I have a vairable
DECLARE #AssignOn nvarchar(20)='0,2,5'
I want to check a condition like this
DECLARE #index int
SET DATEFIRST 7
SELECT #index=DATEPART(DW, GETDATE())-1
IF(CONVERT(nvarchar(2),#index) IN #AssignOn)
IN cannot be used here . Any other methods to do this INLINE
You can use CharIndex to find if you have a match. It returns a non zero value if the first string appears in the second.
IF(CHARINDEX(CONVERT(nvarchar(2),#index), #AssignOn) > 0)
The easiest way to do this is to search for the substring ',needle,' in the csv list string. However, this doesn't work correctly for the first and last elements. This can be overcome by concatenating a comma onto each side of the csv list string.
An example in SQL might be:
SELECT
CHARINDEX(','+ NEEDLE +',', ','+ HAYSTACK +',')
FROM table;
Or using LIKE:
SELECT *
FROM table
WHERE ','+ HAYSTACK +',' LIKE '%,'+ NEEDLE +',';
IF CHARINDEX(','+CONVERT(nvarchar(2),#index)+',', ','+#AssignOn+',') <> 0
As you actually define the values in the code you could instead;
DECLARE #AssignOn TABLE (value int)
INSERT #AssignOn VALUES (0),(2),(5)
... #index IN (SELECT value FROM #AssignOn)
Related
Declare #Temp1 as nvarchar(max) = '[10004120][1100][10033583][1005]'
Declare #Temp2 as nvarchar(max) = '[10004120][1100]'
If #Temp1 like #Temp2 + '%'
Print 'Yup'
Why does this not work? I don't get the "yup" message.
This doesn't work because the brackets in the string have a special function in a LIKE statement - items between the brackets constitute a set of values that the singular character at the specified position matches. Your original pattern looks for a 1,0,4, or 2 followed by a 1 or an 0. To make this work, you should have a pattern like this:
Declare #Temp2 as nvarchar(max) = '[[]10004120][[]1100]'
You have the like backwards. The pattern is the second operand. The logic should be:
#temp1 like #temp2 + '%'
At least in MS Sql Server, the square brackets have meaning in a LIKE.
They are a bit similar to a character class in regex.
For example:
WHERE col LIKE '[0-9]%' will find col values that start with a digit.
WHERE col LIKE '%[^0-9]' will find col values that end with anything but a digit.
So something as a LIKE '[10004120][1100]%' is equivalent to a LIKE '[0-24][01]%' and will actually match with strings like '11X', '40Y'.
But in this case you don't have to use LIKE.
Since the first string just has to start with the second string.
IF #Temp2 = LEFT(#Temp1, LEN(#Temp2))
PRINT 'yup'
And to check if the second string is a part of the first string.
You could use CHARINDEX. Which will return 1 if it's found at the start.
IF CHARINDEX(#Temp2, #Temp1) > 0
BEGIN
PRINT 'yup'
END
Rather than add a specific pattern for an open bracket, I would use the optional ESCAPE clause for LIKE expressions:
Declare #Temp1 as nvarchar(max) = '[10004120][1100][10033583][1005]'
Declare #Temp2 as nvarchar(max) = '[10004120][1100]'
If #Temp1 LIKE REPLACE(#Temp2, '[', '\[') + '%' ESCAPE '\'
Print 'Yup'
To me that's cleaner and more explicit about what you're trying to accomplish. Obviously, you've got to pick an escape character that's guaranteed not to be in your data or otherwise use #Temp1 LIKE REPLACE(REPLACE(#Temp2, '\', '\\'), '[', '\[') + '%' ESCAPE '\' to escape embedded escape characters first.
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.
How would you advise to find out in Sql Server 2010/2012 if a query contains a substring equal to a 6 digits number?
e.g. "agh123456 dfsdfdf" matches the requirements
"x123 ddd456" doesn't match the requirements because the 6 digits are not consecutive
"lm123" doesn't match the requirements because only 3 digits are found (out of the required 6)
The problem I encountered so far: is that SUBSTRING as a function requires parameters (position where the number presumably starts and this is random)
while PATINDEX returns the location of a pattern in a string, but we don't know the exact pattern (it can be any 6 digit number)
Any pointers or advice, much appreciated.
Thank you
You can use the LIKE operator:
SELECT *
FROM MyTable
WHERE Mycolumn LIKE '%[0-9][0-9][0-9][0-9][0-9][0-9]%'
Even this should work.. considering you don't have a string like this
abc123 abc123456
Try this
DECLARE #str varchar(max) = 'abcxyz123456'
SELECT ISNUMERIC(SUBSTRING(#str,(SELECT PATINDEX('%[0-9]%',#str)),6))
If you want to select all rows in the table and mask the first 6-digit substring in each row:
DECLARE #mask varchar(max) = '######'
DECLARE #pattern varchar(max) = '%'+REPLACE(#mask,'#','[0-9]')+'%'
SELECT
ISNULL(STUFF(col1,PATINDEX(#pattern,col1),LEN(#mask),#mask),col1)
FROM Table1
Table FOO has a column FILEPATH of type VARCHAR(512). Its entries are absolute paths:
FILEPATH
------------------------------------------------------------
file://very/long/file/path/with/many/slashes/in/it/foo.xml
file://even/longer/file/path/with/more/slashes/in/it/baz.xml
file://something/completely/different/foo.xml
file://short/path/foobar.xml
There's ~50k records in this table and I want to know all distinct filenames, not the file paths:
foo.xml
baz.xml
foobar.xml
This looks easy, but I couldn't find a DB2 scalar function that allows me to search for the last occurrence of a character in a string. Am I overseeing something?
