
SUBSTRING(PRIO, 2, LEN(PRIO) - 2))
END
Try this:
SELECT left(right(cast(SampleText as nVarchar),LEN(cast(sampleText as nVarchar))-1),LEN(cast(sampleText as nVarchar))-2)
FROM TableName