Related
I got the following entry in my database:
images/test.jpg
I want to trim the entry so I get: test
So basically, I want everything after / and before .
How can I solve it?
use the following function
left(#test, charindex('/', #test) - 1)
If you want to get this out of your table using SQL, take a look at the following functions that will help you: SUBSTRING and CHARINDEX. You can use those to trim your entries.
A possible query will look like this (where col is the name of the column that contains your image directories:
SELECT SUBSTRING(col, LEN(SUBSTRING(col, 0, LEN(col) - CHARINDEX ('/', col))) + 1,
LEN(col) - LEN(SUBSTRING(col, 0, LEN(col) - CHARINDEX ('/', col))) - LEN(SUBSTRING(
col, CHARINDEX ('.', col), LEN(col))));
Bit of an ugly beast. It also depends on the standard format of 'dir/name.ext'.
Edit:
This one (inspired by praveen) is more generic and deals with extensions of different length:
SELECT SUBSTRING(col, LEN(LEFT(col, CHARINDEX ('/', col))) + 1, LEN(col) - LEN(LEFT(col,
CHARINDEX ('/', col))) - LEN(RIGHT(col, LEN(col) - CHARINDEX ('.', col))) - 1);
Before
SELECT SUBSTRING(ParentBGBU,0,CHARINDEX('/',ParentBGBU,0)) FROM dbo.tblHCMMaster;
After
SELECT SUBSTRING(ParentBGBU,CHARINDEX('-',ParentBGBU)+1,LEN(ParentBGBU)) FROM dbo.tblHCMMaster
----select characters before / including /
select SUBSTRING ('abcde/wxyz',0,CHARINDEX('/','abcde/wxyz')+1)
--select characters after / including /
select SUBSTRING('abcde/wxyz',CHARINDEX('/','abcde/wxyz'),LEN('abcde/wxyz'))
declare #T table
(
Col varchar(20)
)
insert into #T
Select 'images/test1.jpg'
union all
Select 'images/test2.png'
union all
Select 'images/test3.jpg'
union all
Select 'images/test4.jpeg'
union all
Select 'images/test5.jpeg'
Select substring( LEFT(Col,charindex('.',Col)-1),charindex('/',Col)+1,len(LEFT(Col,charindex('.',Col)-1))-1 )
from #T
I have made a method which is much more general :
so :
DECLARE #a NVARCHAR(MAX)='images/test.jpg';
--Touch here
DECLARE #keysValueToSearch NVARCHAR(4000) = '/'
DECLARE #untilThisCharAppears NVARCHAR(4000) = '.'
DECLARE #keysValueToSearchPattern NVARCHAR(4000) = '%' + #keysValueToSearch + '%'
--Nothing to touch here
SELECT SUBSTRING(
#a,
PATINDEX(#keysValueToSearchPattern, #a) + LEN(#keysValueToSearch),
CHARINDEX(
#untilThisCharAppears,
#a,
PATINDEX(#keysValueToSearchPattern, #a) + LEN(#keysValueToSearch)
) -(PATINDEX(#keysValueToSearchPattern, #a) + LEN(#keysValueToSearch))
)
SELECT Substring('ravi1234#gmail.com', 1, ( Charindex('#', 'ravi1234#gmail.com')
- 1 ))
Before,
RIGHT('ravi123#gmail.com', ( Charindex('#', 'ravi123#gmail.com') + 1 ))
After
I just did this in one of my reports and it was very simple.
Try this:
=MID(Fields!.Value,8,4)
Note: This worked for me because the value I was trying to get was a constant not sure it what you are trying to get is a constant as well.
I know this has been a while.. but here is an idea
declare #test varchar(25) = 'images/test.jpg'
select
#test as column_name
, parsename(replace(#test,'/','.'),1) as jpg
,parsename(replace(#test,'/','.'),2) as test
,parsename(replace(#test,'/','.'),3) as images
I found Royi Namir's answer useful but expanded upon it to create it as a function. I renamed the variables to what made sense to me but you can translate them back easily enough, if desired.
Also, the code in Royi's answer already handled the case where the character being searched from does not exist (it starts from the beginning of the string), but I wanted to also handle cases where the character that is being searched to does not exist.
In that case it acts in a similar manner by starting from the searched from character and returning the rest of the characters to the end of the string.
