如何按照在 SQL 服务器中输入的顺序进行排序？
How to SORT in order as entered in SQL Server?
我正在使用 SQL 服务器，我正在尝试查找结果，但我希望获得结果的顺序与我输入条件时的顺序相同。
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.
这是一个内联方法
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
EDIT - the subquery Returns
RetSeq RetVal
1 212345
2 312345
3 145687
4 658975
5 256987
6 365874
7 568974
8 124578
9 125689
您可以将 IN 替换为 JOIN，并设置一个排序字段，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
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;
至少在这个简单的例子中，我总是只得到一条无效的记录...
好吧，经过一些重新思考，我会提供最终的 XML 基于 type-safe 和 sort-safe分离器：
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;
诀窍是创建一个 运行 数字列表和合适的元素数量（数字的 table 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);
元素创建为
<x p="99">TheValue</x>
遗憾的是，XQuery 函数 position() 无法用于 检索 值。但是你可以使用这个技巧来计算给定节点之前的所有元素。这是严重的扩展，因为必须一遍又一遍地执行此计数。元素越多越糟...
UPDATE2：如果元素数量已知，可以使用它（性能更好）
使用XQuery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);
我将提供另一种我刚刚发现的方法，但这需要 v2016。遗憾的是，开发人员忘记将索引包含在 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
文档说的很清楚：
When OPENJSON parses a JSON array, the function returns the indexes of the elements in the JSON text as keys.
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，它也可以 return 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
遗憾的是 SQL Server 2019 中不可用，目前仅在 Azure 中可用，希望它会在 SQL Server 2022 中可用。
我正在使用 SQL 服务器，我正在尝试查找结果，但我希望获得结果的顺序与我输入条件时的顺序相同。
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.
这是一个内联方法
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
EDIT - the subquery Returns
RetSeq RetVal
1 212345
2 312345
3 145687
4 658975
5 256987
6 365874
7 568974
8 124578
9 125689
您可以将 IN 替换为 JOIN，并设置一个排序字段，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
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;
至少在这个简单的例子中，我总是只得到一条无效的记录...
好吧，经过一些重新思考，我会提供最终的 XML 基于 type-safe 和 sort-safe分离器：
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;
诀窍是创建一个 运行 数字列表和合适的元素数量（数字的 table 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);
元素创建为
<x p="99">TheValue</x>
遗憾的是，XQuery 函数 position() 无法用于 检索 值。但是你可以使用这个技巧来计算给定节点之前的所有元素。这是严重的扩展，因为必须一遍又一遍地执行此计数。元素越多越糟...
UPDATE2：如果元素数量已知，可以使用它（性能更好）
使用XQuery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);
我将提供另一种我刚刚发现的方法，但这需要 v2016。遗憾的是，开发人员忘记将索引包含在 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
文档说的很清楚：
When OPENJSON parses a JSON array, the function returns the indexes of the elements in the JSON text as keys.
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，它也可以 return 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
遗憾的是 SQL Server 2019 中不可用，目前仅在 Azure 中可用，希望它会在 SQL Server 2022 中可用。