SQL Server:如何按输入的顺序排序?

下面是一个在线方法

例子

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

编辑-子查询返回

RetSeq  RetVal
1       212345
2       312345
3       145687
4       658975
5       256987
6       365874
7       568974
8       124578
9       125689

你可以替换 IN 用一个 JOIN ,并设置用于排序的字段,如下所示:

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

我会再提供一个我刚刚发现的方法,但这需要v2016。遗憾的是,开发人员忘记将索引包含在结果集中。 STRING_SPLIT() ,但这是可行的,并有记录在案:

解决办法 via FROM OPENJSON() 以下内容:

DECLARE @str VARCHAR(100) = 'val1,val2,val3';

SELECT *
FROM OPENJSON('["' +  REPLACE(@str,',','","') + '"]');

结果

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

文件清楚地说明:

当openjson解析json数组时,函数将返回json文本中元素的索引作为键。

这不是一个答案,只是一些测试代码来检查John Cappelletti的方法。

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;

至少在这个简单的例子中,我总是只得到一条被撤销的记录…

好的,经过一些思考之后,我会提供最终的XML 类型安全 和 分类安全 拆分器:

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;

诀窍是用元素的合适数量(一个数字的表更好)创建一个运行编号列表,并根据元素的位置选择元素。

提示:这相当慢…

更新:更好:

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);

元素创建为

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

遗憾的是 XQuery 功能 position() 不可用于 检索 价值。但是你可以用这个技巧计算所有元素 之前 给定的节点。这是严重的缩放,因为这个计数必须反复执行。元素越多,情况就越糟…

update2:使用已知数量的元素,可以使用这个(性能更好)

使用 函数 迭代字面上给定的列表:

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);