I am using SQL Server 2008
I have sql string in column with ; separated values. How i can trim the below value
Current string:
;145615;1676288;178829;
Output:
145615;1676288;178829;
Please help with sql query to trim the first ; from string
Note : The first char may be or may not be ; but if it is ; then only it should trim.
Edit: What i had tried before, although it doesn't make sense after so many good responses.
DECLARE
#VAL VARCHAR(1000)
BEGIN
SET #VAL =';13342762;1334273;'
IF(CHARINDEX(';',#VAL,1)=1)
BEGIN
SELECT SUBSTRING(#VAL,2,LEN(#VAL))
END
ELSE
BEGIN
SELECT #VAL
END
END
SELECT CASE WHEN col LIKE ';%'
THEN STUFF(col,1,1,'') ELSE col END
FROM dbo.table;
Just check the first character, and if it matches, start from the second character:
SELECT CASE WHEN SUBSTRING(col,1,1) = ';'
THEN SUBSTRING(col,2,LEN(col))
ELSE col
END AS col
Here's an example:
DECLARE #v varchar(10)
SET #v = ';1234'
SELECT
CASE
WHEN LEFT(#v,1) = ';' THEN RIGHT(#v, LEN(#v) - 1)
ELSE #v
END
A further development on #Aaron Bertrand's answer:
SELECT
STUFF(col, 1, PATINDEX(';%', col), '')
FROM ...
PATINDEX is similar to LIKE in that it uses a pattern search, but being a function it also returns the position of the first match. In this case, since we a looking for a ; specifically at the beginning of a string, the position returned is going to be either 1 (if found) or 0 (if not found). If it is 1, the STUFF function will delete 1 character at the beginning of the string, and if the position is 0, STUFF will delete 0 characters.
Related
I want to know a flexible way to extract the string between two '-'. The issue is that '-' may or may not exist in the source string. I have the following code which works fine when '-' exists twice in the source string, marking the start and end of the extracted string. But it throws an error "Invalid length parameter passed to the LEFT or SUBSTRING function" when there is only one '-' or none at all or if the string is blank. Can someone please help? Thanks
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
Desired Output: ExtractThis
If there is one dash only e.g. 'BLAH90-ExtractThisWOW' then the output should be everything after the first dash i.e. ExtractThisWOW. If there are no dashes then the string will have a blank space instead e.g. 'BLAH90 ExtractThisWOW' and should return everything after the blank space i.e. ExtractThisWOW.
You can try something like this.
When there is no dash, it starts at the space if there is one or take the whole string if not.
Then I look if there is only one dash or 2
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
declare #dash_pos integer = CHARINDEX('-',#string)
SELECT CASE
WHEN #dash_pos = 0 THEN
RIGHT(#string,LEN(#string)-CHARINDEX(' ',#string))
ELSE (
CASE
WHEN #dash_pos = LEN(#string)-CHARINDEX('-',REVERSE(#string))+1
THEN RIGHT(#string,LEN(#string)-#dash_pos)
ELSE SUBSTRING(#string,#dash_pos+1, CHARINDEX('-',#string,#dash_pos+1) -
#dash_pos -1)
END
)
END as My_String
Try this. If there are two dashes, it'll take what is inside. If there is only one or none, it'll keep the original string.
declare #string varchar(100) = 'BLAH-90ExtractThisWOW'
declare #dash_index1 int = case when #string like '%-%' then CHARINDEX('-', #string) else -1 end
declare #dash_index2 int = case when #string like '%-%'then len(#string) - CHARINDEX('-', reverse(#string)) + 1 else -1 end
SELECT case
when #dash_index1 <> #dash_index2 then SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1)
else #string end
as My_String
Take your existing code:
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
insert one line, like so:
declare #string varchar(100) = 'BLAH90-ExtractThis-WOW'
SET #string = #string + '--'
SELECT SUBSTRING(#string,CHARINDEX('-',#string)+1, CHARINDEX('-',#string,CHARINDEX('-',#string)+1) -CHARINDEX('-',#string)-1) as My_String
and you're done. (If NULL, you will get NULL returned. Also, this will return all data based on the FIRST dash found in the string, regardless of however many dashes are in the string.)