CREATE FUNCTION [dbo].[getValueBetweenTwoStrings](#inputString
NVARCHAR(4000), #stringToSearchFrom NVARCHAR(4000), #stringToSearchTo
NVARCHAR(4000))
RETURNS NVARCHAR(4000)
AS
BEGIN
DECLARE #retVal NVARCHAR(4000)
DECLARE #stringToSearchFromSearchPattern NVARCHAR(4000) = '%' +
#stringToSearchFrom + '%'
SELECT #retVal = SUBSTRING (
#inputString,
PATINDEX(#stringToSearchFromSearchPattern, #inputString) + LEN(#stringToSearchFrom),
(CASE
CHARINDEX(
#stringToSearchTo,
#inputString,
PATINDEX(#stringToSearchFromSearchPattern, #inputString) + LEN(#stringToSearchFrom))
WHEN
0
THEN
LEN(#inputString) + 1
ELSE
CHARINDEX(
#stringToSearchTo,
#inputString,
PATINDEX(#stringToSearchFromSearchPattern, #inputString) + LEN(#stringToSearchFrom))
END) - (PATINDEX(#stringToSearchFromSearchPattern, #inputString) + LEN(#stringToSearchFrom))
)
RETURN #retVal
END
Usage:
SELECT dbo.getValueBetweenTwoStrings('images/test.jpg','/','.') AS MyResult
I got some invalid length errors. So i made this function, this should not give any length problems. Also when you do not find the searched text it will return a NULL.
CREATE FUNCTION [FN].[SearchTextGetBetweenStartAndStop](#string varchar(max),#SearchStringToStart varchar(max),#SearchStringToStop varchar(max))
RETURNS varchar(max)
BEGIN
SET #string = CASE
WHEN CHARINDEX(#SearchStringToStart,#string) = 0
OR CHARINDEX(#SearchStringToStop,RIGHT(#string,LEN(#string) - CHARINDEX(#SearchStringToStart,#string) + 1 - LEN(#SearchStringToStart))) = 0
THEN NULL
ELSE SUBSTRING(#string
,CHARINDEX(#SearchStringToStart,#string) + LEN(#SearchStringToStart) + 1
,(CHARINDEX(#SearchStringToStop,RIGHT(#string,LEN(#string) - CHARINDEX(#SearchStringToStart,#string) + 1 - LEN(#SearchStringToStart)))-2)
)
END
RETURN #string
END
if Input= pg102a-wlc01s.png.intel.com and Output should be pg102a-wlc01s
we can use below query :
select Substring(pc.name,0,charindex('.',pc.name,0)),pc.name from tbl_name pc
You can try this:
Declare #test varchar(100)='images/test.jpg'
Select REPLACE(RIGHT(#test,charindex('/',reverse(#test))-1),'.jpg','')
Below query gives you data before '-'
Ex- W12345A-4S
SELECT SUBSTRING(Column_Name,0, CHARINDEX('-',Column_Name)) as 'new_name'
from [abc].
Output - W12345A
Inspired by the work of Josien, I wondered about a simplification.
Would this also work? Much shorter:
SELECT SUBSTRING(col, CHARINDEX ('/', col) + 1, CHARINDEX ('.', col) - CHARINDEX ('/', col) - 1);
(I can't test right now because of right issues at my company SQL server, which is a problem in its own right)
Simply Try With LEFT ,RIGHT ,CHARINDEX
select
LEFT((RIGHT(a.name,((CHARINDEX('/', name))+1))),((CHARINDEX('.', (RIGHT(a.name,
((CHARINDEX('/', name))+1)))))-1)) splitstring,
a.name
from
(select 'images/test.jpg' as name)a
declare #searchStart nvarchar(100) = 'search ';
declare #searchEnd nvarchar(100) = ' ';
declare #string nvarchar(4000) = 'This is a string to search (hello) in this text ';
declare #startIndex int = CHARINDEX(#searchStart, #string,0) + LEN(#searchStart);
declare #endIndex int = CHARINDEX(#searchEnd, #string, #startIndex + 1);
declare #length int = #endIndex - #startIndex;
declare #sub nvarchar(4000) = SUBSTRING(#string, #startIndex, #length)
select #startIndex, #endIndex, #length, #sub
This is a little more legible than the one-liners in this answer which specifically answer the question, but not in a generic way that would benefit all readers. This could easily be made into a function as well with a slight modification.
