自加入查詢

是否可以使用連接執行父子查詢而不通過臨時表循環？自加入查詢

數據庫樣本：

menuid name    parent url 
---------------------------------------------------------- 
A0000 Master    A0000 # 
A0001 Rekening   A0000 /master/rekening.aspx 
A0002 Master Nominal  A0001 /master/nominal.aspx 
A0003 Master Satuan Other A0001 /master/satuan.aspx 
A0004 Master Kondisi  A0000 /master/kondisi.aspx 
A0005 Master Tujuan  A0003 /master/tujuan.aspx 
A0006 Master Item   A0003 /master/item.aspx 
A0007 Master Warehouse A0000 /master/warehouse.aspx 
A0008 Master Kapal  A0006 /master/kapal.aspx

期望的結果，如果選用uri = '/master/kapal.aspx'：

menuid name    parent url 
---------------------------------------------------------- 
A0000 Master    A0000 # 
A0001 Rekening   A0000 /master/rekening.aspx 
A0003 Master Satuan Other A0001 /master/satuan.aspx 
A0006 Master Item   A0003 /master/item.aspx 
A0008 Master Kapal  A0006 /master/kapal.aspx

期望的結果，如果選用uri = /master/tujuan.aspx'：

menuid name    parent url 
---------------------------------------------------------- 
A0000 Master    A0000 # 
A0001 Rekening   A0000 /master/rekening.aspx 
A0005 Master Tujuan  A0003 /master/tujuan.aspx

樣品查詢：

declare @menuid varchar(255) = 'menuid' 
declare @parent varchar(255) = 'parent' 
declare @temp_parent varchar(255) 
declare @i smallint = 0 

delete from temp_menu 
while (@menuid <> @parent) 
begin 
    if(@i = 0) 
    begin 
    insert into temp_menu 
    select * from menu where uri = '/master/kapal.aspx' 
    select @menuid = menuid, @parent = parent from menu where uri = '/master/kapal.aspx' 
    set @i = 1; 
    end 
    else 
    begin 
    insert into temp_menu 
    select * from menu where menuid = @parent 
    select @menuid = menuid, @temp_parent = parent from menu where menuid = @parent 
    set @parent = @temp_parent; 
    end 
end 
select * from temp_menu

樣品與hieararchy：

A0000 
|_______________________ 
|    |  | 
A0001   A0004 A0007 
|________ 
|  | 
A0002 A0003 
     |_______ 
     |  | 
     A0005 A0006 
       | 
       A0008

更新： 我想從節點上最長的分支可能parent到menuid，並停止所有行如果parent相同與menuid或沒有menuid與parent匹配。

加入SCRIPT和樣品

IF OBJECT_ID('dbo.menu', 'U') IS NOT NULL 
    DROP TABLE dbo.menu 
GO 

IF OBJECT_ID('dbo.temp_menu', 'U') IS NOT NULL 
    DROP TABLE dbo.temp_menu 
GO 

IF OBJECTPROPERTY(object_id('dbo.sp_get_parent'), N'IsProcedure') = 1 
    DROP PROCEDURE dbo.sp_get_parent 
GO 

create table dbo.menu (
menuid varchar(255) 
, name varchar(255) 
, parent varchar(255) 
, uri varchar(255) 
); 

insert into dbo.menu (menuid, name, parent, uri) 
values ('A0000', 'Master', 'A0000', '#') 
, ('A0001', 'Rekening', 'A0000', '/master/rekening.aspx') 
, ('A0002', 'Master Nominal', 'A0001', '/master/nominal.aspx') 
, ('A0003', 'Master Satuan Other', 'A0001', '/master/satuan.aspx') 
, ('A0004', 'Master Kondisi', 'A0000', '/master/kondisi.aspx') 
, ('A0005', 'Master Tujuan', 'A0003', '/master/tujuan.aspx') 
, ('A0006', 'Master Item', 'A0003', '/master/item.aspx') 
, ('A0007', 'Master Warehouse', 'A0000', '/master/warehouse.aspx') 
, ('A0008', 'Master Kapal', 'A0006', '/master/kapal.aspx'); 

create table dbo.temp_menu (
menuid varchar(255) 
, name varchar(255) 
, parent varchar(255) 
, uri varchar(255) 
); 

SET ANSI_NULLS ON 
GO 

SET QUOTED_IDENTIFIER ON 
GO 

Create PROCEDURE [dbo].[sp_get_parent] 
@uri VARCHAR (255) 
AS 

declare @menuid varchar(255) = 'menuid' 
declare @parent varchar(255) = 'parent' 
declare @temp_parent varchar(255) 
declare @i smallint = 0 

delete from temp_menu 
while (@menuid <> @parent) 
begin 
    if(@i = 0) 
    begin 
    insert into temp_menu 
    select * from menu where uri = @uri 
    select @menuid = menuid, @parent = parent from menu where uri = @uri 
    set @i = 1; 
    end 
    else 
    begin 
    insert into temp_menu 
    select * from menu where menuid = @parent 
    select @menuid = menuid, @temp_parent = parent from menu where menuid = @parent 
    set @parent = @temp_parent; 
    end 
end 
select * from temp_menu order by menuid asc 
GO

