SQL Server 树查询

WITH tree
AS
(
SELECT ParentAssetID, AssetID,1 AS x2level,nodename,
CAST(nodename AS NVARCHAR(max)) x2name,
CAST(+AssetID AS NVARCHAR(max)) x2id
FROM dbo.Assets
WHERE ParentAssetID IS null
UNION ALL
SELECT c.ParentAssetID, c.AssetID, tree.x2level + 1,c.nodename, 
CAST(REPLICATE('-', x2level * 4) + c.nodename AS NVARCHAR(max)) x2name,
tree.x2id +':|:'+ CAST(c.AssetID AS NVARCHAR(max)) x2id
FROM dbo.Assets c
INNER JOIN tree
ON c.ParentAssetID = tree.AssetID
)
SELECT x2name, AssetID, ParentAssetID FROM tree
ORDER BY x2id;

---President
------Vice President
---------CEO
---------CTO
------------Group Project Manager
---------------Project Manager 1
------------------Team Leader 1
---------------------Software Engineer 1
---------------------Software Engineer 2
------------------Test Lead 1
---------------------Tester 1
---------------------Tester 2
---------------Project Manager 2
------------------Team Leader 2
---------------------Software Engineer 3
---------------------Software Engineer 4
------------------Test Lead 2
---------------------Tester 3
---------------------Tester 4
---------------------Tester 5
 
原文地址:https://www.cnblogs.com/mingjing/p/6041523.html