If there are more than one or none occurences of given character use this:
DECLARE #rightidx int = CASE
WHEN 'images/images/test.jpg' IS NULL OR (CHARINDEX('.', 'images/images/test.jpg')) <= 0 THEN LEN('images/images/test.jpg')
ELSE (CHARINDEX('.', REVERSE('images/images/test.jpg')) - 1)
END
SELECT RIGHT('images/images/test.jpg', #rightidx)
This was the approach I took.
CREATE FUNCTION dbo.get_text_before_char(#my_string nvarchar(255),#my_char char(1))
RETURNS nvarchar(255)
AS
BEGIN;
return IIF(#my_string LIKE '%' + #my_char + '%',left (#my_string, IIF(charindex(#my_char, #my_string) - 1<1,1,charindex(#my_char, #my_string) - 1)),'');
END;
CREATE FUNCTION dbo.get_text_after_char(#my_string nvarchar(255),#my_char char(1))
RETURNS nvarchar(255)
AS
BEGIN;
return IIF ( #my_string LIKE '%' + #my_char + '%' ,RIGHT ( #my_string , IIF ( charindex ( #my_char ,reverse(#my_string) )-1 < 1 ,1 ,charindex ( #my_char ,reverse(#my_string) )-1 ) ) , '' )
END;
SELECT
dbo.get_text_before_char('foo-bar','-')
, dbo.get_text_after_char('foo-bar','-')
declare #test varchar(100)='images/test.jpg'
select right(left(#test, charindex('.', #test) - 1),4)
How do I get the nth word in a sentence or a set of strings with space delimiter?
Sorry for the change in the requirement.Thank you.
By using instr.
select substr(help, 1, instr(help,' ') - 1)
from ( select 'hello my name is...' as help
from dual )
instr(help,' ') returns the positional index of the first occurrence of the second argument in the first, inclusive of the string you're searching for. i.e. the first occurrence of ' ' in the string 'hello my name is...' plus the space.
substr(help, 1, instr(help,' ') - 1) then takes the input string from the first character to the index indicated in instr(.... I then remove one so that the space isn't included..
For the nth occurrence just change this slightly:
instr(help,' ',1,n) is the nth occurrence of ' ' from the first character. You then need to find the positional index of the next index instr(help,' ',1,n + 1), lastly work out the difference between them so you know how far to go in your substr(.... As you're looking for the nth, when n is 1 this breaks down and you have to deal with it, like so:
select substr( help
, decode( n
, 1, 1
, instr(help, ' ', 1, n - 1) + 1
)
, decode( &1
, 1, instr(help, ' ', 1, n ) - 1
, instr(help, ' ', 1, n) - instr(help, ' ', 1, n - 1) - 1
)
)
from ( select 'hello my name is...' as help
from dual )
This will also break down at n. As you can see this is getting ridiculous so you might want to consider using regular expressions
select regexp_substr(help, '[^[:space:]]+', 1, n )
from ( select 'hello my name is...' as help
from dual )
Try this. An example of getting the 4th word:
select names from (
select
regexp_substr('I want my two dollars','[^ ]+', 1, level) as names,
rownum as nth
from dual
connect by regexp_substr('I want my two dollars', '[^ ]+', 1, level) is not null
)
where nth = 4;
The inner query is converting the space-delimited string into a set of rows. The outer query is grabbing the nth item from the set.
Try something like
WITH q AS (SELECT 'ABCD EFGH IJKL' AS A_STRING FROM DUAL)
SELECT SUBSTR(A_STRING, 1, INSTR(A_STRING, ' ')-1)
FROM q
Share and enjoy.
And here's the solution for the revised question:
WITH q AS (SELECT 'ABCD EFGH IJKL' AS A_STRING, 3 AS OCCURRENCE FROM DUAL)
SELECT SUBSTR(A_STRING,
CASE
WHEN OCCURRENCE=1 THEN 1
ELSE INSTR(A_STRING, ' ', 1, OCCURRENCE-1)+1
END,
CASE
WHEN INSTR(A_STRING, ' ', 1, OCCURRENCE) = 0 THEN LENGTH(A_STRING)
ELSE INSTR(A_STRING, ' ', 1, OCCURRENCE) - CASE
WHEN OCCURRENCE=1 THEN 0
ELSE INSTR(A_STRING, ' ', 1, OCCURRENCE-1)
END - 1
END)
FROM q;
Share and enjoy.
