比如，我要建立一個1，000，000行的數字表：

CREATE TABLE dbo.Nums(n INT NOT NULL PRIMARY KEY);
DECLARE @max AS INT, @rc AS INT;
SET @max = 1000000;
SET @rc = 1;

INSERT INTO Nums VALUES(1);
WHILE @rc * 2 <= @max
BEGIN
INSERT INTO dbo.Nums SELECT n + @rc FROM dbo.Nums;
SET @rc = @rc * 2;
END

INSERT INTO dbo.Nums
SELECT n + @rc FROM dbo.Nums WHERE n + @rc <= @max;




這種方式非常巧妙，它并不是一個一個的循環插入，而是一次插入很多行，{1}，{2}，{3，4}，{5，6，7，8}。。。



為什么這樣會快呢？

是因為它節省了跟比較其他可用解決方案進行比較和記錄這些日志的時間。



然后，作者給了一個CTE的遞歸的解決方案:

DECLARE @n AS BIGINT;
SET @n = 1000000;

WITH Nums AS
(
SELECT 1 AS n
UNION ALL
SELECT n + 1 FROM Nums WHERE n < @n
)
SELECT n FROM Nums
OPTION(MAXRECURSION 0);--為了移除默認100的遞歸限制


有個更優的CTE的解決方案，就是先生成很多行，然后用ROW_NUMBER進行計算，再選擇ROW_NUMBER這列的值就可以了。

復制代碼 代碼如下:

DECLARE @n AS BIGINT;
SET @n = 1000000;

WITH Base AS
(
SELECT 1 AS n
UNION ALL
SELECT n + 1 FROM Base WHERE n < CEILING(SQRT(@n))
),
Expand AS
(
SELECT 1 AS c
FROM Base AS B1, Base AS B2
),
Nums AS
(
SELECT ROW_NUMBER() OVER(ORDER BY c) AS n
FROM Expand
)
SELECT n FROM Nums WHERE n <= @n
OPTION(MAXRECURSION 0);

利用笛卡爾積進行不斷的累加，達到了22n行。

最后，作者給出了一個函數，用于生成這樣的數字表：

復制代碼 代碼如下:

CREATE FUNCTION dbo.fn_nums(@n AS BIGINT) RETURNS TABLE
AS
RETURN
WITH
L0 AS(SELECT 1 AS c UNION ALL SELECT 1),
L1 AS(SELECT 1 AS c FROM L0 AS A, L0 AS B),
L2 AS(SELECT 1 AS c FROM L1 AS A, L1 AS B),
L3 AS(SELECT 1 AS c FROM L2 AS A, L2 AS B),
L4 AS(SELECT 1 AS c FROM L3 AS A, L3 AS B),
L5 AS(SELECT 1 AS c FROM L4 AS A, L4 AS B),
Nums AS(SELECT ROW_NUMBER() OVER(ORDER BY c) AS n FROM L5)
SELECT n FROM Nums WHERE n <= @n;
GO