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.
Related
How to Convert the SQL data type Numerci(15,2) to string(varchar data type) without adding the trailing zeros in sybase.
Example- in column abc below values are present-
0.025
0.02
NULL
0.025589
5.289
on running the query-
select STR(column,10,4) from table --- produces the results 0.025,0.0200
select CAST(column as CHAR(5)) from table -- produces the results as 0.0250 etc
I can not do it in presentation layer
Can someone please help with query.
Unfortunately Sybase ASE does not have any native support for regex's, nor any out-of-the-box functions for stripping trailing zeros.
An obvious (?) first attempt might consist of a looping construct to strip off the trailing zeros, though it'd probably be easier to reverse() the initial string, strip off leading zeros, then reverse() to get back to the original value. Unfortunately this is not exactly efficient and would need to be encapsulated in a user-defined function (which introduces an additional performance hit each time it's invoked) in order to use it in a query.
The next idea would be to convert the zeros into something that can be (relatively) easily stripped off the end of the string, and it just so happens that ASE does provide the rtrim() function for stripping trailing spaces. This idea would look something like:
convert all zeros to spaces [str_replace('string','0',' ')]
strip off trailing spaces [rtrim('string')]
convert any remaining spaces back to zeros [str_replace('string',' ','0')]
** This obviously assumes the original string does not contain any spaces.
Here's an example:
declare #mystring varchar(100)
select #mystring = '0.025000'
-- here's a breakdown of each step in the process ...
select ':'+ #mystring + ':' union all
select ':'+ str_replace(#mystring,'0',' ') + ':' union all
select ':'+ rtrim(str_replace(#mystring,'0',' ')) + ':' union all
select ':'+ str_replace(rtrim(str_replace(#mystring,'0',' ')),' ','0') + ':'
-- and the final solution sans the visual end markers (:)
select str_replace(rtrim(str_replace(#mystring,'0',' ')),' ','0')
go
----------
:0.025000:
: . 25 :
: . 25:
:0.025:
--------
0.025
If you need to use this code snippet quite often then you may want to consider wrapping it in a user-defined function, though keep in mind there will be a slight performance hit each time the function is called.
Following approaches can be used -
1) It uses the Replace function
select COLUMN,str_replace(rtrim(str_replace(
str_replace(rtrim(str_replace(cast(COLUMN as varchar(15)), '0', ' ')), ' ', '0')
, '.', ' ')), ' ', '.')
from TABLE
Output -
0.0252
0.0252
2) Using Regex-
select COLUMN ,str(COLUMN ,10,3),
reverse(substring( reverse(str(COLUMN ,10,3)), patindex('%[^0]%',reverse(str(COLUMN ,10,3))), 10))
from TABLE
Output -
0.0252
0.0252.
I am doing some data clean-up and need to Capitalize the first letter of City names. How do I capitalize the second word in a City Like Terra Bella.
SELECT UPPER(LEFT([MAIL CITY],1))+
LOWER(SUBSTRING([MAIL CITY],2,LEN([MAILCITY])))
FROM masterfeelisting
My results is this 'Terra bella' and I need 'Terra Bella'. Thanks in advance.
Ok, I know I answered this before, but it bugged me that we couldn't write something efficient to handle an unknown amount of 'text segments'.
So re-thinking it and researching, I discovered a way to change the [MAILCITY] field into XML nodes where each 'text segment' is assigned it's own Node within the xml field. Then those xml fields can be processed node by node, concatenated together, and then changed back to a SQL varchar. It's convoluted, but it works. :)
Here's the code:
CREATE TABLE
#masterfeelisting (
[MAILCITY] varchar(max) not null
);
INSERT INTO #masterfeelisting VALUES
('terra bellA')
,(' terrA novA ')
,('chicagO ')
,('bostoN')
,('porT dE sanTo')
,(' porT dE sanTo pallo ');
SELECT
RTRIM
(
(SELECT
UPPER([xmlField].[xmlNode].value('.', 'char(1)')) +
LOWER(STUFF([xmlField].[xmlNode].value('.', 'varchar(max)'), 1, 1, '')) + ' '
FROM [xmlNodeRecordSet].[nodeField].nodes('/N') as [xmlField]([xmlNode]) FOR
xml path(''), type
).value('.', 'varchar(max)')
) as [MAILCITY]
FROM
(SELECT
CAST('<N>' + REPLACE([MAILCITY],' ','</N><N>')+'</N>' as xml) as [nodeField]
FROM #masterfeelisting
) as [xmlNodeRecordSet];
Drop table #masterfeelisting;
First I create a table and fill it with dummy values.