對於期望的樣品上方可以嘗試此查詢：

sp_get_parent '/master/kapal.aspx'

AND

sp_get_parent '/master/tujuan.aspx'

來源

2017-06-07 Colour Dalnet

https://stackoverflow.com/questions/tagged/ sql-server +遞歸查詢 –

所以你想獲得屬於最長分支的節點？ –

@GiorgosBetsos：我更新層次結構方案以便更好地理解。謝謝。 –

在SQL Server中，有關如何查詢分層數據的每個問題的答案都是使用遞歸公用表表達式。

在你的情況，因爲你想獲得最長的分支，你應該增加一個計數列：

;WITH CTE AS 
(
    SELECT menuid, name, parent, url, 0 as level 
    FROM menu WHERE parent = menuid -- Usually, the parent column is simply nullable 
    UNION ALL 
    SELECT menu.menuid, menu.name, menu.parent, menu.url, level + 1 
    FROM menu 
    INNER JOIN CTE ON menu.parent = CTE.menuid 
    AND menu.parent <> CTE.parent -- This is why parent column is nullable :-) 
) 

SELECT TOP 1 * 
FROM CTE 
ORDER BY Level DESC

此查詢將讓你這是最遠的從它的頂部父葉。

更新
基於您的評論，我覺得這是你在找什麼：

;WITH CTERecursion AS 
(
    SELECT menuid, 
      name, 
      parent, 
      url, 
      0 as level, 
      menuid as TopLevelParent 
    FROM menu WHERE parent = menuid -- Usually, the parent column is simply nullable 

    UNION ALL 
    SELECT menu.menuid, 
      menu.name, 
      menu.parent, 
      menu.url, 
      level + 1, 
      TopLevelParent 
    FROM menu 
    INNER JOIN CTERecursion CTE ON menu.parent = CTE.menuid 
    AND menu.menuid <> CTE.menuid -- This is why parent column is nullable :-) 

), CTELongestPath AS 
(
    SELECT TOP 1 TopLevelParent 
    FROM CTERecursion 
    ORDER BY Level DESC 
) 

SELECT menuid, name, parent, url 
FROM CTERecursion r 
INNER JOIN CTELongestPath l ON r.TopLevelParent = r.TopLevelParent

更新＃2
現在，你的問題是改變了，你似乎只是想從葉子到頂端父母進行遍歷。在這種情況下，你的遞歸CTE應該是這樣的：

DECLARE @url varchar(100) = '/master/kapal.aspx'; 

;WITH CTERecursion AS 
(
    SELECT menuid, 
      name, 
      parent, 
      url 
    FROM menu 
    WHERE url = @url 

    UNION ALL 
    SELECT menu.menuid, 
      menu.name, 
      menu.parent, 
      menu.url 
    FROM menu 
    INNER JOIN CTERecursion CTE ON menu.menuid = CTE.parent 
    AND menu.menuid <> CTE.menuid -- This is why parent column is nullable :-) 
) 

SELECT menuid, name, parent, url 
FROM CTERecursion

來源

2017-06-07 07:13:52

你好，我得到的邏輯，但它只顯示最長的分支最高的行，而不是所需的結果。有前途的答案，仍然找出由此造成的。 –

現在我編輯了我的答案後，我看到你再次改變了你的問題。請不要讓我們太辛苦。在發佈問題之前，確保你知道你想要得到什麼。如果您可以將您的示例數據發佈爲DDL + DML，它將非常有幫助。它很難閱讀，但很容易複製並粘貼到測試環境。 –

我再次更新了我的答案，以提供您在問題中指定的所需結果。 –

drop table if exists dbo.Menu; 

create table dbo.Menu (
menuid varchar(100) 
, name varchar(100) 
, parent varchar(100) 
, url varchar(100) 
); 

insert into dbo.Menu (menuid, name, parent, url) 
values ('A0000', 'Master', 'A0000', '#') 
, ('A0001', 'Rekening', 'A0000', '/master/rekening.aspx') 
, ('A0002', 'Master Nominal', 'A0001', '/master/nominal.aspx') 
, ('A0003', 'Master Satuan Other', 'A0001', '/master/satuan.aspx') 
, ('A0004', 'Master Kondisi', 'A0000', '/master/kondisi.aspx') 
, ('A0005', 'Master Tujuan', 'A0003', '/master/tujuan.aspx') 
, ('A0006', 'Master Item', 'A0003', '/master/item.aspx') 
, ('A0007', 'Master Warehouse', 'A0000', '/master/warehouse.aspx') 
, ('A0008', 'Master Kapal', 'A0006', '/master/kapal.aspx'); 

