1

我有一个简单的相关项目表，就像这样（SQL Server db）

id   Item   Parent
1     2      5
2     4      5
3     5      12
4     6      2
5     10     6

我想输出一个表格，显示每个项目的所有相互关联项目的完整路径（最多 4 个“级别”），就像这样

id  Item  ParentL1  ParentL2 ParentL3 ParentL4
1    2      5         12
2    4      5         12
3    5      12
4    6      2         5         12
5    10     6         2          5     12

谢谢！

4

3 回答 3

1

The follwoing query should do the trick

SELECT t1.id, t1.Item, t1.Parent [ParentL1], t2.Parent [ParentL2], t3.Parent [ParentL3], t4.Parent [ParentL4]
FROM MyTable t1
LEFT JOIN MyTable t2
    ON t1.Parent = t2.Item
    LEFT JOIN MyTable t3
        ON t2.Parent = t3.Item
        LEFT JOIN MyTable t4
            ON t3.Parent = t4.Item

Used the following to create the test table, MyTable to confirm the resultset

CREATE TABLE MyTable
(
id      Int IDENTITY,
Item    Int,
Parent  Int
)

INSERT MyTable
VALUES (2, 5),
        (4, 5),
        (5, 12),
        (6, 2),
        (10, 6)
于 2013-06-10T16:29:24.540 回答
1

好的，即使LEFT JOIN在这种情况下 s 是最简单的方法（当只需要 4 级递归时），这是使用递归 CTE（SQL Server 2005+）的另一个选项：

;WITH CTE AS 
(
    SELECT *, 1 RecursionLevel
    FROM YourTable
    UNION ALL
    SELECT B.id, A.Item, B.Parent, RecursionLevel + 1
    FROM CTE A
    INNER JOIN YourTable B
        ON A.Parent = B.Item
)
SELECT  Item,
        MIN(CASE WHEN RecursionLevel = 1 THEN Parent END) ParentL1,
        MIN(CASE WHEN RecursionLevel = 2 THEN Parent END) ParentL2,
        MIN(CASE WHEN RecursionLevel = 3 THEN Parent END) ParentL3,
        MIN(CASE WHEN RecursionLevel = 4 THEN Parent END) ParentL4
FROM CTE
WHERE RecursionLevel <= 4
GROUP BY Item

这是结果：

╔══════╦══════════╦══════════╦══════════╦══════════╗
║ Item ║ ParentL1 ║ ParentL2 ║ ParentL3 ║ ParentL4 ║
╠══════╬══════════╬══════════╬══════════╬══════════╣
║    2 ║        5 ║ 12       ║ NULL     ║ NULL     ║
║    4 ║        5 ║ 12       ║ NULL     ║ NULL     ║
║    5 ║       12 ║ NULL     ║ NULL     ║ NULL     ║
║    6 ║        2 ║ 5        ║ 12       ║ NULL     ║
║   10 ║        6 ║ 2        ║ 5        ║ 12       ║
╚══════╩══════════╩══════════╩══════════╩══════════╝

这是一个带有演示的sqlfiddle。

于 2013-06-10T16:48:57.610 回答
1

这是简单的方法。

SELECT id, t1.Item as Item, 
   t1.Parent as ParentL1,  
   t2.Parent as ParentL2, 
   t3.Parent as ParentL3, 
   t4.Parent as ParentL4
FROM Items t1
LEFT JOIN Items t2 ON t1.Parent = t2.Id
LEFT JOIN Items t3 ON t2.Parent = t3.Id
LEFT JOIN Items t4 ON t3.Parent = t4.Id
于 2013-06-10T16:25:09.677 回答