Now here is the beauty of the code:
For each record in #masterfeelisting, we are going to create an xml field with a node for each 'text segment'.
ie. '<N></N><N>terrA</N><N>novA</N><N></N>'
(This is built from the varchar ' terrA novA ')
1) The way this is done is by using the REPLACE function.
The string starts with a '<N>' to designate the beginning of the node. Then:
REPLACE([MAILCITY],' ','</N><N>')
This effectively goes through the whole [MAILCITY] string and replaces each
' ' with '</N><N>'
and then the string ends with a '</N>'. Where '</N>' designates the end of each node.
So now we have a beautiful XML string with a couple of empty nodes and the 'text segments' nicely nestled in their own node. All the 'spaces' have been removed.
2) Then we have to CAST the string into xml. And we will name that field [nodeField]. Now we can use xml functions on our newly created record set. (Conveniently named [xmlNodeRecordSet].)
3) Now we can read the [xmlNodeRecordSet] into the main sub-Select by stating:
FROM [xmlNodeRecordSet].[nodeField].nodes('/N')
This tells us we are reading the [nodeField] as nodes with a '/N' delimiter.
This table of node fields is then parsed by stating:
as [xmlField]([xmlNode]) FOR xml path(''), type
This means each [xmlField] will be parsed for each [xmlNode] in the xml string.
4) So in the main sub-select:
Each blank node '<N></N>' is discarded. (Or not processed.)
Each node with a 'text segment' in it will be parsed. ie <N>terrA</N>
UPPER([xmlField].[xmlNode].value('.', 'char(1)')) +
This code will grab each node out of the field and take its contents '.' and only grab the first character 'char(1)'. Then it will Upper case that character. (the plus sign at the end means it will concatenate this letter with the next bit of code:
LOWER(STUFF([xmlField].[xmlNode].value('.', 'varchar(max)'), 1, 1, ''))
Now here is the beauty... STUFF is a function that will take a string, from a position, for a length, and substitute another string.
STUFF(string, start position, length, replacement string)
So our string is:
[xmlField].[xmlNode].value('.', 'varchar(max)')
Which grabs the whole string inside the current node since it is 'varchar(max)'.
The start position is 1. The length is 1. And the replacement string is ''. This effectively strips off the first character by replacing it with nothing. So the remaining string is all the other characters that we want to have lower case. So that's what we do... we use LOWER to make them all lower case. And this result is concatenated to our first letter that we already upper cased.
But wait... we are not done yet... we still have to append a + ' '. Which adds a blank space after our nicely capitalized 'text segment'. Just in case there is another 'text segment' after this node is done.
This main sub-Select will now parse each node in our [xmlField] and concatenate them all nicely together.
5) But now that we have one big happy concatenation, we still have to change it back from an xml field to a SQL varchar field. So after the main sub-select we need:
.value('.', 'varchar(max)')
This changes our [MAILCITY] back to a SQL varchar.
6) But hold on... we still are not done. Remember we put an extra space at the end of each 'text segment'??? Well the last 'text segment still has that extra space after it. So we need to Right Trim that space off by using RTRIM.
7) And dont forget to rename the final field back to as [MAILCITY]
8) And that's it. This code will take an unknown amount of 'text segments' and format each one of them. All using the fun of XML and it's node parsers.
Hope that helps :)
Here's one way to handle this using APPLY. Note that this solution supports up to 3 substrings (e.g. "Phoenix", "New York", "New York City") but can easily be updated to handle more.