CREATE PROC spGetCharactersInAStrings
(
#S VARCHAR(100) = '^1402 WSN NI^AMLAB^tev^e^^rtS htimS 0055518',
#Char VARCHAR(100) = '8'
)
AS
-- exec spGetCharactersInAStrings '^1402 WSN NI^AMLAB^tev^e^^rtS htimS 0055518', '5'
BEGIN
DECLARE #i INT = 1,
#c INT,
#pos INT = 0,
#NewStr VARCHAR(100),
#sql NVARCHAR(100),
#ParmDefinition nvarchar(500) = N'#retvalOUT int OUTPUT'
DECLARE #D TABLE
(
ID INT IDENTITY(1, 1),
String VARCHAR(100),
Position INT
)
SELECT #c = LEN(#S), #NewStr = #S
WHILE #i <= #c
BEGIN
SET #sql = ''
SET #sql = ' SELECT #retvalOUT = CHARINDEX(''' + + #Char + ''',''' + #NewStr + ''')'
EXEC sp_executesql #sql, #ParmDefinition, #retvalOUT=#i OUTPUT;
IF #i > 0
BEGIN
set #pos = #pos + #i
SELECT #NewStr = SUBSTRING(#NewStr, #i + 1, LEN(#S))
--SELECT #NewStr '#NewStr', #Char '#Char', #pos '#pos', #sql '#sql'
--SELECT #NewStr '#NewStr', #pos '#pos'
INSERT INTO #D
SELECT #NewStr, #pos
SET #i = #i + 1
END
ELSE
BREAK
END
SELECT * FROM #D
END
If you're using MySQL and cannot use the instr function that accepts four parameters or regexp_substr, you can do this way:
select substring_index(substring_index(help, ' ', 2), ' ', -1)
from (select 'hello my name is...' as help) h
Result: "my".
Replace "2" in the code above with the number of the word you want.
If you are using SQL Server 2016+ then you can take advantage of the STRING_SPLIT function. It returns rows of string values and if you aim to get nth value, then you can use Row_Number() window function.
Here there is a little trick as you don't want to really order by something so that you have to "cheat" the row_number function and allow its value in the natural order which is the STRING_SPLIT() function will spit out.
Below is a code snippet if you want to find the third word of the string
Declare #_intPart INT = 3; -- change nth work here, start # from 1 not 0
SELECT value FROM(
SELECT value,
ROW_NUMBER()OVER(ORDER BY (SELECT 1)) AS rowno
FROM STRING_SPLIT('hello world this is amazing', ' ')
) AS o1 WHERE o1.rowno = #_intPart;
You can also make a scalar function to retrieve values.
I'm looking to pull floats out of some varchars, using PATINDEX() to spot them. I know in each varchar string, I'm only interested in the first float that exists, but they might have different lengths.
e.g.
'some text 456.09 other text'
'even more text 98273.453 la la la'
I would normally match these with a regex
"[0-9]+[.][0-9]+"
However, I can't find an equivalent for the + operator, which PATINDEX accepts. So they would need to be matched (respectively) with:
'[0-9][0-9][0-9].[0-9][0-9]' and '[0-9][0-9][0-9][0-9][0-9].[0-9][0-9][0-9]'
Is there any way to match both of these example varchars with one single valid PATINDEX pattern?
I blogged about this a while ago.
Extracting numbers with SQL server
Declare #Temp Table(Data VarChar(100))
Insert Into #Temp Values('some text 456.09 other text')
Insert Into #Temp Values('even more text 98273.453 la la la')
Insert Into #Temp Values('There are no numbers in this one')
Select Left(
SubString(Data, PatIndex('%[0-9.-]%', Data), 8000),
PatIndex('%[^0-9.-]%', SubString(Data, PatIndex('%[0-9.-]%', Data), 8000) + 'X')-1)
From #Temp
Wildcards.
