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I've below query:
DECLARE #url varchar (max)='http://v.mercola.com/blogs/public_blog/New-Diet-Pill-Expands-1-000-Times-in-Your-Stomach-24728.aspx'
SELECT replace(replace(RIGHT(#URL , CHARINDEX ('/' ,REVERSE(#URL))-1),'.aspx',''),'-',' ') as abc
Which returns below output:
Actual output -
Expected output
i.e i want to eliminate the string after last occurrence of -.
What changes do i have to make to get the expected output..
In all i want a substring after last occurence of / and before last occurence of - as shown above.
Please help and thanks in advance...!
Try this
DECLARE #url VARCHAR (max)='http://v.mercola.com/blogs/public_blog/New-Diet-Pill-Expands-1-000-Times-in-Your-Stomach-24728.aspx'
SELECT Reverse(LEFT(mid, Charindex('/', mid) - 1))
FROM (SELECT Substring(Reverse(#url), Charindex('-', Reverse(#url)) + 1, Len(#url)) AS mid) a
Something like this:
DECLARE #url varchar (max)='http://v.mercola.com/blogs/public_blog/New-Diet-Pill-Expands-1-000-Times-in-Your-Stomach-24728.aspx'
declare #suffix varchar(max)
select #suffix = RIGHT(#URL , CHARINDEX ('/' ,REVERSE(#URL))-1)
select left(#suffix, len(#suffix) - charindex('-', reverse(#suffix)))
Output:
New-Diet-Pill-Expands-1-000-Times-in-Your-Stomach
Another option:
DECLARE #url varchar (max)='http://v.mercola.com/blogs/public_blog/New-Diet-Pill-Expands-1-000-Times-in-Your-Stomach-24728.aspx'
DECLARE #LastSlash int = LEN(#URL) - CHARINDEX('/', REVERSE(#URL)) + 2,
#LastMinus int = LEN(#URL) - CHARINDEX ('-', REVERSE(#URL)) + 1
SELECT SUBSTRING(#URL, #LastSlash, #LastMinus-#LastSlash)
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)
I have the following string.
Input
--------------
2030031469-NAI
To get the numeric part, I am using the following script
declare #str varchar(50)= '2030031469-NAI'
Select
#str
,SUBSTRING(#str, 1, NULLIF(CHARINDEX('-', #str) - 1, -1))
,Left(#str,PATINDEX('%-%',#str)-1)
to get the following output
Output:
----------
2030031469
Is there any other easy/elegant way of doing the same?
select left(#str, patindex('%[^0-9]%', #str+'.') - 1)
In case your string start with alphabet and end with number like ERT-123
you can use this query:
(select substring(#str,patindex('%[0-9]%', #str),len(#str)))
Please check with this, i used in my project for extracting phone numbers
CREATE Function [dbo].[RemoveNonNumericCharacters](#Temp VarChar(1000))
Returns VarChar(1000)
AS
Begin
While PatIndex('%[^0-9]%', #Temp) > 0
Set #Temp = Stuff(#Temp, PatIndex('%[^0-9]%', #Temp), 1, '')
Return #TEmp
End
To extract number from an unformatted string
DECLARE #Text NVARCHAR(100)= 'extract only 23124R integer ##%%'
SELECT SUBSTRING(#Text, PATINDEX('%[0-9]%',#Text), PATINDEX('%[^0-9]%',SUBSTRING(#Text, PATINDEX('%[0-9]%',#Text), LEN(#Text)))-1) [ONLY_INT]
What is the correct way to get the last word of a part of a varchar?
DECLARE #desc varchar(100)
SET #desc='EXCHANGEUNIT P1i / SILVERBLACK/ CYRILLIC'
SELECT RTRIM(LEFT(#desc, CHARINDEX('/', #desc) - 1))
This returns: EXCHANGEUNIT P1i
I need to get only P1i with the query.
Thanks in advance.
Use a combination of REVERSE, LEFT and CHARINDEX - like so:
DECLARE #desc varchar(100)
SET #desc='EXCHANGEUNIT P1i / SILVERBLACK/ CYRILLIC'
SET #subdesc=RTRIM(LEFT(#desc, CHARINDEX('/', #desc) - 1))
SELECT REVERSE( LEFT( REVERSE(#subdesc), CHARINDEX(' ', REVERSE(#subdesc))-1 ) )
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