with cteMenu as (
select 
    m.menuid, m.name, m.parent, m.url 
    , convert(varchar(max), '.' + m.menuid + '.') as Hierarchy 
    , 0 as Lvl 
from dbo.Menu m 
where m.menuid = m.parent 

union all 

select 
    m.menuid, m.name, m.parent, m.url 
    , cm.Hierarchy + m.menuid + '.' as Hierarchy 
    , cm.Lvl + 1 as Lvl 
from dbo.Menu m 
    inner join cteMenu cm on m.parent = cm.menuid 
where m.menuid <> m.parent 
) 
select 
cm.menuid, cm.name, cm.parent, cm.url 
from (
select 
    top(1) 
    cm.* 
from cteMenu cm 
order by cm.Lvl desc 
) t 
    inner join cteMenu cm on t.Hierarchy like cm.Hierarchy + '%'

來源

2017-06-07 07:13:11

使用公用表表達式（CTE）是這樣的：

WITH cte_name AS 
(
    SELECT <base_elements> FROM <table_name> WHERE <root_condition> 
    UNION ALL 
    SELECT <child_elements> 
    FROM cte_name 
    JOIN <table_name> ON cte_name.id = <table_name>.parentid 
) 
SELECT DISTINCT * FROM cte_name

這樣做是會選擇那些根的所有元素元素。在你的情況下，這將是像menuid = 'A0000'。

然後它將所有這些重新組合到自身上，只要它找到parentId與（child）匹配的再次加入原始表，在您的情況下，條件將爲menu.parent = cte.menuid。

然後它會選擇不同於遞歸查詢返回的所有內容。如果你想獲得最大深度，你將不得不在像root（0）這樣的根選擇上添加一個常量值，然後隨着每個union都增加它。然後，您可以從最終的不同查詢中選擇max(nesting_level)。

在你的情況是這樣的可以工作

WITH cteMenu AS 
(
    SELECT menuid, name, parent, url, 0 as nesting_level FROM menu WHERE menuid = 'A0000' 
    UNION ALL 
    SELECT menu.menuid, menu.name, menu.parent, menu.url, nesting_level + 1 
     FROM menu 
     JOIN cteMenu ON menu.parent = cteMenu.menuid 
     WHERE menu.id <> 'A0000' 
) 
SELECT DISTINCT * FROM cteMenu WHERE nesting_level = (SELECT MAX(nesting_level) FROM cteMenu)

/更新：增加了WHERE子句WHERE menu.id <> 'A0000'到內部的選擇，這將刪除選擇的根元素，並停止無限遞歸。

來源

2017-06-07 07:16:17 Adwaenyth

嗨，那裏，我得到這個錯誤'消息530，級別16，狀態1，行1 聲明終止。樣本數據庫在語句完成之前已經耗盡了最大遞歸100。 –

@ColourDalnet剛剛重讀你的問題。這種情況下的問題是根菜單項已將其自身設置爲父項。添加一個where子句（我會更新我的答案）。 – Adwaenyth

謝謝adwaenyth –

declare @uri nvarchar(256) = '/master/kapal.aspx' 

; with cte as (
    select MenuId, Name, Parent, Url 
    , 1 reverseOrder 
    from MyTable 
    where url = @uri 

    union all 

    select b.MenuId, b.Name, b.Parent, b.Url 
    , a.reverseOrder + 1 
    from cte a 
    inner join MyTable b 
    where b.MenuId = a.Parent 
    and b.MenuId = a.MenuId --don't repeat ourselves after reaching the root 
) 
select MenuId, Name, Parent, Url 
from cte 
ordre by reverseOrder desc

更新

對於你的問題（即最長的分支）的第二部分，嘗試：

; with cte as (
    select MenuId, Name, Parent, Url 
    , 1 branchLength 
    , cast(MenuId as nvarchar(max)) branchPath 
    from MyTable 
    where Parent = menuid --i.e. top level/root elements 

    union all 

    select b.MenuId, b.Name, b.Parent, b.Url 
    , a.branchLength + 1 
    , a.branchPath + '\' + cast(b.MenuId as nvarchar(max)) branchPath 
    from cte a 
    inner join MyTable b 
    where b.Parent = a.MenuId --i.e. branch of previous result 
    and b.MenuId != a.MenuId --ensure we don't repeat ourselves 
) 
select a.MenuId, a.Name, a.Parent, a.Url 
from cte a 
inner join (
    select top 1 branchPath 
    from cte 
    order by branchLength Desc 
) b 
on b.branchPath like a.branchPath + '%' --gives us an easy way to traverse back up the tree, without recording every possible paths' inheritance/by using a directory structure 
ordre by a.branchLength desc

來源

2017-06-07 07:57:40 JohnLBevan

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