SELECT PATINDEX('%[0-9]%[0-9].[0-9]%[0-9]%','some text 456.09 other text')
SELECT PATINDEX('%[0-9]%[0-9].[0-9]%[0-9]%','even more text 98273.453 la la la')
Yes you need to link to the clr to get regex support. But if PATINDEX does not do what you need then regex was designed exactly for that.
http://msdn.microsoft.com/en-us/magazine/cc163473.aspx
Should be checked for robustness (what if you only have an int, for example), but this is just to put you on a track:
if exists (select routine_name from information_schema.routines where routine_name = 'GetFirstFloat')
drop function GetFirstFloat
go
create function GetFirstFloat (#string varchar(max))
returns float
as
begin
declare #float varchar(max)
declare #pos int
select #pos = patindex('%[0-9]%', #string)
select #float = ''
while isnumeric(substring(#string, #pos, 1)) = 1
begin
select #float = #float + substring(#string, #pos, 1)
select #pos = #pos + 1
end
return cast(#float as float)
end
go
select dbo.GetFirstFloat('this is a string containing pi 3.14159216 and another non float 3 followed by a new fload 5.41 and that''s it')
select dbo.GetFirstFloat('this is a string with no float')
select dbo.GetFirstFloat('this is another string with an int 3')
Given that the pattern is going to be varied in length, you're not going to have a rough time getting this to work with PATINDEX. There is another post that I wrote, which I've modified to accomplish what you're trying to do here. Will this work for you?
CREATE TABLE #nums (n INT)
DECLARE #i INT
SET #i = 1
WHILE #i < 8000
BEGIN
INSERT #nums VALUES(#i)
SET #i = #i + 1
END
CREATE TABLE #tmp (
id INT IDENTITY(1,1) not null,
words VARCHAR(MAX) null
)
INSERT INTO #tmp
VALUES('I''m looking for a number, regardless of length, even 23.258 long'),('Maybe even pi which roughly 3.14159265358,'),('or possibly something else that isn''t a number')
UPDATE #tmp SET words = REPLACE(words, ',',' ')
;WITH CTE AS (SELECT ROW_NUMBER() OVER (ORDER BY ID) AS rownum, ID, NULLIF(SUBSTRING(' ' + words + ' ' , n , CHARINDEX(' ' , ' ' + words + ' ' , n) - n) , '') AS word
FROM #nums, #tmp
WHERE ID <= LEN(' ' + words + ' ') AND SUBSTRING(' ' + words + ' ' , n - 1, 1) = ' '
AND CHARINDEX(' ' , ' ' + words + ' ' , n) - n > 0),
ids AS (SELECT ID, MIN(rownum) AS rownum FROM CTE WHERE ISNUMERIC(word) = 1 GROUP BY id)
SELECT CTE.rownum, cte.id, cte.word
FROM CTE, ids WHERE cte.id = ids.id AND cte.rownum = ids.rownum
The explanation and origin of the code is covered in more detail in the origional post
PATINDEX is not powerful enough to do that. You should use regular expressions.
SQL Server has Regular expression support since SQL Server 2005.
I am trying to count words of text that is written in a column of table. Therefor I am using the following query.
SELECT LEN(ExtractedText) -
LEN(REPLACE(ExtractedText, ' ', '')) + 1 from EDDSDBO.Document where ID='100'.
I receive a wrong result that is much to high.
On the other hand, if I copy the text directly into the statement then it works, i.e.
SELECT LEN('blablabla text') - LEN(REPLACE('blablabla text', ' ', '')) + 1.
Now the datatype is nvarchar(max) since the text is very long. I have already tried to convert the column into text or ntext and to apply datalength() instead of len(). Nevertheless I obtain the same result that it does work as a string but does not work from a table.
You're counting spaces not words. That will typically yield an approximate answer.
e.g.
' this string will give an incorrect result '
Try this approach: http://www.sql-server-helper.com/functions/count-words.aspx
CREATE FUNCTION [dbo].[WordCount] ( #InputString VARCHAR(4000) )
RETURNS INT
AS
BEGIN
DECLARE #Index INT
DECLARE #Char CHAR(1)
DECLARE #PrevChar CHAR(1)
DECLARE #WordCount INT
SET #Index = 1
SET #WordCount = 0
WHILE #Index <= LEN(#InputString)
BEGIN
SET #Char = SUBSTRING(#InputString, #Index, 1)
SET #PrevChar = CASE WHEN #Index = 1 THEN ' '
ELSE SUBSTRING(#InputString, #Index - 1, 1)
END
IF #PrevChar = ' ' AND #Char != ' '
SET #WordCount = #WordCount + 1
SET #Index = #Index + 1
END
RETURN #WordCount
END
GO
usage
DECLARE #String VARCHAR(4000)
SET #String = 'Health Insurance is an insurance against expenses incurred through illness of the insured.'