I am working in SQL Server 2008. I am trying to test whether a string (varchar) has only digit characters (0-9). I know that the IS_NUMERIC function can give spurious results. (My data can possibly have $ signs, which should not pass the test.) So, I'm avoiding that function.
I already have a test to see if a string has any non-digit characters, i.e.,
some_column LIKE '%[^0123456789]%'
I would think that the only-digits test would be something similar, but I'm drawing a blank. Any ideas?
Use Not Like
where some_column NOT LIKE '%[^0-9]%'
Demo
declare #str varchar(50)='50'--'asdarew345'
select 1 where #str NOT LIKE '%[^0-9]%'
There is a system function called ISNUMERIC for SQL 2008 and up. An example:
SELECT myCol
FROM mTable
WHERE ISNUMERIC(myCol)<> 1;
I did a couple of quick tests and also looked further into the docs:
ISNUMERIC returns 1 when the input expression evaluates to a valid numeric data type; otherwise it returns 0.
Which means it is fairly predictable for example
-9879210433 would pass but 987921-0433 does not.
$9879210433 would pass but 9879210$433 does not.
So using this information you can weed out based on the list of valid currency symbols and + & - characters.
Solution:
where some_column NOT LIKE '%[^0-9]%'
Is correct.
Just one important note: Add validation for when the string column = '' (empty string). This scenario will return that '' is a valid number as well.
Method that will work. The way it is used above will not work.
declare #str varchar(50)='79136'
select
case
when #str LIKE replicate('[0-9]',LEN(#str)) then 1
else 0
end
declare #str2 varchar(50)='79D136'
select
case
when #str2 LIKE replicate('[0-9]',LEN(#str)) then 1
else 0
end
DECLARE #x int=1
declare #exit bit=1
WHILE #x<=len('123c') AND #exit=1
BEGIN
IF ascii(SUBSTRING('123c',#x,1)) BETWEEN 48 AND 57
BEGIN
set #x=#x+1
END
ELSE
BEGIN
SET #exit=0
PRINT 'string is not all numeric -:('
END
END
I was attempting to find strings with numbers ONLY, no punctuation or anything else. I finally found an answer that would work here.
Using PATINDEX('%[^0-9]%', some_column) = 0 allowed me to filter out everything but actual number strings.
The selected answer does not work.
declare #str varchar(50)='79D136'
select 1 where #str NOT LIKE '%[^0-9]%'
I don't have a solution but know of this potential pitfall. The same goes if you substitute the letter 'D' for 'E' which is scientific notation.
I have a table with only 1 column that contains a string. I am trying to only get the email address. How can I do that? I looked at Substring/Ltrim, etc, but I haven't been able to piece together how to extract only part of the string. I am fairly new to SQL. Thank you for your help!
Column1:
John Smith Email: John.Smith#987456email.com Terminate:
Jacqueline Smith Email: Jacqueline.Smith#987456email.com Terminate:
Assuming the email is prefixed by Email: and does not contain spaces, you can just take all characters after Email: and before the next space (or end of string);
SELECT CASE WHEN CHARINDEX(' ', a.em) <> 0
THEN SUBSTRING(a.em, 1, CHARINDEX(' ', a.em) - 1)
ELSE a.em END email
FROM (
SELECT SUBSTRING(column1, CHARINDEX('Email: ', column1) + 7, LEN(column1)) em
FROM mytable
) a
The subquery keeps anything after Email: and the outer query cuts everything trailing the next space (or end of string).
The query assumes that there is an Email: tag, if that's not guaranteed, you'll want to use a WHERE to make sure that only rows that have will be returned.
An SQLfiddle to test with.