DECLARE #string varchar(100) = 'nEW yoRk ciTY';
WITH DELIMCOUNT(String, DC) AS
(
SELECT #string, LEN(RTRIM(LTRIM(#string)))-LEN(REPLACE(RTRIM(LTRIM(#string)),' ',''))
),
CIPOS AS
(
SELECT *
FROM DELIMCOUNT
CROSS APPLY (SELECT CHARINDEX(char(32), string, 1)) CI1(CI1)
CROSS APPLY (SELECT CHARINDEX(char(32), string, CI1.CI1+1)) CI2(CI2)
)
SELECT
OldString = #string,
NewString =
CASE DC
WHEN 0 THEN UPPER(SUBSTRING(string,1,1))+LOWER(SUBSTRING(string,2,8000))
WHEN 1 THEN UPPER(SUBSTRING(string,1,1))+LOWER(SUBSTRING(string,2,CI1-1)) +
UPPER(SUBSTRING(string,CI1+1,1))+LOWER(SUBSTRING(string,CI1+2,100))
WHEN 2 THEN UPPER(SUBSTRING(string,1,1))+LOWER(SUBSTRING(string,2,CI1-1)) +
UPPER(SUBSTRING(string,CI1+1,1))+LOWER(SUBSTRING(string,CI1+2,CI2-(CI1+1))) +
UPPER(SUBSTRING(string,CI2+1,1))+LOWER(SUBSTRING(string,CI2+2,100))
END
FROM CIPOS;
Results:
OldString NewString
--------------- --------------
nEW yoRk ciTY New York City
This will only capitalize the first letter of the second word. A shorter but less flexible approach. Replace #str with [Mail City].
DECLARE #str AS VARCHAR(50) = 'Los angelas'
SELECT STUFF(#str, CHARINDEX(' ', #str) + 1, 1, UPPER(SUBSTRING(#str, CHARINDEX(' ', #str) + 1, 1)));
This is a way to use imbedded Selects for three City name parts.
It uses CHARINDEX to find the location of your separator character. (ie a space)
I put an 'if' structure around the Select to test if you have any records with more than 3 parts to the city name. If you ever get the warning message, you could add another sub-Select to handle another city part.
Although... just to be clear... SQL is not the best language to do complicated formatting. It was written as a data retrieval engine with the idea that another program will take that data and massage it into a friendlier look and feel. It may be easier to handle the formatting in another program. But if you insist on using SQL and you need to account for city names with 5 or more parts... you may want to consider using Cursors so you can loop through the variable possibilities. (But Cursors are not a good habit to get into. So don't do that unless you've exhausted all other options.)
Anyway, the following code creates and populates a table so you can test the code and see how it works. Enjoy!
CREATE TABLE
#masterfeelisting (
[MAILCITY] varchar(30) not null
);
Insert into #masterfeelisting select 'terra bella';
Insert into #masterfeelisting select ' terrA novA ';
Insert into #masterfeelisting select 'chicagO ';
Insert into #masterfeelisting select 'bostoN';
Insert into #masterfeelisting select 'porT dE sanTo';
--Insert into #masterfeelisting select ' porT dE sanTo pallo ';
Declare #intSpaceCount as integer;
SELECT #intSpaceCount = max (len(RTRIM(LTRIM([MAILCITY]))) - len(replace([MAILCITY],' ',''))) FROM #masterfeelisting;
if #intSpaceCount > 2
SELECT 'You need to account for more than 3 city name parts ' as Warning, #intSpaceCount as SpacesFound;
else
SELECT
cThird.[MAILCITY1] + cThird.[MAILCITY2] + cThird.[MAILCITY3] as [MAILCITY]
FROM
(SELECT
bSecond.[MAILCITY1] as [MAILCITY1]
,SUBSTRING(bSecond.[MAILCITY2],1,bSecond.[intCol2]) as [MAILCITY2]
,UPPER(SUBSTRING(bSecond.[MAILCITY2],bSecond.[intCol2] + 1, 1)) +
SUBSTRING(bSecond.[MAILCITY2],bSecond.[intCol2] + 2,LEN(bSecond.[MAILCITY2]) - bSecond.[intCol2]) as [MAILCITY3]
FROM
(SELECT
SUBSTRING(aFirst.[MAILCITY],1,aFirst.[intCol1]) as [MAILCITY1]
,UPPER(SUBSTRING(aFirst.[MAILCITY],aFirst.[intCol1] + 1, 1)) +
SUBSTRING(aFirst.[MAILCITY],aFirst.[intCol1] + 2,LEN(aFirst.[MAILCITY]) - aFirst.[intCol1]) as [MAILCITY2]
,CHARINDEX ( ' ', SUBSTRING(aFirst.[MAILCITY],aFirst.[intCol1] + 1, LEN(aFirst.[MAILCITY]) - aFirst.[intCol1]) ) as intCol2
FROM
(SELECT
UPPER (LEFT(RTRIM(LTRIM(mstr.[MAILCITY])),1)) +
LOWER(SUBSTRING(RTRIM(LTRIM(mstr.[MAILCITY])),2,LEN(RTRIM(LTRIM(mstr.[MAILCITY])))-1)) as [MAILCITY]
,CHARINDEX ( ' ', RTRIM(LTRIM(mstr.[MAILCITY]))) as intCol1
FROM
#masterfeelisting as mstr -- Initial Master Table
) as aFirst -- First Select Shell
) as bSecond -- Second Select Shell
) as cThird; -- Third Select Shell
Drop table #masterfeelisting;
How can I remove any white space from substring of string?
