How to SORT in order as entered in SQL Server?

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猫巷女王i
猫巷女王i 2020-12-04 00:44

I\'m using SQL Server and I\'m trying to find results but I would like to get the results in the same order as I had input the conditions.

My code:

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5条回答
  • 2020-12-04 01:17

    Here is an in-line approach

    Example

    Declare @List varchar(max)='212345, 312345, 145687, 658975, 256987, 365874, 568974, 124578, 125689'
    
    Select A.AccountNumber 
          ,A.EndDate
     From  Accounts A
     Join (
            Select RetSeq = Row_Number() over (Order By (Select null))
                  ,RetVal = v.value('(./text())[1]', 'int')
            From  (values (convert(xml,'<x>' + replace(@List,',','</x><x>')+'</x>'))) x(n)
            Cross Apply n.nodes('x') node(v)
          ) B on A.AccountNumber = B.RetVal
     Order By B.RetSeq
    

    EDIT - the subquery Returns

    RetSeq  RetVal
    1       212345
    2       312345
    3       145687
    4       658975
    5       256987
    6       365874
    7       568974
    8       124578
    9       125689
    
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  • 2020-12-04 01:21

    Okay, after some re-thinking I'll offer the ultimative XML based type-safe and sort-safe splitter:

    Declare @List varchar(max)='212345, 312345, 145687, 658975, 256987, 365874, 568974, 124578, 125689';
    DECLARE @delimiter VARCHAR(10)=', ';
    
    WITH Casted AS
    (
        SELECT (LEN(@List)-LEN(REPLACE(@List,@delimiter,'')))/LEN(REPLACE(@delimiter,' ','.')) + 1 AS ElementCount
               ,CAST('<x>' + REPLACE((SELECT @List AS [*] FOR XML PATH('')),@delimiter,'</x><x>')+'</x>' AS XML) AS ListXml
    )
    ,Tally(Nmbr) As
    (
        SELECT TOP((SELECT ElementCount FROM Casted)) ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) FROM master..spt_values v1 CROSS JOIN master..spt_values v2
    )
    SELECT Tally.Nmbr AS Position
          ,(SELECT ListXml.value('(/x[sql:column("Tally.Nmbr")])[1]','int') FROM Casted) AS Item 
    FROM Tally;
    

    The trick is to create a list of running numbers with the fitting number of element (a number's table was even better) and to pick the elements according to their position.

    Hint: This is rather slow...

    UPDATE: even better:

    WITH Casted AS
    (
        SELECT (LEN(@List)-LEN(REPLACE(@List,@delimiter,'')))/LEN(REPLACE(@delimiter,' ','.')) + 1 AS ElementCount
               ,CAST('<x>' + REPLACE((SELECT @List AS [*] FOR XML PATH('')),@delimiter,'</x><x>')+'</x>' AS XML)
               .query('
                       for $x in /x
                       return <x p="{count(/x[. << $x])}">{$x/text()[1]}</x>
                      ') AS ListXml
    )
    SELECT x.value('@p','int') AS Position
          ,x.value('text()[1]','int') AS Item 
    FROM Casted
    CROSS APPLY Casted.ListXml.nodes('/x') AS A(x);
    

    Elements are create as

    <x p="99">TheValue</x>
    

    Regrettfully the XQuery function position() is not available to retrieve the value. But you can use the trick to count all elements before a given node. this is scaling badly, as this count must be performed over and over. The more elements the worse it goes...

    UPDATE2: With a known count of elements one might use this (much better performance)

    Use XQuery to iterate a literally given list:

    WITH Casted AS
    (
        SELECT (LEN(@List)-LEN(REPLACE(@List,@delimiter,'')))/LEN(REPLACE(@delimiter,' ','.')) + 1 AS ElementCount
               ,CAST('<x>' + REPLACE((SELECT @List AS [*] FOR XML PATH('')),@delimiter,'</x><x>')+'</x>' AS XML)
               .query('
                       for $i in (1,2,3,4,5,6,7,8,9)
                       return <x p="{$i}">{/x[$i]/text()[1]}</x>
                      ') AS ListXml
    )
    SELECT x.value('@p','int') AS Position
          ,x.value('text()[1]','int') AS Item 
    FROM Casted
    CROSS APPLY Casted.ListXml.nodes('/x') AS A(x);
    
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  • 2020-12-04 01:23

    You can replace IN with a JOIN, and set a field for ordering, like this:

    SELECT AccountNumber , EndDate
    FROM Accounts a
    JOIN (
        SELECT 212345 AS Number, 1 AS SeqOrder
    UNION ALL
        SELECT 312345 AS Number, 2 AS SeqOrder
    UNION ALL
        SELECT 145687 AS Number, 3 AS SeqOrder
    UNION ALL
        ... -- and so on
    ) AS inlist ON inlist.Number = a.AccountNumber
    ORDER BY inlist.SeqOrder
    
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  • 2020-12-04 01:24

    This is not an answer, just some test-code to check John Cappelletti's approach.

    DECLARE @tbl TABLE(ID INT IDENTITY,SomeGuid UNIQUEIDENTIFIER);
    
    
    --Create more than 6 mio rows with an running number and a changing Guid
    WITH tally AS (SELECT ROW_NUMBER()OVER(ORDER BY (SELECT NULL)) AS Nmbr 
                   FROM master..spt_values v1 
                   CROSS JOIN master..spt_values v2)
    INSERT INTO @tbl 
    SELECT NEWID() from tally;
    
    SELECT COUNT(*) FROM @tbl; --6.325.225 on my machine
    
    --Create an XML with nothing more than a list of GUIDs in the order of the table's ID
    DECLARE @xml XML=
    (SELECT SomeGuid FRom @tbl ORDER BY ID FOR XML PATH(''),ROOT('root'),TYPE);
    
    --Create one invalid entry
    UPDATE @tbl SET SomeGuid = NEWID() WHERE ID=10000;
    
    --Read all GUIDs out of the XML and number them
    DECLARE @tbl2 TABLE(Position INT,TheGuid UNIQUEIDENTIFIER);
    INSERT INTO @tbl2(Position,TheGuid)
    SELECT ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
          ,g.value(N'text()[1]',N'uniqueidentifier')
    FROM @xml.nodes(N'/root/SomeGuid') AS A(g);
    
    --then JOIN them via "Position" and check, 
    --if there are rows, where not the same values get into the same row.
    SELECT *
    FROM @tbl t
    INNER JOIN @tbl2 t2 ON t2.Position=t.ID
    WHERE t.SomeGuid<>t2.TheGuid;
    

    At least in this simple case I always get exactly only the one record back which was invalidated...

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  • 2020-12-04 01:28

    I will offer one more approach I just found out, but this needs v2016. Regrettfully the developers forgot to include the index into the resultset of STRING_SPLIT(), but this would work and is documented:

    A solution via FROM OPENJSON():

    DECLARE @str VARCHAR(100) = 'val1,val2,val3';
    
    SELECT *
    FROM OPENJSON('["' +  REPLACE(@str,',','","') + '"]');
    

    The result

    key value   type
    0   val1    1
    1   val2    1
    2   val3    1
    

    The documentation tells clearly:

    When OPENJSON parses a JSON array, the function returns the indexes of the elements in the JSON text as keys.

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