为什么这个 sql 脚本这么慢，如何让它快如闪电？答案

【问题标题】：Why is this sql script so slow and how to make it lightning fast?为什么这个 sql 脚本这么慢，如何让它快如闪电？
【发布时间】：2014-04-15 15:13:10
【问题描述】：

这是我的 sql 脚本

CREATE TABLE dbo.calendario (
    datacal DATETIME NOT NULL PRIMARY KEY,
    horautil BIT NOT NULL DEFAULT 1
);

-- DELETE FROM dbo.calendario;

DECLARE @dtmin DATETIME, @dtmax DATETIME, @dtnext DATETIME;

SELECT
    @dtmin = '2014-03-11 00:00:00'
    , @dtmax = '2030-12-31 23:50:00'
    , @dtnext = @dtmin;

WHILE (@dtnext <= @dtmax) BEGIN
    INSERT INTO dbo.calendario(datacal) VALUES (@dtnext);
    SET @dtnext = DATEADD(MINUTE, 10, @dtnext);
END;

基本上，我想创建一个日期间隔为 10 分钟的表格。循环插入了很多记录，但我认为执行它会很快。需要几分钟...

我使用的是 sql server 2008 r2。

感谢任何帮助。

【问题讨论】：

也许在循环中引入显式事务会有所帮助。
您的示例是一种程序/迭代方法。 SQL Server 通常使用基于集合的方法更有效。
我同意明确的交易可能会有所帮助。但是，这是一个需要几分钟的问题吗？这只是一个一次性的设置脚本，对吧？如果您在生产系统上定期运行此程序，您可能需要重新考虑您的方法。
是的，这将是生产中的一次性设置，但我仍在开发中 :)
啊，“几个”在 20 分钟内，而不是 2-3 分钟。是的，我知道这会变得多么烦人。 :-)

标签： sql sql-server optimization sql-server-2008-r2

【解决方案1】：

您应该避免循环等，并尝试使用基于集合的方法。（谷歌搜索“RBAR SQL”）

无论如何，这在我的笔记本电脑上运行只需 1 秒：

DROP TABLE dbo.calendario 
GO

CREATE TABLE dbo.calendario (
    datacal DATETIME NOT NULL PRIMARY KEY,
    horautil BIT NOT NULL DEFAULT 1
);

-- DELETE FROM dbo.calendario;

DECLARE @dtmin DATETIME, @dtmax DATETIME, @intervals int

SELECT @dtmin = '2014-03-11 00:00:00'
     , @dtmax = '2030-12-31 23:50:00'


SELECT @intervals = DateDiff(minute, @dtmin, @dtmax) / 10

;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),
  L6   AS(SELECT 1 AS c FROM L5 AS A, L5 AS B),
  Nums AS(SELECT ROW_NUMBER() OVER(ORDER BY c) AS n FROM L6)

INSERT INTO dbo.calendario(datacal)
SELECT DateAdd(minute, 10 * (n - 1), @dtmin)
  FROM Nums
 WHERE n BETWEEN 1 AND @intervals + 1

-- SELECT * FROM dbo.calendario ORDER BY datacal

【讨论】：

这就是我喜欢stackoverflow的原因。谢谢。

【解决方案2】：

这段代码在我的机器上需要 23 秒（大部分是在排序中）

DECLARE @DateMin AS datetime = '2014-03-11 00:00:00';
DECLARE @DateMax AS datetime = '2030-12-31 23:50:00';

DECLARE @Test AS Table (
   datacal DATETIME NOT NULL PRIMARY KEY
);

WITH Counter AS (
    SELECT ROW_NUMBER() OVER (ORDER BY a.object_id) -1 AS Count
    FROM sys.all_objects AS a
         CROSS JOIN sys.all_objects AS b
)
INSERT INTO @Test (datacal)
SELECT DATEADD(minute, 10 * Count, @DateMin)
FROM Counter
WHERE DATEADD(minute, 10 * Count, @DateMin) <= @DateMax

【讨论】：

【解决方案3】：

已经有了一些有趣的答案。这是另一个基于集合的选项，它也使用公用表表达式。

DECLARE @StartDate DATETIME = '2014-03-11 00:00:00';
DECLARE @EndDate DATETIME = '2030-12-31 00:00:00';

--CTE of days in the Start/End date range.
WITH DaysTable AS
(
    SELECT @StartDate AS CalendarDate
    UNION ALL
    SELECT DATEADD(dd, 1, CalendarDate)
    FROM DaysTable
    WHERE CalendarDate < @EndDate
) 
--CTE of 10 min intervals for one day.
,MinsTable AS
(
    SELECT CAST(0 AS DATETIME) MinutesDate
    UNION ALL
    SELECT DATEADD(MINUTE, 10, mt.MinutesDate)
    FROM MinsTable mt
    WHERE MinutesDate < DATEADD(MINUTE, -10, CAST(0 AS DATETIME) + 1)
) 
SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, CAST(0 AS DATETIME), mt.MinutesDate), dt.CalendarDate) Every10Min
INTO #TestTable
FROM MinsTable mt
--Cross join for all 10 min intervals of all days in date range.
CROSS JOIN DaysTable dt
ORDER BY DATEADD(MINUTE, DATEDIFF(MINUTE, CAST(0 AS DATETIME), mt.MinutesDate), dt.CalendarDate)
OPTION (MAXRECURSION 32767)

SELECT *
FROM #TestTable

【讨论】：