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递归排序的T-SQL查询

[英]Recursive sort of T-SQL query

Based on the following table 根据下表

ID    Path       
---------------------------------------
1  \\Root
2  \\Root\Node0
3  \\Root\Node0\Node1
4  \\Root\Node0\Node2
5  \\Root\Node3
6  \\Root\Node3\Node4
7  \\Root\Node5
...
N  \\Root\Node5\Node6\Node7\Node8\Node9\Node10

so on... 等等...

There are around 1000 rows in this table. 该表中大约有1000行。 I want to display individual nodes in separate columns. 我想在单独的列中显示各个节点。 Maximum columns to be displayed 5 (ie node till 5 level deep). 要显示的最大列数为5(即直到5级深度的节点)。 So the output will look as below 因此输出将如下所示

ID    Path           Level 0   Level 1  Level 2  Level 3  Level 4  Level 5
----------------------------------------------------------------------------------------
1  \\Root                    Root      Null     Null     Null     Null     Null
2  \\Root\Node0              Root      Node 0   Null     Null     Null     Null
3  \\Root\Node0\Node1        Root      Node 0   Node 1   Null     Null     Null
4  \\Root\Node0\Node2        Root      Node 0   Node 2   Null     Null     Null
5  \\Root\Node3              Root      Node 3   Null     Null     Null     Null
6  \\Root\Node3\Node4        Root      Node 3   Node 4   Null     Null     Null
7  \\Root\Node5              Root      Node 5   Null     Null     Null     Null
...
N  (see in above table)      Root      Node 5   Node 6   Node 7   Node 8 Node 9

The only way I can think of is to open a cursor, loop through each row and perform a string split, just fetch the first 5 nodes and then insert into a temp table. 我能想到的唯一方法是打开游标,遍历每一行并执行字符串拆分,仅获取前5个节点,然后插入临时表中。

Please suggest. 请提出建议。

Thanks 谢谢

What you need is a table-valued split function akin to the following: 您需要的是一个类似于以下内容的表值拆分函数:

CREATE FUNCTION [dbo].[udf_Split] (@DelimitedList nvarchar(max), @Delimiter nvarchar(2) = ',')
RETURNS @SplitResults TABLE (Position int NOT NULL PRIMARY KEY, Value nvarchar(max))
AS
Begin
    Declare @DelimiterLength int
    Set @DelimiterLength = DataLength(@Delimiter) / 2

    If Left(@DelimitedList, @DelimiterLength) <> @Delimiter
        Set @DelimitedList = @Delimiter + @DelimitedList

    If Right(@DelimitedList, @DelimiterLength) <> @Delimiter
        Set @DelimitedList = @DelimitedList + @Delimiter

    Insert @SplitResults(Position, Value)
    Select CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength            
        , Substring (
                    A.List
                    , CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength         
                    , CharIndex(@Delimiter, A.list, N.Value + 1)                            
                        - ( CharIndex(@Delimiter, A.list, N.Value) + @DelimiterLength ) 
                    )
    From dbo.Numbers As N
        Cross Join (Select @DelimitedList As list) As A
    Where N.Value > 0
        And N.Value < LEN(A.list)
        And Substring(A.list, N.Value, @DelimiterLength) = @Delimiter
    Order By N.Value

    Return
End

This function relies on the existence of a Numbers table which contains a sequential list of integers. 此函数依赖于Numbers表的存在,该表包含顺序的整数列表。 Now, you could take your original data and do something like: 现在,您可以获取原始数据并执行以下操作:

With TableData As
    (
    Select 1 As Id, '\\Root' As [Path]
    Union Select All 2, '\\Root\Node0'
    Union Select All 3, '\\Root\Node0\Node1'
    Union Select All 4, '\\Root\Node0\Node2'
    Union Select All 5, '\\Root\Node3'
    Union Select All 6, '\\Root\Node3\Node4'
    Union Select All 7, '\\Root\Node5'
    )
    , SplitData As
    (
    Select T.Id, T.[Path], S.Value
        , Row_Number() Over ( Partition By T.Id Order By S.Position ) As Level
    From TableData As T
        Cross Apply dbo.udf_Split( (Substring(T.[Path],2,Len(T.[Path])) + '\') , '\') As S
    )
Select Id, [Path]
    , Min( Case When Level = 1 Then S.Value End ) As Level0
    , Min( Case When Level = 2 Then S.Value End ) As Level1
    , Min( Case When Level = 3 Then S.Value End ) As Level2
    , Min( Case When Level = 4 Then S.Value End ) As Level3
    , Min( Case When Level = 5 Then S.Value End ) As Level4
From SplitData As S
Group By Id, [Path]

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