There's already an answer for this question in SO with a MySQL tag. So I just decided to make your lives easier and put the answer below for SQL Server users. Always happy to see different answers perhaps with a better performance.
Happy coding!
SELECT SUBSTRING(#YourString, 1, LEN(#YourString) - CHARINDEX(' ', REVERSE(#YourString)))
Edit: Make sure #YourString is trimmed first as Alex M has pointed out:
SET #YourString = LTRIM(RTRIM(#YourString))
Just an addition to answers.
The doc for LEN function in MSSQL:
LEN excludes trailing blanks. If that is a problem, consider using the DATALENGTH (Transact-SQL) function which does not trim the string. If processing a unicode string, DATALENGTH will return twice the number of characters.
The problem with the answers here is that trailing spaces are not accounted for.
SELECT SUBSTRING(#YourString, 1, LEN(#YourString) - CHARINDEX(' ', REVERSE(#YourString)))
As an example few inputs for the accepted answer (above for reference), which would have wrong results:
INPUT -> RESULT
'abcd ' -> 'abc' --last symbol removed
'abcd 123 ' -> 'abcd 12' --only removed only last character
To account for the above cases one would need to trim the string (would return the last word out of 2 or more words in the phrase):
SELECT SUBSTRING(RTRIM(#YourString), 1, LEN(#YourString) - CHARINDEX(' ', REVERSE(RTRIM(LTRIM(#YourString)))))
The reverse is trimmed on both sides, that is to account for the leading as well as trailing spaces.
Or alternatively, just trim the input itself.
DECLARE #Sentence VARCHAR(MAX) = 'Hi This is Pavan Kumar'
SELECT SUBSTRING(#Sentence, 1, CHARINDEX(' ', #Sentence) - 1) AS [First Word],
REVERSE(SUBSTRING(REVERSE(#Sentence), 1,
CHARINDEX(' ', REVERSE(#Sentence)) - 1)) AS [Last Word]
DECLARE #String VARCHAR(MAX) = 'One two three four'
SELECT LEFT(#String,LEN(#String)-CHARINDEX(' ', REVERSE(#String),0)+1)
All the answers so far are actually about removing a character, not a word as the OP wanted.
In my case I was building a dynamic SQL statement with UNION'd SELECT statements and wanted to remove the last UNION:
DECLARE #sql NVARCHAR(MAX) = ''
/* populate #sql with something like this:
SELECT 1 FROM dbo.T1 WHERE condition
UNION
SELECT 1 FROM dbo.T2 WHERE condition
UNION
SELECT 1 FROM dbo.T3 WHERE condition
UNION
SELECT 1 FROM dbo.T4 WHERE condition
UNION
*/
-- remove the last UNION
SET #sql = SUBSTRING(#sql, 1, LEN(#sql) - PATINDEX(REVERSE('%UNION%'), REVERSE(#sql)) - LEN('UNION'))
SELECT LEFT(username , LEN(json_path) - CHARINDEX('/', REVERSE(username ))+1)
FROM Login_tbl
UPDATE Login_tbl
SET username = LEFT(username , LEN(json_path) - CHARINDEX('/', REVERSE(username ))+1)
DECLARE #String VARCHAR(MAX) = 'One two three four'
SELECT LEFT(#String,LEN(#String)-CHARINDEX(' ', REVERSE(#String),0)+1)
Related
I would like to remove the brackets, single quote before and after the comma from a variable in a elegant way in SQL
DECLARE #str NVARCHAR(MAX);
SET #str = N'(''202102'',''202104'',''202105'',''202106'',''202107'')'
Expected output
'202102,202104,202105,202106,202107'
I have tried and managed to remove the brackets but couldn't progress further for the single quotes.
SELECT SUBSTRING(#str, 2, LEN(#str) - 2); -- '202102','202104','202105','202106','202107'
Also tried the following which removes the whole lot after the first value.
SELECT LEFT(#str, CHARINDEX(',', #str) - 1) -- '202102'
I think this does what you want:
select replace(replace(replace(str, ''',''', ','), '(''', ''), ''')', '')
Here is a db<>fiddle.
