SQL集合运算参考及案例(二)：树形节点数量逐级累计汇总

CREATE TABLE tMaterial
(
      Id        INT PRIMARY KEY
    , ParentId  INT
    , Qty       INT
    , Qty_Sum   INT
)

INSERT INTO tMaterial
            SELECT  1, 0, 1, 0
UNION ALL   SELECT  2, 1, 2, 0
UNION ALL   SELECT  3, 1, 3, 0
UNION ALL   SELECT  4, 2, 4, 0
UNION ALL   SELECT  5, 4, 5, 0
GO

CREATE FUNCTION fn_getQty_Sum(@Id INT)
RETURNS INT
AS
BEGIN
    DECLARE @Qty_Sum INT
    SELECT @Qty_Sum = Qty FROM tMaterial WHERE Id = @Id
    
    DECLARE   @OID  INT, @Qty INT
    DECLARE cursor1 CURSOR FOR 
        SELECT t.ID from tMaterial AS t WHERE t.ParentId = @Id
    OPEN cursor1
    
    FETCH NEXT FROM cursor1 INTO @OID
    
    WHILE @@FETCH_STATUS = 0 
    BEGIN
        SET @Qty = dbo.fn_getQty_Sum(@OID)
        SET @Qty_Sum = @Qty_Sum + @Qty
    
        FETCH NEXT FROM cursor1 INTO @OID
    END
    
    CLOSE cursor1
    DEALLOCATE cursor1

    RETURN @Qty_Sum
END

UPDATE tMaterial
SET Qty_Sum = dbo.fn_getQty_Sum(Id)

SELECT *
FROM tMaterial

WITH tmp AS 
(
    SELECT  t1.*, CAST(CAST(t1.Id AS NVARCHAR) + ‘.‘ AS NVARCHAR(100)) AS node_path
    FROM    tMaterial t1
    WHERE   t1.ParentId = 0
    UNION ALL 
    SELECT  t1.*, CAST(t2.node_path + CAST(t1.Id AS NVARCHAR) + ‘.‘ AS NVARCHAR(100))
    FROM    tMaterial t1
        JOIN tmp AS t2 ON t1.ParentId = t2.Id
)
, T2 AS 
(
    SELECT  t1.Id, t1.ParentId, t1.Qty, sum(t2.qty) AS Qty_Sum
    FROM    tmp t1
        JOIN tmp t2 ON t2.node_path LIKE t1.node_path + ‘%‘ 
    GROUP BY t1.Id, t1.ParentId, t1.Qty, t1.Qty_Sum
)

UPDATE T1
SET T1.Qty_Sum = T2.Qty_Sum
FROM tMaterial T1
    JOIN T2 ON T1.Id = T2.Id

SELECT * 

FROM tMaterial

WITH tmp AS (
    SELECT t.Id tm, * FROM tMaterial t
    UNION ALL
    SELECT t2.tm tm, t1.* FROM tMaterial t1 JOIN tmp t2 ON t1.ParentId = t2.Id
)

SELECT  tm,  sum(Qty) 

FROM    tmp 

GROUP BY tm