SELECT [dbo].[WordCount] ( #String )
Leading spaces, trailing spaces, two or more spaces between the neighbouring words – these are the likely causes of the wrong results you are getting.
The functions LTRIM() and RTRIM() can help you eliminate the first two issues. As for the third one, you can use REPLACE(ExtractedText, ' ', ' ') to replace double spaces with single ones, but I'm not sure if you do not have triple ones (in which case you'd need to repeat the replacing).
UPDATE
Here's a UDF that uses CTEs and ranking to eliminate extra spaces and then counts the remaining ones to return the quantity as the number of words:
CREATE FUNCTION fnCountWords (#Str varchar(max))
RETURNS int
AS BEGIN
DECLARE #xml xml, #res int;
SET #Str = RTRIM(LTRIM(#Str));
WITH split AS (
SELECT
idx = number,
chr = SUBSTRING(#Str, number, 1)
FROM master..spt_values
WHERE type = 'P'
AND number BETWEEN 1 AND LEN(#Str)
),
ranked AS (
SELECT
idx,
chr,
rnk = idx - ROW_NUMBER() OVER (PARTITION BY chr ORDER BY idx)
FROM split
)
SELECT #res = COUNT(DISTINCT rnk) + 1
FROM ranked
WHERE chr = ' ';
RETURN #res;
END
With this function your query will be simply like this:
SELECT fnCountWords(ExtractedText)
FROM EDDSDBO.Document
WHERE ID='100'
UPDATE 2
The function uses one of the system tables, master..spt_values, as a tally table. The particular subset used contains only values from 0 to 2047. This means the function will not work correctly for inputs longer than 2047 characters (after trimming both leading and trailing spaces), as #t-clausen.dk has correctly noted in his comment. Therefore, a custom tally table should be used if longer input strings are possible.
Replace the spaces with something that never occur in your text like ' $!' or pick another value.
then replace all '$! ' and '$!' with nothing this way you never have more than 1 space after a word. Then use your current script. I have defined a word as a space followed by a non-space.
This is an example
DECLARE #T TABLE(COL1 NVARCHAR(2000), ID INT)
INSERT #T VALUES('A B C D', 100)
SELECT LEN(C) - LEN(REPLACE(C,' ', '')) COUNT FROM (
SELECT REPLACE(REPLACE(REPLACE(' ' + COL1, ' ', ' $!'), '$! ',''), '$!', '') C
FROM #T ) A
Here is a recursive solution
DECLARE #T TABLE(COL1 NVARCHAR(2000), ID INT)
INSERT #T VALUES('A B C D', 100)
INSERT #T VALUES('have a nice day with 7 words', 100)
;WITH CTE AS
(
SELECT 1 words, col1 c, col1 FROM #t WHERE id = 100
UNION ALL
SELECT words +1, right(c, len(c) - patindex('% [^ ]%', c)), col1 FROM cte
WHERE patindex('% [^ ]%', c) > 0
)
SELECT words, col1 FROM cte WHERE patindex('% [^ ]%', c) = 0
You should declare the column using the varchar data type, like:
create table emp(ename varchar(22));
insert into emp values('amit');
select ename,len(ename) from emp;
output : 4
How do I remove the last character in a string in T-SQL?
For example:
'TEST STRING'
to return:
'TEST STRIN'
e.g.
DECLARE #String VARCHAR(100)
SET #String = 'TEST STRING'
-- Chop off the end character
SET #String =
CASE #String WHEN null THEN null
ELSE (
CASE LEN(#String) WHEN 0 THEN #String
ELSE LEFT(#String, LEN(#String) - 1)
END
) END
SELECT #String
If for some reason your column logic is complex (case when ... then ... else ... end), then the above solutions causes you to have to repeat the same logic in the len() function. Duplicating the same logic becomes a mess. If this is the case then this is a solution worth noting. This example gets rid of the last unwanted comma. I finally found a use for the REVERSE function.
select reverse(stuff(reverse('a,b,c,d,'), 1, 1, ''))
Try this:
select substring('test string', 1, (len('test string') - 1))
If your string is empty,
DECLARE #String VARCHAR(100)
SET #String = ''
SELECT LEFT(#String, LEN(#String) - 1)
then this code will cause error message 'Invalid length parameter passed to the substring function.'