I'm making a few assumptions about your data, namely that the characters 'Name:' don't appear before the name and that each line includes the substring 'Terminate:'
In SQL Server, use a combination of PATINDEX, CHARINDEX and SUBSTRING to parse the address from the string in each row. The cursor lets you loop through your table. This will print out all the e-mail addresses in your table. It needs formatting and if you want to search for a particular person's email, you will have to modify the select statement with a WHERE clause. I hope this helps:
declare #strung as nvarchar(255)
,#start as int
,#end as int
,#result as int
,#emailCursor Cursor
set #emailCursor = CURSOR FORWARD_ONLY STATIC READ_ONLY FOR
Select yourColumnName
from yourTableName
OPEN #emailCursor
FETCH NEXT FROM #emailCursor INTO #strung
WHILE ##FETCH_STATUS = 0
BEGIN
set #start = (select charindex(':',#strung)+1);
set #end = (SELECT PATINDEX('%Terminate:%', #strung)-1)
set #result = #end-#start
set #address = (select SUBSTRING(#strung, #start, #result ) AS eMailAddress)
print #address
FETCH NEXT FROM #emailCursor INTO #strung
END
CLOSE #emailCursor
DEALLOCATE #emailCursor
CHARINDEX returns the position of the first ':' character in your string (the one after EMAIL). I add one to that value to move you past the ':'
PATINDEX returns the beginning position of the substring 'Terminate'
SUBSTRING returns all the character between the starting position [CHARNINDEX(':', #strung)] and the space before 'Terminate' [PATINDEX('%Terminate:%', #strung)]
http://sqlfiddle.com/#!6/5ce48/8/0
select ltrim(rtrim(substring(column1,patindex(column1,':')+1, len(column1)-patindex(column1,':')-11))) as email
from t;
assumes Terminate: is consistent and first : denotes end of first characters to remove.
What it does:
It uses SUBSTRING() to take out a string based on a specific starting character and end character.
To determine the start character we look for the first occurrence of : using PATINDEX() and add 2 1 for the space, 1 to move to the starting character. This gives us the starting potion for SUBSTRING()
To determine how many characters in the SUBSTRING() we take the LEN() of the entire string subtract from it the length of everything before the first ':' + 1 and the length of ' Terminate:' (11)
Again this HEAVILY assumes consistent formatting. If it's not ': space' and ' Terminate:' isn't 11 with a space, then this doesn't work.
This also works...
declare #data varchar(100) = 'John Smith Email: John.Smith#987456email.com
Terminate:'
select SUBSTRING(#data, PATINDEX('%Email%', #data)+7, PATINDEX('%com%', #data)-
PATINDEX('%Email%', #data)-4)
I currently have a table Telephone it has entries like the following:
9073456789101
+773456789101
0773456789101
What I want to do is remove only the 9 from the start of all the entries that have a 9 there but leave the others as they are.
any help would be greatly appreciated.
While all other answer are probably also working, I'd suggest to try and use STUFF function to easily replace a part of the string.
UPDATE Telephone
SET number = STUFF(number,1,1,'')
WHERE number LIKE '9%'
SQLFiddle DEMO
Here is the code and a SQLFiddle
SELECT CASE
WHEN substring(telephone_number, 1, 1) <> '9'
THEN telephone_number
ELSE substring(telephone_number, 2, LEN(telephone_number))
END
FROM Telephone
Update Telephone set number = RIGHT(number,LEN(number)-1) WHERE number LIKE '9%';
I recently solved a similar problem with a combination of RIGHT(), LEN() & PATINDEX(). PATINDEX will return the integer 1 when it finds a 9 as the first character and 0 otherwise. This method allows all records to be returned at once without a CASE WHEN statement.
SELECT
RIGHT(number, LEN(number) - PATINDEX('9%', number))
FROM Telephone
UPDATE dbo.Telephone
SET column_name = SUBSTRING(column_name, 2, 255)
WHERE column_name LIKE '9%';
Stuff is a great function for this. However, using it with an update statement with a where clause is great, but what if I was doing an insert, and I needed all of the rows inserted in one pass. The below will remove the first character if it is a period, does not use the slower case statement, and converts nulls to an empty string.
DECLARE #Attachment varchar(6) = '.GIF',
#Attachment2 varchar(6)
SELECT
#Attachment2 = ISNULL(ISNULL(NULLIF(LEFT(#Attachment, 1), '.'), '') + STUFF(#Attachment, 1, 1, ''), '')
SELECT
#Attachment2
DECLARE #STR nvarchar(200) = 'TEST'
SET #STR = STUFF(#STR,1,1,'')
PRINT #STR
Result will be "EST"
You can use replace in select statement instead of where or update
SELECT REPLACE(REPLACE('_'+number,'_9',''),'_','') FROM #tbl
I am working on a SQL query that reads from a SQLServer database to produce an extract file. One of the requirements to remove the leading zeroes from a particular field, which is a simple VARCHAR(10) field. So, for example, if the field contains '00001A', the SELECT statement needs to return the data as '1A'.
