SQL服务器如何正确地创建依赖表的视图?

SQL Server How to properly create a view upon dependent tables?

鉴于以下 tables(见下文),我需要创建一个视图来显示来自 table MedDRARelations 的每个代码(SOC、HLGT、HLT、PT 和 LLT)以及来自 table MedDRANames.

我试过多次加入 MedDRANames table 并创建一个标量函数,给定代码(键)returns 值。

Table 定义:

CREATE TABLE dbo.MedDRANames 
(
    [Key]   INT NOT NULL,
    [Value] VARCHAR(100) NULL,
    CONSTRAINT [PK_MedDRANames] PRIMARY KEY CLUSTERED ([Key])
);

CREATE TABLE dbo.MedDRARelations 
(
    [SOC]  INT NOT NULL,
    [HLGT] INT NOT NULL,
    [HLT]  INT NOT NULL,
    [PT]   INT NOT NULL,
    [LLT]  INT NOT NULL,    
    CONSTRAINT [PK_MedDRARelations] 
        PRIMARY KEY CLUSTERED ([SOC] ASC, [HLGT] ASC, [HLT] ASC, [PT] ASC, [LLT] ASC),
    CONSTRAINT [FK_MedDRANameSOC] 
        FOREIGN KEY ([SOC]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameHLGT] 
        FOREIGN KEY ([HLGT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameHLT] 
        FOREIGN KEY ([HLT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANamePT] 
        FOREIGN KEY ([PT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameLLT] 
        FOREIGN KEY ([LLT]) REFERENCES dbo.MedDRANAmes([Key])
);

观看次数:

此视图执行时间为 22.1 秒 (SD 1.25),影响 156867 行

SELECT 
    [X].[SOC] AS [Código SOC],
    [SOC].[Value] AS [Término SOC],
    [X].[HLGT] AS [Código HLGT],
    [HLGT].[Value] AS [Término HLGT],
    [X].[HLT] AS [Código HLT],
    [HLT].[Value] AS [Término HLT],
    [X].[PT] AS [Código PT],
    [PT].[Value] AS [Término PT],
    [X].[LLT] AS [Código LLT],
    [LLT].[Value] AS [Término LLT]
FROM 
    dbo.MedDRARelations AS [X]
INNER JOIN 
    dbo.MedDRANames AS [SOC] ON [X].[SOC] = [SOC].[Key]
INNER JOIN 
    dbo.MedDRANames AS [HLGT] ON [X].[HLGT] = [HLGT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [HLT] ON [X].[HLT] = [HLT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [PT] ON [X].[PT] = [PT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [LLT] ON [X].[LLT] = [LLT].[Key]

此视图执行时间为 35.3 秒 (SD 2.1),影响 156867 行

SELECT 
    [X].[SOC] AS [Código SOC],
    dbo.FindMedDRA(x.SOC) AS [Término SOC],
    [X].[HLGT] AS [Código HLGT],
    dbo.FindMedDRA(x.HLGT) AS [Término HLGT],
    [X].[HLT] AS [Código HLT],
    dbo.FindMedDRA(x.HLT) AS [Término HLT],
    [X].[PT] AS [Código PT],
    dbo.FindMedDRA(x.PT) AS [Término PT],
    [X].[LLT] AS [Código LLT],
    dbo.FindMedDRA(x.LLT) AS [Término LLT]
FROM 
    dbo.MedDRARelations AS [X]

标量函数:

CREATE FUNCTION [dbo].[FindMedDRA]
    (@code INT)
RETURNS VARCHAR(100)
AS
BEGIN
    DECLARE @returning VARCHAR(100)

    SELECT @returning = dbo.MedDRANames.Value 
    FROM dbo.MedDRANames 
    WHERE dbo.MedDRANames.[Key] = @code

    RETURN @returning
END

我想知道实现此目的的正确方法是什么,因为创建连接的 table 耗时 12.2 秒(SD 0.1),但成本是原始 space 的 5 倍。

提前致谢。

更新:执行计划

  1. 使用函数:https://www.brentozar.com/pastetheplan/?id=rkso0aKnV
  2. 在 Ixs 创作之前:https://www.brentozar.com/pastetheplan/?id=HkFlyRth4
  3. 在 Ixs 创作之后(如 Laughing Vergil 所建议):https://www.brentozar.com/pastetheplan/?id=B13ZW0t3V

第一种方法是正确的。 INNER JOINs 在单列整数主键上非常快,通常通过 MERGE JOIN 运算符执行。正如您的结果所显示的那样,肯定比创建函数更快。

这里最大的问题是 MedRARelations 中的值没有索引。注意:一个常见的误解是创建外键会创建索引。它没有。

所以,试试运行这个:

CREATE INDEX Idx_MedDRARelations_SOC ON MedDRARelations (SOC);
CREATE INDEX Idx_MedDRARelations_HGLT ON MedDRARelations (HGLT);
CREATE INDEX Idx_MedDRARelations_HLT ON MedDRARelations (HLT);
CREATE INDEX Idx_MedDRARelations_PT ON MedDRARelations (PT);
CREATE INDEX Idx_MedDRARelations_LLT ON MedDRARelations (LLT);

然后再次尝试您的视图和其他方法。