You can handle it this way:
SELECT LEFT(#String, NULLIF(LEN(#String)-1,-1))
It will always return result, and NULL in case of empty string.
This will work even when source text/var is null or empty:
SELECT REVERSE(SUBSTRING(REVERSE(#a), 2, 9999))
select left('TEST STRING', len('TEST STRING')-1)
#result = substring(#result, 1, (LEN(#result)-1))
This is quite late, but interestingly never mentioned yet.
select stuff(x,len(x),1,'')
ie:
take a string x
go to its last character
remove one character
add nothing
If your coloumn is text and not varchar, then you can use this:
SELECT SUBSTRING(#String, 1, NULLIF(DATALENGTH(#String)-1,-1))
If you want to do this in two steps, rather than the three of REVERSE-STUFF-REVERSE, you can have your list separator be one or two spaces. Then use RTRIM to trim the trailing spaces, and REPLACE to replace the double spaces with ','
select REPLACE(RTRIM('a b c d '),' ', ', ')
However, this is not a good idea if your original string can contain internal spaces.
Not sure about performance. Each REVERSE creates a new copy of the string, but STUFF is a third faster than REPLACE.
also see this
I can suggest this -hack- ;).
select
left(txt, abs(len(txt + ',') - 2))
from
t;
SQL Server Fiddle Demo
you can create function
CREATE FUNCTION [dbo].[TRUNCRIGHT] (#string NVARCHAR(max), #len int = 1)
RETURNS NVARCHAR(max)
AS
BEGIN
IF LEN(#string)<#len
RETURN ''
RETURN LEFT(#string, LEN(#string) - #len)
END
Get the last character
Right(#string, len(#String) - (len(#String) - 1))
Try this
DECLARE #String VARCHAR(100)
SET #String = 'TEST STRING'
SELECT LEFT(#String, LEN(#String) - 1) AS MyTrimmedColumn
I encountered this problem and this way my problem was solved:
Declare #name as varchar(30)='TEST STRING'
Select left(#name, len(#name)-1) as AfterRemoveLastCharacter
My answer is similar to the accepted answer, but it also check for Null and Empty String.
DECLARE #String VARCHAR(100)
SET #String = 'asdfsdf1'
-- If string is null return null, else if string is empty return as it is, else chop off the end character
SET #String = Case #String when null then null else (case LEN(#String) when 0 then #String else LEFT(#String, LEN(#String) - 1) end ) end
SELECT #String
declare #string varchar(20)= 'TEST STRING'
Select left(#string, len(#string)-1) as Tada
output:
Tada
--------------------
TEST STRIN
I love #bill-hoenig 's answer; however, I was using a subquery and I got caught up because the REVERSE function needed two sets of parentheses. Took me a while to figure that one out!
SELECT
-- Return comma delimited list of all payment reasons for this Visit
REVERSE(STUFF(REVERSE((
SELECT DISTINCT
CAST(CONVERT(varchar, r1.CodeID) + ' - ' + c.Name + ', ' AS VARCHAR(MAX))
FROM VisitReason r1
LEFT JOIN ReasonCode c ON c.ID = r1.ReasonCodeID
WHERE p.ID = r1.PaymentID
FOR XML PATH('')
)), 1, 2, '')) ReasonCode
FROM Payments p
To update the record by trimming the last N characters of a particular column:
UPDATE tablename SET columnName = LEFT(columnName , LEN(columnName )-N) where clause
Try It :
DECLARE #String NVARCHAR(100)
SET #String = '12354851'
SELECT LEFT(#String, NULLIF(LEN(#String)-1,-1))
declare #x varchar(20),#y varchar(20)
select #x='sam'
select
case when #x is null then #y
when #y is null then #x
else #x+','+#y
end
go
declare #x varchar(20),#y varchar(20)
select #x='sam'
--,#y='john'
DECLARE #listStr VARCHAR(MAX)
SELECT #listStr = COALESCE(#x + ', ' ,'') +coalesce(#y+',','')
SELECT left(#listStr,len(#listStr)-1)
Try this,
DECLARE #name NVARCHAR(MAX) SET #name='xxxxTHAMIZHMANI****'SELECT Substring(#name, 5, (len(#name)-8)) as UserNames
And the output will be like, THAMIZHMANI