Is there a way in SQL to easily remove the leading zeroes in this way? I know there is an RTRIM function, but this seems only to remove spaces.
select substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
select replace(ltrim(replace(ColumnName,'0',' ')),' ','0')
You can use this:
SELECT REPLACE(LTRIM(REPLACE('000010A', '0', ' ')),' ', '0')
I had the same need and used this:
select
case
when left(column,1) = '0'
then right(column, (len(column)-1))
else column
end
select substring(substring('B10000N0Z', patindex('%[0]%','B10000N0Z'), 20),
patindex('%[^0]%',substring('B10000N0Z', patindex('%[0]%','B10000N0Z'),
20)), 20)
returns N0Z, that is, will get rid of leading zeroes and anything that comes before them.
If you want the query to return a 0 instead of a string of zeroes or any other value for that matter you can turn this into a case statement like this:
select CASE
WHEN ColumnName = substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
THEN '0'
ELSE substring(ColumnName, patindex('%[^0]%',ColumnName), 10)
END
In case you want to remove the leading zeros from a string with a unknown size.
You may consider using the STUFF command.
Here is an example of how it would work.
SELECT ISNULL(STUFF(ColumnName
,1
,patindex('%[^0]%',ColumnName)-1
,'')
,REPLACE(ColumnName,'0','')
)
See in fiddler various scenarios it will cover
https://dbfiddle.uk/?rdbms=sqlserver_2012&fiddle=14c2dca84aa28f2a7a1fac59c9412d48
You can try this - it takes special care to only remove leading zeroes if needed:
DECLARE #LeadingZeros VARCHAR(10) ='-000987000'
SET #LeadingZeros =
CASE WHEN PATINDEX('%-0', #LeadingZeros) = 1 THEN
#LeadingZeros
ELSE
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
END
SELECT #LeadingZeros
Or you can simply call
CAST(CAST(#LeadingZeros AS INT) AS VARCHAR(10))
Here is the SQL scalar value function that removes leading zeros from string:
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
-- =============================================
-- Author: Vikas Patel
-- Create date: 01/31/2019
-- Description: Remove leading zeros from string
-- =============================================
CREATE FUNCTION dbo.funRemoveLeadingZeros
(
-- Add the parameters for the function here
#Input varchar(max)
)
RETURNS varchar(max)
AS
BEGIN
-- Declare the return variable here
DECLARE #Result varchar(max)
-- Add the T-SQL statements to compute the return value here
SET #Result = #Input
WHILE LEFT(#Result, 1) = '0'
BEGIN
SET #Result = SUBSTRING(#Result, 2, LEN(#Result) - 1)
END
-- Return the result of the function
RETURN #Result
END
GO
To remove the leading 0 from month following statement will definitely work.
SELECT replace(left(Convert(nvarchar,GETDATE(),101),2),'0','')+RIGHT(Convert(nvarchar,GETDATE(),101),8)
Just Replace GETDATE() with the date field of your Table.
To remove leading 0, You can multiply number column with 1
Eg: Select (ColumnName * 1)
select CASE
WHEN TRY_CONVERT(bigint,Mtrl_Nbr) = 0
THEN ''
ELSE substring(Mtrl_Nbr, patindex('%[^0]%',Mtrl_Nbr), 18)
END
you can try this
SELECT REPLACE(columnname,'0','') FROM table
I borrowed from ideas above. This is neither fast nor elegant. but it is accurate.
CASE
WHEN left(column, 3) = '000' THEN right(column, (len(column)-3))
WHEN left(column, 2) = '00' THEN right(a.column, (len(column)-2))
WHEN left(column, 1) = '0' THEN right(a.column, (len(column)-1))
ELSE
END
select ltrim('000045', '0') from dual;
LTRIM
-----
45
This should do.