Can SQL Server Hierarchy type method IsDescendantO

2019-06-11 18:51发布

I am using the HierarchyId data type for the storage of locations. A user may be limited by location (LocationId). If the user has more than 1 location limit the IsDescendantOf method on the HierarchyId data type has to be invoked again with an OR.

Example(filter Employees by LocationId 5 and 6):

SELECT * FROM Employee
INNER JOIN Location ON Employee.LocationId = Location.LocationId
WHERE Location.Node.IsDescendantOf((SELECT TOP 1 Node
    FROM Location 
    WHERE LocationId = 5)) = 1
OR 
Location.Node.IsDescendantOf((SELECT TOP 1 Node
    FROM Location 
    WHERE LocationId=6)) = 1`

This works fine for 2 LocationId filters but what if this grows and a person has say 10 filters. Can IsDescendantOf work like the sql IN clause?

Tables used:

CREATE TABLE Location (
LocationId int NOT NULL PRIMARY KEY IDENTITY(1,1),
Name       nvarchar(100) NOT NULL,
[Node]  hierarchyid    NOT NULL,
[ParentNode]  AS ([Node].[GetAncestor]((1))) PERSISTED,
[Level]  AS ([Node].[GetLevel]()) PERSISTED,
);
CREATE TABLE [dbo].[Employee] (
[EmployeeId] [int] PRIMARY KEY IDENTITY(1,1) NOT NULL,
[LocationId] [int] NULL,
[Name] [nvarchar](50) NULL
) ;

1条回答
走好不送
2楼-- · 2019-06-11 19:40

Note: I addes the second solution (point 6).

1) You may use a table variable to store all searched locations (ex. DECLARE @SearchedAncestorLocation TABLE(LocationId INT PRIMARY KEY)).

2) You have to find HIERARCHYID's nodes for every location ID from @SearchedAncestorLocation.

3) You have to do an INNER JOIN with employee's location using this filter: employee_location.Node.IsDescendantOf(searched_location.Node) = 1.

4) I think you should add an UNIQUE(Node) constraint to Location table to prevent duplicated locations (duplicated nodes).

5) First solution: demo here.

DECLARE @Location TABLE(
LocationId  int NOT NULL PRIMARY KEY,
Name        nvarchar(100) NOT NULL,
[Node]      hierarchyid    NOT NULL,
UNIQUE ([Node])
);

DECLARE @Employee TABLE (
[EmployeeId] [int] PRIMARY KEY,
[LocationId] [int] NULL,
[Name] [nvarchar](50) NULL
);

INSERT  @Location(LocationId, Name, [Node])
VALUES  ( 1, N'A',     '/1/'),
        ( 2, N'AA',    '/1/1/'),
        ( 3, N'AA-1',  '/1/1/1/'), -- <-- First employee  @ AA-1
        ( 4, N'AA-2',  '/1/1/2/'),
        ( 5, N'AA-3',  '/1/1/3/'),
        ( 6, N'AB',    '/1/2/'),
        ( 7, N'AA-1',  '/1/2/1/'),
        ( 8, N'AB-2',  '/1/2/2/'),

        ( 9, N'B',     '/2/'),
        (10, N'BA',    '/2/1/'),
        (11, N'BA-1',  '/2/1/1/'), -- <-- Second employee @ BA-1
        (12, N'BA-2',  '/2/1/2/'),
        (13, N'BA-3',  '/2/1/3/'),
        (14, N'BB',    '/2/2/'),
        (15, N'BB-1',  '/2/2/1/');

INSERT  @Employee(EmployeeId, [Name], LocationId)
VALUES  (1,  N'Ion Ionescu',   3), -- AA-1
        (2, N'Geo Georgescu', 11); -- BA-1

DECLARE @SearchedAncestorLocation TABLE(LocationId INT PRIMARY KEY);
INSERT  @SearchedAncestorLocation 
VALUES  (1), --A 
        (2), --AA
        (3), --AA-1
        (9), --B
       (10), --BA
       (14); --BB

SELECT  e.*, 
        el.Name             AS EmpLocationName,
        el.Node.ToString()  AS EmpLocationHID,
        s.LocationId        AS SearchedLocationId,
        sl.Name             AS SearchedLocationName,
        sl.Node.ToString()  AS SearchedLocationHID
FROM    @Employee e
INNER JOIN  @Location el ON e.LocationId = el.LocationId
INNER JOIN  @Location sl ON el.Node.IsDescendantOf(sl.Node) = 1
INNER JOIN  @SearchedAncestorLocation s ON sl.LocationId = s.LocationId 
--AND           sl.Node <> el.Node

Results:

EmployeeId LocationId  Name          EmpLocationName EmpLocationHID SearchedLocationId SearchedLocationName SearchedLocationHID
---------- ----------- ------------- --------------- -------------- ------------------ -------------------- -------------------
1          3           Ion Ionescu   AA-1            /1/1/1/        1                  A                    /1/
1          3           Ion Ionescu   AA-1            /1/1/1/        2                  AA                   /1/1/
1          3           Ion Ionescu   AA-1            /1/1/1/        3                  AA-1                 /1/1/1/
2          11          Geo Georgescu BA-1            /2/1/1/        9                  B                    /2/
2          11          Geo Georgescu BA-1            /2/1/1/        10                 BA                   /2/1/

Results if you uncomment the last line (AND sl.Node <> el.Node):

EmployeeId LocationId  Name          EmpLocationName EmpLocationHID SearchedLocationId SearchedLocationName SearchedLocationHID
---------- ----------- ------------- --------------- -------------- ------------------ -------------------- -------------------
1          3           Ion Ionescu   AA-1            /1/1/1/        1                  A                    /1/
1          3           Ion Ionescu   AA-1            /1/1/1/        2                  AA                   /1/1/
2          11          Geo Georgescu BA-1            /2/1/1/        9                  B                    /2/
2          11          Geo Georgescu BA-1            /2/1/1/        10                 BA                   /2/1/

6) Second solution.

SELECT  e.EmployeeId,
        e.LocationId,
        e.Name
FROM    @Employee e
INNER JOIN  @Location el ON e.LocationId = el.LocationId
WHERE EXISTS 
(
        SELECT  *
        FROM    @SearchedAncestorLocation s
        INNER JOIN @Location sl ON s.LocationId = sl.LocationId 
        WHERE   el.Node.IsDescendantOf(sl.Node) = 1
        --AND       el.Node <> sl.Node 
);
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