For example I have this number '+370 650 12345'. I need all numbers to have this format country_code rest_of_the_number or in that example: +370 65012345. How could you achieve that with PostgreSQL?
I could use trim() function, but then it would remove all whitespace.
Assuming the column is named phone_number:
left(phone_number, strpos(phone_number, ' '))
||regexp_replace(substr(phone_number, strpos(phone_number, ' ') + 1), ' ', '', 'g')
It first takes everything up to the first space and then concatenates it with the result of replacing all spaces from the rest of the string.
If you also need to deal with other whitespace than just a space, you could use '\s' for the search value in regexp_replace()
If you are able to assume that a country code will always be present, you could try using a regular expression to capture the parts of interest. Assuming that your phone numbers are stored in a column named content in a table named numbers, you could try something like the following:
SELECT parts[1] || ' ' || parts[2] || parts[3]
FROM (
SELECT
regexp_matches(content, E'^\\s*(\\+\\d+)\\s+(\\d+)\\s+(\\d+)\\s*$') AS parts
FROM numbers
) t;
The following will work even if the country code is absent (see SQL Fiddle Demo here):
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('+370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: +370 65012345
SELECT TRIM(REPLACE(REPLACE(REGEXP_REPLACE('370 650 12345', '^((\+\d+)\s+)?(.*)$', '\1|\3'), ' ', ''), '|', ' '));
Returns: 37065012345
It looks for a country code (a set of numbers starting with a + sign) at the beginning, and replaces any whitespace following that code with a pipe |. It then replaces all the spaces in the resulting string with the empty string, then replaces occurrences of | with spaces! The choice of | is arbitrary, I suppose it could be any non-digit character.
All I have what seems to be a pretty straight forward question, that I haven't been able to figure out. For example, if I have a text string like T6L 7H5. Using SQL I need to remove the inner white space from this string so that it displays like T6L7H5.
Things to consider:
Teradata (v.13.10) is my RDBMS, so REPLACE('T6L 7H5', ' ', '') is not an
option here.
On this particular server I am a business user w/ very limited
permissions so creating a UDF is not an option either.
Can't test this, so spit-balling, but you should be able to leverage Substring and Position:
SELECT SUBSTRING('T6L 7H5', 1, POSITION (' ' IN 'T6L 7H5')-1) || SUBSTRING('T6L 7H5', POSITION (' ' IN 'T6L 7H5')+1, CHARACTER_LENGTH('T6L 7H5') - POSITION (' ' IN 'T6L 7H5') )
If the field is consistently formatted like your example then:
substring('T6L 7H5',1,3)||substring('T6L 7H5',4,3)
For single white space you can use POSITION() and SUBSTRING(). You have to offset for the whitespace location that is returned by the POSITION() function.
WITH CTE(FieldName) AS
(SELECT 'TB7 TCH' AS FieldName)
SELECT SUBSTRING(FieldName FROM 1 FOR (POSITION(' ' IN FieldName) - 1))
|| SUBSTRING(FieldName FROM (POSITION(' ' IN FieldName) + 1))
FROM CTE;
try using substr( );
Ex : 4= start at position 4 && 1 = delete one char.
substr(Table1.fieldname, 4,1)
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