I could do this with a recursive query, but this appears to be overkill for such a simple task and (oh wonder) is extremely slow:
WITH PATHFRAGMENTS (POS, PATHFRAGMENT) AS (
SELECT
1,
FILEPATH
FROM FOO
UNION ALL
SELECT
POSITION('/', PATHFRAGMENT, OCTETS) AS POS,
SUBSTR(PATHFRAGMENT, POSITION('/', PATHFRAGMENT, OCTETS)+1) AS PATHFRAGMENT
FROM PATHFRAGMENTS
)
SELECT DISTINCT PATHFRAGMENT FROM PATHFRAGMENTS WHERE POS = 0
I think what you're looking for is the LOCATE_IN_STRING() scalar function. This is what Info Center has to say if you use a negative start value:
If the value of the integer is less than zero, the search begins at
LENGTH(source-string) + start + 1 and continues for each position to
the beginning of the string.
Combine that with the LENGTH() and RIGHT() scalar functions, and you can get what you want:
SELECT
RIGHT(
FILEPATH
,LENGTH(FILEPATH) - LOCATE_IN_STRING(FILEPATH,'/',-1)
)
FROM FOO
One way to do this is by taking advantage of the power of DB2s XQuery engine. The following worked for me (and fast):
SELECT DISTINCT XMLCAST(
XMLQuery('tokenize($P, ''/'')[last()]' PASSING FILEPATH AS "P")
AS VARCHAR(512) )
FROM FOO
Here I use tokenize to split the file path into a sequence of tokens and then select the last of these tokens. The rest is only conversion from SQL to XML types and back again.
I know that the problem from the OP was already solved but I decided to post the following anyway to hopefully help others like me that land here.
I came across this thread while searching for a solution to my similar problem which had the exact same requirement but was for a different kind of database that was also lacking the REVERSE function.
In my case this was for a OpenEdge (Progress) database, which has a slightly different syntax. This made the INSTR function available to me that most Oracle typed databases offer.
So I came up with the following code:
SELECT
SUBSTRING(
foo.filepath,
INSTR(foo.filepath, '/',1, LENGTH(foo.filepath) - LENGTH( REPLACE( foo.filepath, '/', '')))+1,
LENGTH(foo.filepath))
FROM foo
However, for my specific situation (being the OpenEdge (Progress) database) this did not result into the desired behaviour because replacing the character with an empty char gave the same length as the original string. This doesn't make much sense to me but I was able to bypass the problem with the code below:
SELECT
SUBSTRING(
foo.filepath,
INSTR(foo.filepath, '/',1, LENGTH( REPLACE( foo.filepath, '/', 'XX')) - LENGTH(foo.filepath))+1,
LENGTH(foo.filepath))
FROM foo
Now I understand that this code won't solve the problem for T-SQL because there is no alternative to the INSTR function that offers the Occurence property.
Just to be thorough I'll add the code needed to create this scalar function so it can be used the same way like I did in the above examples.
-- Drop the function if it already exists
IF OBJECT_ID('INSTR', 'FN') IS NOT NULL
DROP FUNCTION INSTR
GO
-- User-defined function to implement Oracle INSTR in SQL Server
CREATE FUNCTION INSTR (#str VARCHAR(8000), #substr VARCHAR(255), #start INT, #occurrence INT)
RETURNS INT
AS
BEGIN
DECLARE #found INT = #occurrence,
#pos INT = #start;
WHILE 1=1
BEGIN
-- Find the next occurrence
SET #pos = CHARINDEX(#substr, #str, #pos);
-- Nothing found
IF #pos IS NULL OR #pos = 0
RETURN #pos;
-- The required occurrence found
IF #found = 1
BREAK;
-- Prepare to find another one occurrence
SET #found = #found - 1;
SET #pos = #pos + 1;
END
RETURN #pos;
END
GO
To avoid the obvious, when the REVERSE function is available you do not need to create this scalar function and you can just get the required result like this:
SELECT
SUBSTRING(
foo.filepath,
LEN(foo.filepath) - CHARINDEX('\', REVERSE(foo.filepath))+2,
LEN(foo.filepath))
FROM foo
You could just do it in a single statement:
select distinct reverse(substring(reverse(FILEPATH), 1, charindex('/', reverse(FILEPATH))-1))
from filetable
Using T-SQL, how would I go about getting the last 3 characters of a varchar column?
So the column text is IDS_ENUM_Change_262147_190 and I need 190
SELECT RIGHT(column, 3)
That's all you need.
You can also do LEFT() in the same way.
Bear in mind if you are using this in a WHERE clause that the RIGHT() can't use any indexes.
You can use either way:
SELECT RIGHT(RTRIM(columnName), 3)
OR
SELECT SUBSTRING(columnName, LEN(columnName)-2, 3)
Because more ways to think about it are always good:
select reverse(substring(reverse(columnName), 1, 3))
declare #newdata varchar(30)
set #newdata='IDS_ENUM_Change_262147_190'
select REVERSE(substring(reverse(#newdata),0,charindex('_',reverse(#newdata))))
=== Explanation ===
I found it easier to read written like this:
SELECT
REVERSE( --4.
SUBSTRING( -- 3.
REVERSE(<field_name>),
0,
CHARINDEX( -- 2.
'<your char of choice>',
REVERSE(<field_name>) -- 1.
)
)
)
FROM
<table_name>
Reverse the text
Look for the first occurrence of a specif char (i.e. first occurrence FROM END of text). Gets the index of this char
Looks at the reversed text again. searches from index 0 to index of your char. This gives the string you are looking for, but in reverse
Reversed the reversed string to give you your desired substring
if you want to specifically find strings which ends with desired characters then this would help you...
select * from tablename where col_name like '%190'