Try this char(39) based approach
replace(replace(replace(#str, char(39), ''),'(',char(39)),')',char(39))
I am using this an an sql query to return just the first numbers out of a string and nothing else. I need it to do that for the first match that starts with a space.
so
hello world56 12345
would return only 12345
right now I get the 56
SUBSTRING(s.Description, PATINDEX('%[0-9]%',s.Description), PATINDEX('%[^0-9]%',SUBSTRING(s.Description, PATINDEX('%[0-9]%',s.Description), LEN(s.Description)))-1);
--test
DECLARE #Str nvarchar(1000)
SET #Str = 'ANDERSON, LEILANI M - MEDICAL ONCOLOGY 40225 (DFCI)'
SELECT SUBSTRING(#str,
PATINDEX(' %[0-9]%',#str)+1,
PATINDEX('%[^0-9]%',
SUBSTRING(#str,
PATINDEX(' %[0-9]%',#str)+1,
LEN(#str)))-1);
Put a space at the beginning of the pattern, and then add 1 to the index that it returns to get the position of the digit after the space.
SUBSTRING(s.Description,
PATINDEX('% [0-9]%',s.Description)+1,
PATINDEX('%[^0-9]%',
SUBSTRING(s.Description,
PATINDEX('% [0-9]%',s.Description)+1,
LEN(s.Description))
+ ' ')-1);
I append a space in the second PATINDEX() call so that it will work correctly if the number is at the end of the string.
DEMO
You may use the following select statement with substring, charindex, and patindex :
select substring(q.str, charindex(' ',q.str)+1,len(q.str)) as "Result"
from
(
select substring(Description,patindex('%[0-9]%',Description),len(Description)) str
from tab
) q;
Result
------
12345
Rextester Demo
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. :)
We have the below in row in MS SQL:
Got event with: 123.123.123.123, event 34, brown fox
How can we extract the 2nd number ie the 34 reliable in one line of SQL?
Here's one way to do it using SUBSTRING and PATINDEX -- I used a CTE just so it wouldn't look so awful :)
WITH CTE AS (
SELECT
SUBSTRING(Data,CHARINDEX(',',Data)+1,LEN(Data)) data
FROM Test
)
SELECT LEFT(SUBSTRING(Data, PATINDEX('%[0-9]%', Data), 8000),
PATINDEX('%[^0-9]%',
SUBSTRING(Data, PATINDEX('%[0-9]%', Data), 8000) + 'X')-1)
FROM CTE
And here is some sample Fiddle.
As commented, CTEs will only work with 2005 and higher. If by chance you're using 2000, then this will work without the CTE:
SELECT LEFT(SUBSTRING(SUBSTRING(Data,CHARINDEX(',',Data)+1,LEN(Data)),
PATINDEX('%[0-9]%', SUBSTRING(Data,CHARINDEX(',',Data)+1,LEN(Data))), 8000),
PATINDEX('%[^0-9]%',
SUBSTRING(SUBSTRING(Data,CHARINDEX(',',Data)+1,LEN(Data)),
PATINDEX('%[0-9]%', SUBSTRING(Data,CHARINDEX(',',Data)+1,LEN(Data))), 8000) + 'X')-1)
FROM Test
Simply replace #s with your column name to apply this to a table. Assuming that number is between last comma and space before the last comma. Sql-Fiddle-Demo
declare #s varchar(100) = '123.123.123.123, event 34, brown fox'
select right(first, charindex(' ', reverse(first),1) ) final
from (
select left(#s,len(#s) - charindex(',',reverse(#s),1)) first
--from tableName
) X
OR if it is between first and second commas then try, DEMO
select substring(first, charindex(' ',first,1),
charindex(',', first,1)-charindex(' ',first,1)) final
from (
select right(#s,len(#s) - charindex(',',#s,1)-1) first
) X
I've thought of another way that's not been mentioned yet. Presuming the following are true:
Always one comma before the second "part"
It's always the word "event" with the number in the second part
You are using SQL Server 2005+
Then you could use the built in ParseName function meant for parsing the SysName datatype.
--Variable to hold your example
DECLARE #test NVARCHAR(50)
SET #test = 'Got event with: 123.123.123.123, event 34, brown fox'
SELECT Ltrim(Rtrim(Replace(Parsename(Replace(Replace(#test, '.', ''), ',', '.'), 2), 'event', '')))
Results:
34
ParseName parses around dots, but we want it to parse around commas. Here's the logic of what I've done:
Remove all existing dots in the string, in this case swap them with empty string.
Swap all commas for dots for ParseName to use
Use ParseName and ask for the second "piece". In your example this gives us the value
" event 34".
Remove the word "event" from the string.
Trim both ends and return the value.
I've no comments on performance vs. the other solutions, and it looks just as messy. Thought I'd throw the idea out there anyway!
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