This is the easiest answer to retrieve the 67 (type-safe!!):
SELECT CAST('<x>' + REPLACE('1,222,2,67,888,1111',',','</x><x>') + '</x>' AS XML).value('/x[4]','int')
In the following you will find examples how to use this with variables for the string, the delimiter and the position (even for edge-cases with XML-forbidden characters)
##The easy one
This question is not about a string split approach, but about how to get the nth element. The easiest, fully inlineable way would be this IMO:
This is a real one-liner to get part 2 delimited by a space:
DECLARE @input NVARCHAR(100)=N'part1 part2 part3';
SELECT CAST(N'<x>' + REPLACE(@input,N' ',N'</x><x>') + N'</x>' AS XML).value('/x[2]','nvarchar(max)')
##Variables can be used with sql:variable()
or sql:column()
Of course you can use variables for delimiter and position (use sql:column
to retrieve the position directly from a query's value):
DECLARE @dlmt NVARCHAR(10)=N' ';
DECLARE @pos INT = 2;
SELECT CAST(N'<x>' + REPLACE(@input,@dlmt,N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)')
##Edge-Case with XML-forbidden characters
If your string might include forbidden characters, you still can do it this way. Just use FOR XML PATH
on your string first to replace all forbidden characters with the fitting escape sequence implicitly.
It's a very special case if - additionally - your delimiter is the semicolon. In this case I replace the delimiter first to '#DLMT#', and replace this to the XML tags finally:
SET @input=N'Some <, > and &;Other äöü@€;One more';
SET @dlmt=N';';
SELECT CAST(N'<x>' + REPLACE((SELECT REPLACE(@input,@dlmt,'#DLMT#') AS [*] FOR XML PATH('')),N'#DLMT#',N'</x><x>') + N'</x>' AS XML).value('/x[sql:variable("@pos")][1]','nvarchar(max)');
##UPDATE for SQL-Server 2016+
Regretfully the developers forgot to return the part's index with STRING_SPLIT
. But, using SQL-Server 2016+, there is JSON_VALUE
and OPENJSON
.
With JSON_VALUE
we can pass in the position as the index' array.
For OPENJSON
the documentation states clearly:
When OPENJSON parses a JSON array, the function returns the indexes of the elements in the JSON text as keys.
A string like 1,2,3
needs nothing more than brackets: [1,2,3]
.
A string of words like this is an example
needs to be ["this","is","an"," example"]
.
These are very easy string operations. Just try it out:
DECLARE @str VARCHAR(100)='Hello John Smith';
DECLARE @position INT = 2;
--We can build the json-path '$[1]' using CONCAT
SELECT JSON_VALUE('["' + REPLACE(@str,' ','","') + '"]',CONCAT('$[',@position-1,']'));
--See this for a position safe string-splitter (zero-based):
SELECT JsonArray.[key] AS [Position]
,JsonArray.[value] AS [Part]
FROM OPENJSON('["' + REPLACE(@str,' ','","') + '"]') JsonArray
In this post I tested various approaches and found, that OPENJSON
is really fast. Even much faster than the famous "delimitedSplit8k()" method...
##UPDATE 2 - Get the values type-safe
We can use an array within an array simply by using doubled [[]]
. This allows for a typed WITH
-clause:
DECLARE @SomeDelimitedString VARCHAR(100)='part1|1|20190920';
DECLARE @JsonArray NVARCHAR(MAX)=CONCAT('[["',REPLACE(@SomeDelimitedString,'|','","'),'"]]');
SELECT @SomeDelimitedString AS TheOriginal
,@JsonArray AS TransformedToJSON
,ValuesFromTheArray.*
FROM OPENJSON(@JsonArray)
WITH(TheFirstFragment VARCHAR(100) '$[0]'
,TheSecondFragment INT '$[1]'
,TheThirdFragment DATE '$[2]') ValuesFromTheArray