如何按在 SQL Server 中输入的顺序排序？

Question

我正在使用 SQL Server，我正在尝试查找结果，但我希望以与输入条件相同的顺序获取结果。

我的代码：

SELECT 
    AccountNumber, EndDate
FROM 
    Accounts
WHERE 
    AccountNumber IN (212345, 312345, 145687, 658975, 256987, 365874, 568974, 124578, 125689)   -- I would like the results to be in the same order as these numbers.

Answer 1

这是一种内联方法

示例

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 - 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

Answer 2

您可以将IN 替换为JOIN，并设置一个用于排序的字段，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

Answer 3

我将提供我刚刚发现的另一种方法，但这需要 v2016。遗憾的是，开发人员忘记将索引包含到 STRING_SPLIT() 的结果集中，但这会起作用并记录在案：

一个解决方案via FROM OPENJSON()：

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

文档说得很清楚：

OPENJSON 解析 JSON 数组时，函数返回 JSON 文本中元素的索引作为键。

Answer 4

这不是答案，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;

至少在这种简单的情况下，我总是只得到一条无效的记录...

Answer 5

好的，经过重新思考后，我将提供基于 XML 的终极 type-safe 和 sort-safe 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;

诀窍是用合适的元素数量创建一个运行数字列表（数字表更好），并根据它们的位置选择元素。

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);

元素被创建为

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

遗憾的是，XQuery 函数 position() 无法检索该值。但是您可以使用该技巧来计算给定节点之前的所有元素。这是严重的扩展，因为这个计数必须一遍又一遍地执行。元素越多越糟糕……

UPDATE2：使用已知数量的元素可能会使用它（性能要好得多）

使用XQuery 迭代一个字面上给定的列表：

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);

Answer 6

在 Azure SQL 中，现在有 STRING_SPLIT 的扩展版本，如果第三个可选参数 enable_ordinal 设置为 1，它还可以返回项目的顺序。

那么这个简单的任务终于简单了：

DECLARE @string AS varchar(200) = 'a/b/c/d/e'
DECLARE @position AS int = 3

SELECT value FROM STRING_SPLIT(@string, '/', 1) WHERE ordinal = @position

很遗憾，SQL Server 2019 中不可用，目前仅在 Azure 中可用，希望它会在 SQL Server 2022 中。